I was testing some very simple circuit's today,and found one that has my up most attention. As the frequency is around 127KHz,i don't trust my DMMs to give me an accurate amp average reading-although all 3 show the same current draw. I have drawn a basic circuit below,which we will use to answer my question.
Using the 3 ohm resistor in the circuit,can i use my scope across that 3 ohm resistor to work out the circuit's power draw,or will the rms voltage across that resistor only tell me what the resistor is dissipating?. (I tried a 1 ohm resistor in it's place,but i get a very noisy signal on my scope-the 3 ohm is much smoother on the scope.)
SG setting's.
Square wave at tuned frequency (around 127KHz)
Duty cycle-38%
VPP-8.2
Off set-4.1v-so as we have 0 volts at the bottom of the wave(62% off time)
If my calculations are correct,then my circuit is driving the same LED 100% more efficiently than a straight DC current.But there is a bonus with this circuit which i will leave for a video demo. But first i need to get an accurate P/in.
So if some one can tell if and how i can use that 3 ohm resistor to get an accurate P/in,i can then work out if i have what i think i have.
Brad,
A quick review of power measurement is perhaps in order.
P=VxI, so you need to measure the voltage across, and current through the device you want to measure. If you want to know the input power "Pin" for your circuit, then you need to measure the voltage across the SG (for Vin), and across the 3 Ohm for "Iin". This is measuring the power supplied by the SG (yes, measuring the rms voltage across the resistor only gives you the resistor dissipation).
If you can measure and multiply those two signals simultaneously in your scope then you have it made. if not, you could try a manual calculation, but the inductor may skew things too much. Can you show the scope trace across the resistor?
Quote from: poynt99 on 2014.09.12, 23:42:43
Brad,
Can you show the scope trace across the resistor?
This is where it gets interesting Darren,the trace across the resistor is quite smooth(when using average on the scope) when i use my complete circuit-even though the supply from the SG is a square wave with a38% on time @ 8.2v/pp. Using the circuit as depicted(incomplete circuit) in the first post,the wave form across the resistor is consistant with the SG input signal-A square wave,but with very short voltage spike's-consistant with having the inductor in series.
V/rms across the 3 ohm resistor= 8mV
SG-V/in=8.2vpp @ 38% duty cycle.-->>offset= 4.1v
I will get some scope shots up for you within the next couple oh hours-along with an explanation/schematic for each shot.
@ Poynt
Looking at the scope shot below(across the SG),which voltage do i use to calculate my P/in-as i already have the I/in.
My DMM is showing a very accurate I/in,as i checked it against my scope using a known value resistor across my SG at the same frequency.
In regards to the scope shot,it dosnt seem right that we should be using the RMS voltage to calculate power. As the duty cycle is 38%,and the vpp is 8.2,how is it that the RMS is 4.72?-which is more than half the 8.2 vpp,and yet only a 38% duty cycle.
The mean voltage is the same as the average voltage,and my DMM also shows 2.8 volts across the SG-same as means and average voltage on the scope.
@ all
I have been using a light box (solar pannel in a sealed box) to measure the P/out given from the LED's (i am not taking into account the efficiency of the solar pannel,which at best would be 20% efficient).
I must be doing something wrong some where,as i have 6.6 times more power coming from the light box using my circuit than if i drive the light box with the same P/in using straight DC. :o
Quote TK from OU.com
You have a situation where your input voltage is pulsed. So what you need to do to find the _average_ input power over many cycles, is to compute the VxI value during the pulse, then multiply that by the 0.38 duty cycle. Your voltage varies a little during the pulse, going from 8.2 down to perhaps 7.5 volts. So really you want an "average" here, take 7.8 volts for example. Your scope is computing the "average" and "mean" (here the same thing for this waveform) by doing something like this, taking the "average peak" and multiplying by the duty cycle. So you take the average voltage during the pulse, multiply by the average current during the pulse, and then multiply the result by the duty cycle, and this will be the average power. This won't be as strictly accurate as performing the instantaneous multiplication of the V and I traces and then integrating that resulting trace over time but it will be close.
Doing it this way,the measurements line up exactly with what my DMM's are telling me. So now i will show you the average of all straight DC P/in to my circuit P/in over 5 test carried out-that being 5 of the same tests to make sure.
Straight DC in.
V/in-2.74
I/in-4.35mA
P/in=11.919mW
P/out from light box
149.3mV/4.7k ohms
P/out=.00474mW
My circuit.
P/in- 7.8v average x 38% (38% duty cycle)=2.964volts-almost same as V/average on scope shot.
I/in1.82mA
P/in=5.3944mW
P/out from light box
166mV/4.7k ohms
P/out=.005862mW
My circuit P/in / DC P/in
5.3944mW / 11.919mW x 100
45.258%--> % of power my circuit uses to that of the P/in on straight DC
Light box P/out-My circuit/ DC P/out
.005862/.00474 x 100=123.67%
So my circuit uses only 45.258% of the P/in to that of straight DC,and gives off 23.67% more light power from that of the straight DC power.
After 6 more test of which the messurements were within 1% through out,this is what we have.
Efficiency to that of DC/in
220.951%
Efficiency to that of DC/out-via light box.
123.67%
Looking for confirmation of my average V/in-as i have my average I/in.
Guy's<if this is correct,this may hold the answer as to how the TPU opperates.
@ poynt
Here are two scope shot's.1st across the 3.2 ohm resistor(exact value of resistor)
2nd-across the SG-(circuit)
Could you take a look,and let me know what you think the two average voltages are,or if you can work out the P/in,please post so as i know im very close to that figure.
Cheers\
Brad
Brad,
From the current wave form, it is evident that we can't easily do a manual calculation on the Pin. If the wave form followed the rectangular form of the input, it would be relatively easy.
Let's account for the power dissipated both in the 3.2 Ohm resistor, and the SG's 50 Ohm output resistor. To do that we need the rms voltage across the 3.2 Ohm resistor. According to your scope, the rms voltage is 4mV.
P(3.2) = 4mV2/3.2 = 5uW
Irms = 4mV/3.2 = 1.25mA
P(50) = 1.25mA2 x 50 = 78uW
Here's what we can do manually.
Eyeballing the current wave form, an equivalent pulse wave would be about 4mV (1.25mA) peak. Now we can simply multiply it out:
1.25mAp x 6Vp x 0.38 = 2.85mW. Add our 78uW for the 50 Ohm, and we have 2.93mW for Pin.
WARNING!
Keep in mind that scopes exhibit terrible accuracy (and resolution or precision) down in these low millivolt levels. The offsets in the inputs is often 5mV (or more), so there could be 100% or more error here.
Quote from: poynt99 on 2014.09.13, 12:43:54
WARNING!
Keep in mind that scopes exhibit terrible accuracy down in these low millivolt levels. The offsets in the inputs is often 5mV (or more), so there could be 100% or more error here.
Must be close,as my DMM read's 1.27mA. Also this 50 ohm on the output of the SG,is that the impedance?,as i have two settings on my SG-1=0 and the other = 50 ohm's. I have it set at 0.
So in regards to reading the millivolt's,can i just use a larger resistance-say a 10 ohm resistor.Would this increase the accuracy?
I just went and ran a test using a 15 ohm resistor(exact resistance is 15.2 ohms)
Below are 3 screen captures-hope this helps. Of course,my I/in has gone down a little on the DMM now-1.19mA.
Brad,
Yeah the larger the voltage across the CSR, the better in terms of scope accuracy.
OK, if SG impedance is zero, don't calculate a dissipation for the 50 Ohm.
So,
26.4mV/15.2 x 6Vp x 0.38 = 3.96mW for Pin.
Quote from: poynt99 on 2014.09.13, 13:26:31
So,
26.4mV/15.2 x 6Vp x 0.38 = 3.96mW for Pin.
Darren-if this is correct(and it's within 2% of my DMM amp meter x volts-3.87mW),i am above 220% to that of straight DC P/in.
This is using the very same LED's and light box-as pictured below.
You may be running well below optimal efficiency of the LED while driving at only 2.74VDC, which could be giving your pulsed circuit an unfair advantage.
Quote from: poynt99 on 2014.09.13, 14:13:15
You may be running well below optimal efficiency of the LED while driving at only 2.74VDC, which could be giving your pulsed circuit an unfair advantage.
The LED's are rated at 2.8 volt's. So when testing them on stright DC,i ran test from 2.6 volt's to 3 volt's. Anything after 2.85 volt's DC,the current starts to climb very fast,and the voltage across the LED's remains almost unchanged. 2.83 volt's seems to be the optimal light output(via the light box) for P/in. Even then,my circuit is still over 220% more efficient than DC.
Im going to do a late nighter,and go shoot a video for you guys to see. Maybe from there,you might pick up a mistake im makeing some where.
For what it's worth
Driving LED's in a pulsed mode is more efficient in terms of light output than DC to the human eye. This is well known, and one of the reasons why many led displays are multiplexed. In th 80's and 90's I used to design alpha numeric LED display readouts for our instrumentation that operated with a 10% duty cycle in multiplexed mode. This was always more efficient and brighter to the eye than straight DC drive, watt for watt.
http://www.ledsmagazine.com/articles/2008/05/pulse-driven-leds-have-higher-apparent-brightness.html
How all this computes to using a non biological sensor and getting back to actual power transferred should be interesting, as the eye can be fooled, and eyeballing brightness without considering duty cycle is misleading in terms of average power transferred vs. peak power
The best torches on the market use a blocking oscillator or similar switched inductor to efficiently drive LED's as opposed to straight DC.
The attached paper from Cree has some graphs of interest.
Now, I am interested in how you believe this might be the answer to the TPU.
Quote from: ION on 2014.09.13, 15:13:24
For what it's worth
Driving LED's in a pulsed mode is more efficient in terms of light output than DC to the human eye. This is well known, and one of the reasons why many led displays are multiplexed. In th 80's and 90's I used to design alpha numeric LED display readouts for our instrumentation that operated with a 10% duty cycle in multiplexed mode. This was always more efficient and brighter to the eye than straight DC drive, watt for watt.
How all this computes to using a non biological sensor and getting back to actual power transferred should be interesting, as the eye can be fooled, and eyeballing brightness without considering duty cycle is misleading in terms of average power transferred vs. peak power
The best torches on the market use a blocking oscillator or similar switched inductor to efficiently drive LED's as opposed to straight DC.
Now, I am interested in how you believe this might be the answer to the TPU.
Thanks for the input ION.
Latest test-after fine tuning.
Now here is the interesting thing with this circuit. One LED is not pulsing,it is on 100% of the time, at a steddy 2.76 volt's@ our 2.41mA-this is due to how i have the circuit set up.The other LED has a duty cycle of 41%.This is at a peak voltage of 6.7v.( 6.7 x .41=an average of 2.747
So LED 1 is -->2.76@2.41mA=6.6516mW
LED 2 is-->2.747 @ 2.41mA=6.6202mW.
This is a total of 13.2718mW.
Our P/in for the whole circuit is-->peak voltage=7.6 volt's-this is across the SG,and is what the P/P voltage of the SG is set at.
Our I/in is 2.41mA
7,6 x .40(duty cycle)=3.40 V/average
3.4v x 2.41mA=8.194mW
13.2718/8.194 x 100=161.96%
Now in regards to the TPU,it has 2 or 3 inductors(wires rapped around the steel core), caps,and what i believe to be the drive circuit.1 or 2 of the smaller inductors would be to drive the pulse circuit,while the large one supplies the power to run the circuit,and the extra is to drive the load's. In my case,the extra 61.96 % is driving the LED's harder-this is why it is more efficient than driving the LED's with a DC P/in. Taking Darren's advice about the optimal efficiency at running LED's on a dc current,i spent an hour obtaining the best efficiency for the LED's on DC current. This was done using the light box. I went up in .01 v steps on my power supply until i got the most P/out from the light box,for the least Pin to the LED's. The LED's are rated at 2.8v,and the best efficiency was found at 2.78v-so very close to the spec's of the LED's.This 2.78v was at 4.36mA-so our P/in is 12.1208mW
Summery
DC P/in=12.1208mW
Circuits P/in=8.194mW
12.1208/8.194 x 100=147.92%
The 14.04% difference between the 161.96% and the 147.92,is because the LED's are being driven harder. This is confirmed through the light box,where we can obtain a higher voltage across the light box circuit using the pulsed circuit than we can using a DC current.
TEST 1
I am now running my scope from a UPS. When i unplug the UPS from the main's,i can scope every component of my circuit individually.
First to clarify something. The input current is also the current required to maintain X amount of volt's for a set amount of time over each LED-->is this correct?. If so,then the I/in(along with the average voltage across each LED) is then used to calculate each LED's P/out(or power disipation of each LED)
As you will see,the scope shots are very clean.This was a result of lifting the frequency until i got the cleanest signal across each component.
Test-1
SG settings
Frequency=165.055khZ
voltage P/P=7.5v
Offset=3.75v --> to obtain a 0 volt off time.
Duty cycle=40%
I/in is 2mA for this test -->Test 1
Scope shot 1 is across the SG-thus the circuit.We will be generous and use the max voltage to calculate our P/in-->to eliminate error in calculating V/Iin
V max is 7.35.
V average is 7.35 x 40%=2.94v
2.94v x 2ma=5.88mW
P/in=5.88mW
LED 1-scope shot 2.
LED 1 is on 100% of the time -->duty cycle=100%
V across LED 1 is 2.72v. But once again we will be generous and call it 2.7v
2.7 x 2mA=5.4mW
Power disipated by LED 1 =5.4mW
LED 2-scope shot 3.
The duty cycle of LED 2 is 59%-->once again we will be generous,and use the lowest voltage across LED 2 to calculate P/out
V=2.4 x 59%=1.416v x 2mA=2.832mW
Power disipated by LED 2 =2.832mW
Total P/in=5.88mW
Total P/out=LED 1 + LED 2=8.232mW
8.232/5.88 x 100=140%
TEST 2
In this test,i have raised the P/P voltage on the SG to 7.6v,so as to put more power into the circuit.
SG settings
Frequency=165.055
V P/P=7.6
Offset=3.8v
Duty cycle-40%
I/in is 2.9mA for this test.
Scope shot 1 is across the SG-thus the circuit. Once again we will be generous and use the V/max to calculate P/in
V=7.6 x 40%=3.04
3.04 x 2.9mA=8.816mW
P/in=8.816mW
LED 1-scope shot 2
v=2.8v-->(this is also the recommended rated voltage of the LED's)
I=2.9mA
P/out=8.12mW
LED2-scope shot 3
v=2.6-->once again,i am using the lowest voltage value across LED 2 to calculate P/out.
2.6 x 59%=1.534v
1.534 x 2.9=4.4486mW
LED 1 + LED 2 =12.5686mW
12.5686/8.816 x 100=142.56%
I have carried out the following test to confirm the below.
LED 1-scope shot 2.
LED 1 is on 100% of the time -->duty cycle=100%
V across LED 1 is 2.72v. But once again we will be generous and call it 2.7v
2.7 x 2mA=5.4mW
Power dissipated by LED 1 =5.4MW
Using straight DC from my power supply-set at 2.7volt,and with a DMM also across the LED to confirm voltage from the power supply,it takes exactly 2mA to run the LED at 2.7 volt's =5.4mW (LED 1 in our test). This is one test that confirms the accuracy of my DMM I/in.
Second confirmation test.
I found a non inductive carbon resistor that is rated at 10 watts-10 ohm's. With my circuit running,i have 20mV across the resistor.
This also confirms a 2mA current.
From this we can accurately confirm that my DMM is well within +/- 2%.
The same DMM is used throughout all test carried out for the I/in measurement.
Brad,
How did you determine that Iin is 2mA?
Why is LED1 on 100%? Isn't it also in series with LED2?
Quote from: poynt99 on 2014.09.14, 11:55:18
Brad,
How did you determine that Iin is 2mA?
Why is LED1 on 100%? Isn't it also in series with LED2?
Both with my DMM and the 10ohm 5 watt carbon resistor.-->>please see post 22.
LED 1 is not in series with LED 2
LED 1 is driven 40% of time on by the SG input,and the other 60% is via the inductor flyback. LED 2 is also driven only by the inductor flyback-as you can see in scope shot 3,it is on 60% of the time-->the off time of the SG.
Both LED's are driven via the inductor the 60% off time of the SG. LED 1 is driven by the other 40%(on time of the SG) of each cycle.
Darren
Please remember i was using the lowest voltage value when doing my calculation's.Looking at the scope shot across LED 2 below,we can clearly see that the average voltage is indeed 2.8 volt's(2.9v down to 2.7v over the cycl) Now i know it takes 2mA to drive those LED's at 2.8v-per the test i carried out in post 22.
2.8v x 60% x 2mA=3.36mW
Or-2.8v x 2mA x 60%=3.36mW
EDIT:-In fact Darren,i forgot all about the inductor,so i just went out and checked the current for the inductor and LED 1 combined.I have 3.88mA -->voltage across the two is the SG voltage in-->7.4 volt's x 40%.
I have ordered a few different size inductors,and also some 10mm super white LED's. I will have 2 sets of 2 parallel LED in the next test circuit,so as i can up the P/in P/out of the circuit.This will all be set out on a test board,so as i can messure P/out(dissipation) of each component individually.
@ Darren
My time is limited now I'm back at work,but i managed the following test. I ask if you can take the time,and have a look at the scope shots,and crunch some numbers. All to often we look for the exotic and complicated to try and achieve OU,when there is as much chance at finding it within the simplest of circuit's-maybe?.
I also took the time to test my DMM's accuracy at these frequencies and duty cycle-all my DMM's are the same. Once again they are spot on.
DMM test 1-i placed a 22ohm 1 watt resistor between the circuit and SG(negative side of course) Here are the result's
DMM 1 reads 43.2mV across the 22 ohm resistor,scope reads 44 V/avg.
DMM 2 is reading the current-I/in. It reads 1.96mA.
43.2mV across a 22ohm resistor=1.96mA. This confirms my DMM's are accurate.
With regards to the scope shots below,my I/in is 1.96mA.
So my P/in is-->2.88v(the average voltage)x 1.96mA(the average current)=5.6448mW
With these new LED's(larger ones)It take 3.66mA to run them at 2.8v.This is 10.248mW. Not forgetting that my circuit runs a second LED at a 60% duty cycle@2.8 volts. This is another 3.2928mW.
There is also my light box circuit(as pictured a few post back.I get far more P/out from the light box using my circuit than i do from straight DC-even with a higher DC P/in.
OK Brad,
I'll have a closer look in a day or two.
Thanks Darren
To the rest of the fredom energy guys-your input is more than welcome.
@ Steve
We may have to revisit your ou pulse circuit.
Chet has asked me to post the below result's and setup.
While i am aware that a pulsed LED using less P/in than that driven by a DC can look brighter to the human eye,a solar pannel is not fooled in the same way. The output from a solar pannel is relient on the true amount of light the pannel is exposed to.
Below i have reposted the light box setup. This is a sealed box that houses the two LED's and a small solar pannel from a garden light. the P/out from the solar pannel is hooked in parallel with a 200uf cap and a 4.7k resistor. From this we can measure the light output power from the LED's by measureing the voltage across the resistor/capacitor/solar pannel.
The result's are as follow's.
DC P/in to LED=2.8v @ 3.76mA. This is 10.528mW.
Output from light box(across pannel/resistor/cap) is 251mV/4.7k
SG P/in to circuit. 7.6vpp at 40% duty cycle. This is an average voltage of 3.04v.
I/in average is 2.23mA. This is 6.7792mW.
Output from light box is-325mV/4.7k
So we use 3.7488mW less,and we get 29.48% more P/out from the light box.
The scope shot below is over a 220 ohm 5 watt resistor with an exact value of 216.1 ohm's.
This is as clear as i could get the wave form over a resistor to get our I/in. If some one could tell me the current value from this shot,then i can compair that to what my meter says-which is 2.59mA
Cheers
Brad
This is very interesting, Brad. Especially your numbers are looking good
QuoteWith regards to the scope shots below,my I/in is 1.96mA.
So my P/in is-->2.88v(the average voltage)x 1.96mA(the average current)=5.6448mW
With these new LED's(larger ones)It take 3.66mA to run them at 2.8v.This is 10.248mW. Not forgetting that my circuit runs a second LED at a 60% duty cycle@2.8 volts. This is another 3.2928mW.
There is also my light box circuit(as pictured a few post back.I get far more P/out from the light box using my circuit than i do from straight DC-even with a higher DC P/in.
O0
I've played a lot with LED's as you know, and enjoy all that. I've got tons of LED's now, different colors etc..
Not nearly as fun, but easier to get a solid value on Pout perhaps --
would be to replace the LED's with a diode and a resistor, then place these in water (in an insulated vessel) and measure temp-rise of the water. Simple calorimeter. Clear results.
If it still "works" with the diode+resistor in lieu of LED's.
I'm also interested in your HHO work, Brad.
--Steve
Quote from: PhysicsProf on 2014.09.16, 14:57:38
This is very interesting, Brad. Especially your numbers are looking good
O0
I've played a lot with LED's as you know, and enjoy all that. I've got tons of LED's now, different colors etc..
Not nearly as fun, but easier to get a solid value on Pout perhaps --
would be to replace the LED's with a diode and a resistor, then place these in water (in an insulated vessel) and measure temp-rise of the water. Simple calorimeter. Clear results.
If it still "works" with the diode+resistor in lieu of LED's.
I'm also interested in your HHO work, Brad.
--Steve
I was going to try replacing the LED's with a resistor and diode,but first i must be absolutely sure my DMM's are reading the correct average I/in. Every test i have done so far says they are correct,and if they are,then this system is running at 140%+. This circuit is based around a duel figure 8 current loop,and as soon as i get some feed back on the scope shot's(so as i know I'm reading them correctly). Tonight after work,i will post some scope shot's showing exactly what is happening within the circuit,along with an explanation.
Brad
Brad,
i tried to replicate, but the "ceramic inductor" has a huge impact on the signal.
Do you have any specifications on this inductor, like dc resistance / inductance? (guess the inductance is rather high, as in mH range).
Regards Kees
Quote from: Itsu on 2014.09.17, 07:18:27
Brad,
i tried to replicate, but the "ceramic inductor" has a huge impact on the signal.
Do you have any specifications on this inductor, like dc resistance / inductance? (guess the inductance is rather high, as in mH range).
Regards Kees
Kees
Im not sure how you replicated the circuit,as i havnt posted it yet. The circuit in my first post was just a sample so as i could ask the question on how to use a resistor and scope to accuratly get the I/in for such a circuit. Befor i post the circuit,i want to make sure that my P/in measurements asre correct-other wise im just wasting your time and mine.But i believe that my DMM's are reading correct,and this will be shown in the next few post,along with what the circuit is doing.
Brad
Quote from: TinMan on 2014.09.17, 11:51:11
Kees
Im not sure how you replicated the circuit,as i havnt posted it yet. The circuit in my first post was just a sample so as i could ask the question on how to use a resistor and scope to accuratly get the I/in for such a circuit. Befor i post the circuit,i want to make sure that my P/in measurements asre correct-other wise im just wasting your time and mine.But i believe that my DMM's are reading correct,and this will be shown in the next few post,along with what the circuit is doing.
Brad
Well, i just toke a resistor (in fact 2, one of 3.3 Ohm, and one of 220 Ohm), a led and an inductor and put the SG as you mentioned (square wave, 4.1V all positive offset at 127KHz).
The signals are similar as you showed, only when using a high inductance coil (5.5mH).
All other coils with lower values shown more square wave like signals across the resistors, not the peak like you have.
I do have severe peaks on the SG square wave signal transitions which skew the scopes measurement, so they are NOT in agreement with my Fluke DMM current (rms) value.
Anyway, i will await further posts of you or others, thanks.
Regards Kees
just made a mod to the circuit. I have now reduced the P/in to only 3.42mW to achieve more light power than that of 10.52mW DC. This is 307.6% efficiency to that of DC. All my new 10mm LED's and inductors of all different value's arrived today. So this weekend(starting Friday night) i will be putting together a test bed,so as we can go through each component and get the P/in and P/out of all components.
Below i have a scope shot showing the trace across LED 1 and LED 2. I have dropped the duty cycle to 39% on the SG ,so as you can see on the blue trace(LED 1)where the P/in starts and finishes.This also lines up with the bottom half of the yellow trace(our 39% on time) LED 1 and the inductor receive there power from the SG for 39% of 1 cycle,and the inductor supplies both LED 1 and LED 2 with power the other 60% of the cycle. At a 40% duty cycle,LED 1 is on 100% of the time(as shown in other scope shots here on this thread).
As soon as I'm sure that i have what i think i have,i will post the circuit. I'm happy and wanting others to replicate it and test for them self's,but only with a real device-no sim's results to be taken notice of. I'm sorry it has to be that way,but if we are looking for the unknown,then we wont find it in a simulator based around the known.
Below is the modified circuit. I have added 2 diodes( D1 and D3 ).Because of the voltage loss,you will have to lift the V/in to around 9vpp. Adding these two diodes now eliminates the SG as being the current path during the off period. This halved the P/in needed to achieve the same output result's-for some reason???.
For those who wish to replicate and post there result's,could you set up a simple light box(as i posted),and check the light output power from the circuit,and then that of the LED running from straight DC current.
Thanks.
EDIT:
D1 is around the wrong way in the circuit below.
Looks like we're all jumping in with this ! O0
Chet rang and said to check it out.
Also, sorry for my recent absence, got some sciatica and hip troubles.
"we can obtain a higher voltage across the light box circuit using the pulsed circuit than we can using a DC current"
That is the key quote that i've taken from the writings so far.
FYI, but related to this thread perhaps, a Cool Joule has been running since the middle of June. It's got 2x 1.2V 200mAh Ni-Cd AAA's on it and 'dancing flower' 500ohm coils. The circuit began running when Lasersaber first showed his fancy coil/low energy system and Brad mentioned the Cool Joule...its run 24/7 ever since, just needing a tap on the Emitter+Collector to fire up and run for a week at a time if it stops. No battery swap overs since the beginning of August.
Own tests usually show a 50% duty cycle to be the optimum, so 38% is intriguing. The inductor is another area to explore, maybe a 'dancing flower' coil with ferrite either side...or does the inductor have to be low uH for response timings ?
I'm going to try a variety of LED's.
See, greens are great at generating all on their own in bright light, in common with other exposed silicon....so it's own output may affect P/in. For example, 4x 3mm regular frosted greens will power a cheapo digital watch in sunlight.
Reds for their lower voltage needs, ultrabrights for light, try I/R types too.
A string of them relates well to your fathers device Brad.
In my opinion, we can use a combination of this circuit, Lasersabers efficient LED driver and the effects passed from one LED to another.
Efficient collection and harvesting would then point to practical OU or only a neat curio.
ETA: for the light box, I will be using an amorphous panel, they work best with non natural light.
ETA2: An opto-coupler would seem a good alternative to a light box ?
And, have different waveshapes other than square been tried ? no point ? Just wondering about a ramp type, with a square top to drive the inductor differently.
Great stuff, Brad! I'm excited to get going on this.
Thanks for INVITING replications! O0
Did you state the approx frequency needed from the SG? (sorry if I missed it)
It would be nice to get a simpler SG than my big Cenco SG box... Is some sort of simple circuit to "replace" the SG possible?
PS - I bought an SG from china a while back; will look for it. Its a little solid-state device...
Sorry guy's,i got D1 ass about in the first circuit post.
Very tired.
Brad do you have a part number for those super brights?
Quote from: poynt99 on 2014.09.17, 23:42:09
Brad do you have a part number for those super brights?
No sorry Darren, but I do for the 10mm ones that turned up yesterday-that im yet to try. I will post that part number when I get home tonight.
What is the resistance of your coil?
You pretty certain of the 1500uH spec?
I'm simming this...or trying to. O0
You've not shown your CVR resistor on the input (or gnd side). What value should I use? And where in the circuit is it installed?
btw Brad, try the circuit without D1, I don't think it is required.
Quote from: poynt99 on 2014.09.18, 01:24:33
You've not shown your CVR resistor on the input (or gnd side). What value should I use? And where in the circuit is it installed?
btw Brad, try the circuit without D1, I don't think it is required.
Im using my DMM as my CVR, but I would place it on the ground side of the circuit/SG.The value I used at these low P/in is 220ohms, but of course this reflects on the circuits operation.
D1 is just there to drop the voltage so as D2will be the primary current path during the off period
Looking at the scope shot below that shows voltage and current,we can see the P/in. From this i can also see that my DMM amp meter is reading correctly. The average voltage is quite easy to work out from the trace(blue) which is across a 216.1 ohm 5 watt resistor,as it is almost a straight incline.It seems to me that the average voltage across that resistor is 800mV,and the average voltage across the circuit/SG is 7.7 volts.
So 800mV over 216.1 ohms is 3.70mA x 40%= an average current of 1.48mA. My DMM in this test read 1.51mA-so very close.
7.7v x 40% is 3.08 volt's.
P/in is 4.558mW going by the scope,and 4.65mW going by my DMM. A difference of 0.092mW's.
I have run this test at 5 different voltage levels and frequencies,and the DMM is always within 2% of what ohms law says it should be.
Quote from: poynt99 on 2014.09.18, 01:17:32
What is the resistance of your coil?
You pretty certain of the 1500uH spec?
I'm simming this...or trying to. O0
Looking a bit closer,it may be a violet band-not a brown one(hard to tell)
So that being the case,then the inductor would be 7.5 Millihenrys. The resistance is 9.2 ohm's.
Schematic update below.
Quote from: poynt99 on 2014.09.17, 23:42:09
Brad do you have a part number for those super brights?
These are the ones i will be using from now on,hopefully there will be no change to the way the circuit performs. The one i have been using are from one of those dynamo torches.
Viewing Angle: 25 degrees
Brightness: 5500mcd
Voltage Typical: 3.5V
10mm water clear
Part number-Z0908
Thanks Brad.
I created a model for the LED as PSpice doesn't have one. It is a diode configuration (12 x 1N4007) that yields about 2mA with 2.8V across it.
I will try the new coil inductance and resistance values, and add the CVR. ;)
Ah! a question answered, the coil resistance being 9.2Ohms.
Can build to that spec now :)
Be aware that there is a color shift associated with LED's that changes with applied current. In pulse mode the LED is operating at one end of it's spectrum, at direct current it may be at the other.
The light box should be characterized for frequency of light output also narrow band, and wide band spectral response vs output to account for some minor differences in measurement.
I highly recommend getting familiar with the LED manufacturers curves from the data sheet, not just the box specs.
I have highlighted a few important points in the attached document on page 2.
Poynt: I'd be interested in taking a peek at your LED model and rationale.
Quick and dirty replication of the latest circuit presented by Tinman
CSR = 219.6 Ohm, the inductor = 7.56mH / 9.4 Ohm (extra resistors added)
3 x 1n4148 diodes, 2x 10mm bright leds
input SG is 4.5V ALL positive square wave @ 130KHz @ 40% duty cycle
The led2 is NOT on, the led1 not full.
Screenshot shows:
yellow: input from SG
Blue: signal across ground and led1 anode
purple: signal across ground and led2 cathode
green: signal across the 219.6 Ohm csr
The Fluke meter set to current shows 0.281mA rms, but its not designed to be used at 130KHz and the current calculated via the CSR is 90mV rms / 219.6 Ohm = 0.4mA rms (confirmed lateron (not on video) sort off via my AC/DC current probe working at its lower limits, 1mA/div.)
Video here: https://www.youtube.com/watch?v=bJM7qvI6hBY&feature=youtu.be
Regards Itsu
Quote from: ION on 2014.09.18, 15:37:15
Be aware that there is a color shift associated with LED's that changes with applied current. In pulse mode the LED is operating at one end of it's spectrum, at direct current it may be at the other.
Thanks for your input ION.
This is why i have the LED's running at there rated voltage in the circuit. LED 1 on 100% at the 2.8v and LED 2 on for 60% of the cycle at 2.8 volt's. This is the same voltage applied using DC. Once the new circuit is made with the 10mm LED's,then we can go through a wide voltage range,both with the circuit and a DC supply. From there we can make a graph showing the difference in light output.
Brad
Quote from: poynt99 on 2014.09.18, 13:24:24
Thanks Brad.
I created a model for the LED as PSpice doesn't have one. It is a diode configuration (12 x 1N4007) that yields about 2mA with 2.8V across it.
I will try the new coil inductance and resistance values, and add the CVR. ;)
Darren
Thanks for your time on this,but i must ask-why are you using a sim?. All that i asked was that a real modle be used to do the testing on. The circuit is very simple,and very few part's. Here you have a situation with the sim where as you have to simulate part's to simulate the circuit. This also brings the question-how are you going to use a light box to see the difference in light output from the circuit to that of a DC supply with a string of diodes?.
Brad
It would seem from the link which ION posted that getting more Lumens per watt with a Pulse is a design feature of these LED's
Strait DC will never throw as much LUMEN per Watt as a Properly Pulsed LED [pulsed at its design frequency]
Link below
snip
Power efficiency is another advantage of the pulse width
modulation approach. Because pulse width modulation uses
a constant current level for each pulse, it's possible to select
a pulse level where the LED operates most efficiently, that is,
where the lumen output per watt is the greatest.
to Summarize ,these LED's are very inefficient when Driven with strait DC and a properly pulsed LED will thro more Lumen per watt then an improperly run unpulsed LED
a Manufacturers spec sheet for the particular LED will give its design parameters and its most efficient watts per Lumen spec [pulse freq]
thx
Chet
Quote from: Chet K on 2014.09.18, 22:29:58
Brad
It would seem from the link which ION posted that getting more Lumens per watt with a Pulse is a design feature of these LED's
Strait DC will never throw as much LUMEN per Watt as a Properly Pulsed LED [pulsed at its design frequency]
Link below
snip
Power efficiency is another advantage of the pulse width
modulation approach. Because pulse width modulation uses
a constant current level for each pulse, it's possible to select
a pulse level where the LED operates most efficiently, that is,
where the lumen output per watt is the greatest.
to Summarize ,these LED's are very inefficient when Driven with strait DC and a properly pulsed LED will thro more Lumen per watt then an improperly run unpulsed LED
a Manufacturers spec sheet for the particular LED will give its design parameters and its most efficient watts per Lumen spec [pulse freq]
thx
Chet
Thanks for the info Chet.
In regards to the constant current pulse-as you will see from the last scope shot I posted, the current is anything but constant. The reason is that the current is already flowing in the right direction when the 40% on time starts. The only thing we change at that instant is the voltage polarity. The current starts to build up in a steddy ramp, and peaks at the switch off of the cycle.
ION,
Here is my model in schematic form. This is what is inside the LED symbol on my diagram (next post).
Also shown is the I-V characteristic of the LED model.
Apologies to anyone who may have thought I was trying to model the optical characteristics of the LED. No, simply the current vs. voltage (I-V) curve. Why? As a first step so I can try to match the scope shots.
Here is the schematic and resulting voltage plots across D1 and D2 LED1 and LED2.
Fairly close I think to your scope shots Brad.
Correction for voltage plot: I really mean voltage across LED1 and LED2, not D1 and D2.
Quote from: poynt99 on 2014.09.19, 02:21:47
Here is the schematic and resulting voltage plots across D1 and D2 LED1 and LED2.
Fairly close I think to your scope shots Brad.
Correction for voltage plot: I really mean voltage across LED1 and LED2, not D1 and D2.
Yes it is close Darren. The peak voltage slope is incorrect, but not bad at all. Could you put a current trace in there as well, like I showed? .
Cheers
Quote from: Itsu on 2014.09.18, 21:12:58
Quick and dirty replication of the latest circuit presented by Tinman
CSR = 219.6 Ohm, the inductor = 7.56mH / 9.4 Ohm (extra resistors added)
3 x 1n4148 diodes, 2x 10mm bright leds
input SG is 4.5V ALL positive square wave @ 130KHz @ 40% duty cycle
The led2 is NOT on, the led1 not full.
Screenshot shows:
yellow: input from SG
Blue: signal across ground and led1 anode
purple: signal across ground and led2 cathode
green: signal across the 219.6 Ohm csr
The Fluke meter set to current shows 0.281mA rms, but its not designed to be used at 130KHz and the current calculated via the CSR is 90mV rms / 219.6 Ohm = 0.4mA rms (confirmed lateron (not on video) sort off via my AC/DC current probe working at its lower limits, 1mA/div.)
Video here: https://www.youtube.com/watch?v=bJM7qvI6hBY&feature=youtu.be
Regards Itsu
Itsu
The voltage should be 9vpp , with a 4.5v offset to get a 0 volt 60% off time
Brad,
I'll not only add the current trace, but power dissipation as well. Do you have a recent scope shot of the SG voltage as well?
Quote from: TinMan on 2014.09.19, 05:21:32
Yes it is close Darren. The peak voltage slope is incorrect, but not bad at all.
I've seen this in other of my sims as well where one slope is opposite. I suspect it might be because I am using an air-core (ideal) inductor, whereas I'm sure yours has a ferrite core? I've also not included some small capacitance parallel to the inductor.
This article might be of interest, not exactly the same but raises some interesting application ideas.
http://www.edn.com/design/led/4368392/An-LED-s-intrinsic-capacitance-works-in-a-650-mV-LRC-circuit
Quote from: poynt99 on 2014.09.19, 12:56:20
I've seen this in other of my sims as well where one slope is opposite. I suspect it might be because I am using an air-core (ideal) inductor, whereas I'm sure yours has a ferrite core? I've also not included some small capacitance parallel to the inductor.
I'm using one of those small inductors that look a bit like a 1/2 watt resistor. I bought a heap of different ones to try.
I'm not sure why,but the best i can get with these 10mm LED's is only 125.4% better than with a DC power input. I might have to have a dig around my old salvaged TV board's,and see if i can find a larger inductor.
I found one strange thing while trying different inductors-with one inductor (not sure of the value's yet)i can turn the SG off,and by touching one of the LED legs,i can get it to light up??? -not very bright,but definitely visible.
Oh,by the way Darren,these 10mm LED's draw 25mA @ 3.2 volt's. I dont have a screen shot of the new setup yet,but will get one tomorrow for you,as it's bed time for me after the long week i just had.
Quote from: TinMan on 2014.09.19, 16:34:24
I found one strange thing while trying different inductors-with one inductor (not sure of the value's yet)i can turn the SG off,and by touching one of the LED legs,i can get it to light up??? -not very bright,but definitely visible.
Now you understand why many of us have been saying for quite some time that superbrights are not a good choice for barometers of power output. It takes next to no power to illuminate them.
Quote from: TinMan on 2014.09.19, 05:42:38
Itsu
The voltage should be 9vpp , with a 4.5v offset to get a 0 volt 60% off time
Ok, done, now both leds are on.
With an "isolated from ground" SG i measured directly across the both leds, like seen in point99 his drawing only i had to reverse
the ground leads so that both ground leads are connected on the junction between the leds.
I inverted the both channels, so the screenshot 1 below shows the correct signals across led1 (blue) and led2 (purple).
I disconnected both probes and now measured the input voltage from the SG (yellow) and the input current with my current probe (controller set to 10mA/div. so
similar as the scope setting) (green), see screenshot 2.
The math function (red) shows the ch1 * ch4 = input power = 18.9mW
Regards Itsu
Itsu,
Can you also do a power dissipation measurement for each LED?
I've changed to a new LED model (much more realistic), and my Pin is about 15mW. Power dissipation in each LED is about 6mW. I will post my results in a day or so.
i will be going back to the 5mm LED's ,as these 10mm ones seem to behave in a very different way-- just cant get the efficiency i was with the 5mm LED's for some reason???. Maybe i need a larger inductor(more robust) for these 10mm ones. The best i can achieve is 126.$% efficiency over the DC supply.
Before i change back to the 5mm LED's,i will plot a graph between the circuit and DC output from the 10mm LED's. Hopefully i will have this done tomorrow-along with that screen shot Darren.
Cheers
Brad
Quote from: poynt99 on 2014.09.19, 21:52:34
Itsu,
Can you also do a power dissipation measurement for each LED?
I've changed to a new LED model (much more realistic), and my Pin is about 15mW. Power dissipation in each LED is about 6mW. I will post my results in a day or so.
Sure,
screenshot 1 is from led1, screenshot 2 from led2:
Regards Itsu
One observation is that with the SF "isolated from ground", as been used in the above measurement/screenshots, the LED1 (only) is dimly on without
a signal from the SG, infact even with this SG powered off it still is dimly on!!
So there is a strong 50Hz signal inserted somewhere, not sure this will skew the above measurements.
See screenshot taken with no signal from the SG across the dimly lid led1, but with current controller set to 1mA/div, so the Ch4 rms value and the power math value need to be divided by 10, so we have ch4 0.281mA rms and 0.244mW power.
Regards Itsu
Thanks Itsu.
My wave forms now look as per the attached.
Brad, I think your inductor is as you originally had thought; 1.5mH. That is the only value that will match the current wave form and peak level of 1.5V across the 200R CVR resistor (7.5mA peak). Note the value in my new schematic here.
Brad,
You were wondering about your input current and Pin calculation. This is how I would eyeball my wave forms to do a rough manual calculation.
So I have a peak current of 7.5mA. I know my duty cycle is 40%. I estimate the current wave form to be approximately an equilateral right triangle, plus a bit more (we are trying to determine area under the curve there). I would estimate a factor of about 0.6 then. Now to multiply it all out:
7.5mA x 0.4 x 0.6 x 9V = 16.2mW Pin
This is close to my estimate (15mW) from looking at the averaged wave form in my plot. If I let the sim run with more cycles on the screen, the actual level would be more accurate, but you get a pretty good sense on where it is going to settle, even with as few as 4 or 5 cycles on the screen, as I have.
Quote from: TinMan on 2014.09.16, 13:56:33
The scope shot below is over a 220 ohm 5 watt resistor with an exact value of 216.1 ohm's.
This is as clear as i could get the wave form over a resistor to get our I/in. If some one could tell me the current value from this shot,then i can compair that to what my meter says-which is 2.59mA
Cheers
Brad
(http://www.overunityresearch.com/index.php?action=dlattach;topic=2603.0;attach=14947)
So Brad,
Using the same method I used above, I would estimate the following:
Ipeak is 6.94mA (1.5V / 216.1)
Area factor about 0.6
duty cycle 40%
VSG = 9V
So to calculate Pin we have: 6.94mA x 0.6 x 0.4 x 9V =
15mW PinFrom this, is the way you are computing your Pin resulting in the same value?
Here is the average power in each LED.
LED1 = 5.2mW
LED2 = 5.65mW
Quote from: poynt99 on 2014.09.20, 15:53:38
(http://www.overunityresearch.com/index.php?action=dlattach;topic=2603.0;attach=14947)
So Brad,
Using the same method I used above, I would estimate the following:
Ipeak is 6.94mA (1.5V / 216.1)
Area factor about 0.6
duty cycle 40%
VSG = 9V
So to calculate Pin we have: 6.94mA x 0.6 x 0.4 x 9V = 15mW Pin
From this, is the way you are computing your Pin resulting in the same value?
What is the x .6 for?
If a complete cycle is 100%,and we know our duty cycle is 40%,then the 60% is the off time. So i averaged out the voltage over the 216.1 ohm resistor for the 40% on time (800mV),and the average voltage is 7.7 volts.
Below is the scope shot i calculated this from.
At this point in time,i would like to know what the scope means as in-
1-V avg=320mV
2-V mean=320mV
I'm guessing that V/avg is making this calculation over 1 complete cycle. if this is the case,then my estimate of 800mV over the 216.1 ohm resistor is accurate,as 800x40% is indeed 320mV average-as the scope says.As there is a P/P spike on the voltage(yellow trace),the scopes calculation of V/avg will not be accurate,so this part we must do manually. I estimate the voltage at 7.7 volts over that 40% on time. I calculate the P/in using these values,and quote TK:-->taking the "average peak" and multiplying by the duty cycle. So you take the average voltage during the pulse, multiply by the average current during the pulse, and then multiply the result by the duty cycle, and this will be the average power.
800mV/216.1 ohms=3.7mA x 7.7v=28.49mW average over that 40% on time.
28.49 x 40%=11.396mW
P/in = 11.396mW.
I am not sure what this .6 area you talk about is ?.
From here on in Darren,you will need an actual circuit,so as you can messure light output power against that of straight DC,as i dont think your diodes will put out much light on your sim lol.
My point is,that the P/in value of the circuit,is far less than that required by DC to achieve the same output from the light box.
Brad,
I explained the process (perhaps poorly) in post #73 (http://www.overunityresearch.com/index.php?topic=2603.msg41665#msg41665).
I think you misinterpreted what I was explaining. Here is a detailed explanation that hopefully helps.
What we are trying to do (when doing this eyeball calculation) is convert both wave forms (voltage and current) to a nice flat pulse so they can be multiplied together for the on-time duration. The SG is already close enough to a pulse that we can say it is a 9Vp (or whatever is measured) pulse with 40% duty cycle. The current trace however is a sawtooth shape, but this is relatively easy to convert to pulse using the approach shown in the attached diagram. This is where the "0.6" factor comes from. So the sawtooth current trace converted to an equivalent pulse will be about 0.9Vp in amplitude (orange rectangle). When all factored together it looks like this:
[(1.5Vp x 0.6) / 216.1] x 9Vp x 0.4 = 15mW
Quote from: TinMan on 2014.09.16, 13:56:33
The scope shot below is over a 220 ohm 5 watt resistor with an exact value of 216.1 ohm's.
This is as clear as i could get the wave form over a resistor to get our I/in. If some one could tell me the current value from this shot,then i can compair that to what my meter says-which is 2.59mA
Cheers
Brad
So to answer this question, we have the following:
[(1.5Vp x 0.6) / 216.1] =
4.16mANote that if the current trace was already a pulse with 1.5Vp, we would have this:
(1.5Vp / 216.1) =
6.94mANote that duty cycle is not present yet as it is factored in when you multiply the current and voltage together.
So if you have the 4.16mA current pulse and 7.7Vp SG voltage, we would have this:
4.16mA x 7.7Vp x 0.4 =
12.8mW for Pin
NOTE: the 0.6 factor for the sawtooth current trace may be a bit high and closer to 0.55 or something.
Here some current/power plots of one of my 10mm LEDs:
EDIT:
I added lateron the output plot of my blackbox (photo cell from garden light, 4.7KOhm resistor and 330uF cap parallel)
The values are in mV, the 10mm LED was 9.5cm away from the PC (measured from the back of the led to front of PC).
Regards Itsu
Thank you Itsu and Darren for your time on this.
Darren.
Could you provide me with the following info from your sim.
First up, please remove D1 and D3 from the circuit.
1-what is the average P/in over the full cycles
1A-what is the average V/in over the 40% on time
1B-what is the average I/in over the 40% on time
2-what is the average voltage across LED1 for the 40% on time
3-what is the average current flowing into LED1 for that 40% on time
4-what is the average voltage across LED1 for the 60% off time
5-what is the average current across LED 1 for the 60% off time
6-same as q4 and q5 , but for LED2
Cheers
Brad
Why remove D1 and D3?
Quote from: poynt99 on 2014.09.22, 03:11:51
Why remove D1 and D3?
So as we dont have to calculate the power disipation for these two components.
Quote from: poynt99 on 2014.09.21, 13:28:44
Brad,
I explained the process (perhaps poorly) in post #73 (http://www.overunityresearch.com/index.php?topic=2603.msg41665#msg41665).
I think you misinterpreted what I was explaining. Here is a detailed explanation that hopefully helps.
What we are trying to do (when doing this eyeball calculation) is convert both wave forms (voltage and current) to a nice flat pulse so they can be multiplied together for the on-time duration. The SG is already close enough to a pulse that we can say it is a 9Vp (or whatever is measured) pulse with 40% duty cycle. The current trace however is a sawtooth shape, but this is relatively easy to convert to pulse using the approach shown in the attached diagram. This is where the "0.6" factor comes from. So the sawtooth current trace converted to an equivalent pulse will be about 0.9Vp in amplitude (orange rectangle). When all factored together it looks like this:
[(1.5Vp x 0.6) / 216.1] x 9Vp x 0.4 = 15mW
Ah,ok,got it now. TK missed that bit about the .6 factor. This is one of the reasons i hassle you all the time lol-i get the full answer to my question's,and so now i know how to calculate the current trace correctly.;
Looking forward to those measurements i requested,as i think something is widely missed when calculating P/out of the components in these types of circuit's.
Quote from: TinMan on 2014.09.22, 05:07:17
So as we dont have to calculate the power disipation for these two components.
I see, so you're thinking of comparing Pin to a tally of the total dissipation in the circuit components?
Quote from: poynt99 on 2014.09.22, 12:40:34
I see, so you're thinking of comparing Pin to a tally of the total dissipation in the circuit components?
Yes,and also to tally something i believe is being missed,and has been in many of these type of flyback circuits.
Quote from: TinMan on 2014.09.22, 10:43:42
Ah,ok,got it now. TK missed that bit about the .6 factor. This is one of the reasons i hassle you all the time lol-i get the full answer to my question's,and so now i know how to calculate the current trace correctly.;
Looking forward to those measurements i requested,as i think something is widely missed when calculating P/out of the components in these types of circuit's.
I think TK got it correct. He said this:
QuoteSo you take the average voltage during the pulse, multiply by the average current during the pulse, and then multiply the result by the duty cycle, and this will be the average power.
The key phrase here is "during the pulse". This is what I've shown you with the current trace; we are taking the average of the sawtooth, which converts it to a lower amplitude pulse. A DMM meter does not do this; it takes the average over the entire cycle, not just during the ON time.
OK, this is interesting, and useful!
The DMM meter appears to take the average during the pulse, AND account for the duty cycle! (which does make sense)
SO?
If you take the DMM reading, divide it by the resistor value (which gives you current x duty cycle), then multiply it by your SG voltage, you get the correct Pin measurement. Like this:
(Vdmm/216.1) x 9V = 15mW (Vdmm being the voltage across the CVR as measured with a DMM) (Isg = 1.67mA in this case)
In this case you DO NOT multiply by the duty cycle as it is already done by the DMM.
Cool 8)
Note: this only works when the current and voltage are unipolar, i.e. above or below 0V.
You guys are doing a fine job on the measurement end. Now comes the crux of the problem.
If the PV cell should operate at a max of 10% efficiency, you will need a 900% efficient LED to achieve a COP equal to 1.0 for the pair.
While some have claimed an efficiency of 230% for LED's, (electrical power input vs. photon power output) it is in the picowatt range and far short of the required efficiency.
http://www.wired.co.uk/news/archive/2012-03/09/230-percent-efficient-leds
Unfortunately no closed loop operation yet. Anyone think of a way around this?
Agreed ION.
However at this point I think Tinman is comparing his Pin measurement to what he is measuring as an output power via the PV cell. With the above information regarding the DMM reading, it may become evident that the Pin calculations are too low due to a reduction via the extra multiplication of the duty cycle, 40% in this case.
As the DMM is already accounting for the duty cycle, it becomes not only unnecessary, but incorrect to again multiply by the duty cycle with this method.
Quote from: ION on 2014.09.22, 13:20:06
You guys are doing a fine job on the measurement end. Now comes the crux of the problem.
If the PV cell should operate at a max of 10% efficiency, you will need a 900% efficient LED to achieve a COP equal to 1.0 for the pair.
While some have claimed an efficiency of 230% for LED's, it is in the picowatt range and far short of the required efficiency.
http://www.wired.co.uk/news/archive/2012-03/09/230-percent-efficient-leds
Unfortunately no closed loop operation yet. Anyone think of a way around this?
Hi ION
The solar cell was never intended for looping the system,but more to show an increase or decrease of light output from the LED's. Our eyes can be tricked,but a solar pannel only converts actual ligh to power-cant be tricked like our eyes.If i see an increase of voltage over the resistor that is across the solar pannel,i know this means an increase in real light from the LED's.
Quote from: poynt99 on 2014.09.22, 13:32:58
Agreed ION.
However at this point I think Tinman is comparing his Pin measurement to what he is measuring as an output power via the PV cell. With the above information regarding the DMM reading, it may become evident that the Pin calculations are too low due to a reduction via the extra multiplication of the duty cycle, 40% in this case.
The solar pannel thing explaind in above post.
The P/in,and P/out's(disipated power) are now to be made with the scope-now that i know how it's done. But there is something else i wish to explore with this circuit now,and see what you guys have to say about it. This requires the measurements i requested from Darren. I also ask that going by your own measurements Darren,that you to calculate the P/in,and total disipation (P/out) of the circuit. I will then post what i believe to be the total disipation(P/out) of the circuit,and see how close we are to each others measurements.
Cheers
Brad
Quote from: TinMan on 2014.09.22, 13:33:42
Hi ION
The solar cell was never intended for looping the system,but more to show an increase or decrease of light output from the LED's. Our eyes can be tricked,but a solar pannel only converts actual ligh to power-cant be tricked like our eyes.If i see an increase of voltage over the resistor that is across the solar pannel,i know this means an increase in real light from the LED's.
All well and good, however for the work to be scientific would require data sheets be posted of both the PV cell and the light emitters so that we have a working knowledge of the spectral response vs power output of the PV cell and frequency of light output vs. applied current of the LED's. Without this, important information might be hidden from the experiment and wrong conclusions arrived at. I attempted to point this out earlier with a few pdf's .
Quote from: ION on 2014.09.22, 13:49:42
All well and good, however for the work to be scientific would require data sheets be posted of both the PV cell and the light emitters so that we have a working knowledge of the spectral response vs power output of the PV cell and frequency of light output vs. applied current of the LED's. Without this, important information might be hidden from the experiment and wrong conclusions arrived at. I attempted to point this out earlier with a few pdf's .
Well the PV cell data sheet may be a bit of a problem,as it's from an el'cheapo garden light-yep,no brand name on it. I posted the spec's for the LED's im using(the 10mm ones),and i believe Itsu posted a spec sheet as well. Now the solar pannel and LED's are really not that important ATM,as we are looking at Darrens sim result's-->his sim LED's wont put out much light at all.
Now im back at work,time is short my end. I start at 5am,and dont get home to around 8-9pm. Today i only did 13 hours,so i was home early. But this dosnt happen often. last weekend was preaty much taken aswell-work saterday,and grandsons bday sunday-family first. But im hoping to get a good day at it this weekend,but first there is a few questions that will revolve around Darrens P/in P/out measurements.
Quote from: TinMan on 2014.09.22, 13:39:10
The solar pannel thing explaind in above post.
The P/in,and P/out's(disipated power) are now to be made with the scope-now that i know how it's done. But there is something else i wish to explore with this circuit now,and see what you guys have to say about it. This requires the measurements i requested from Darren. I also ask that going by your own measurements Darren,that you to calculate the P/in,and total disipation (P/out) of the circuit. I will then post what i believe to be the total disipation(P/out) of the circuit,and see how close we are to each others measurements.
Cheers
Brad
Brad, up to now I believe you have been encouraged by your measurements because the output power (on the output of the PVC) have been greater than your measured and/or calculated Pin, correct? If this is incorrect, then I've made an erroneous assumption. If this is correct, then before going any further, I believe your previous measurements and conclusions deserve a revisit on your part.
Quote from: poynt99 on 2014.09.22, 14:01:21
Brad, up to now I believe you have been encouraged by your measurements because the output power (on the output of the PVC) have been greater than your measured and/or calculated Pin, correct? If this is incorrect, then I've made an erroneous assumption. If this is correct, then before going any further, I believe your previous measurements and conclusions deserve a revisit on your part.
No Darren,the P/out from the PVC are no where near my P/in.
OK, i will try again to explain what the PVC was for.
I have a PVC,cap and resistor in parallel(the light box)-as per diagram posted some time back.The light box is only to give me an indication of an increase or decrease of light from the LED's.
If i have x amount of DC power going into the LED's,the voltage across the light box will be Y amount. If i use my circuit,and i get a greater voltage from the light box than the Y amount ,using less P/in than that of what i used of DC,then i start looking into why that is the case. The light box is my lux meter of types-as i dont have a lux meter to measure differences in light output.
The power from the light box is less than 1/20th of that of the P/in. So no-no mistake made there.The light box is just my way of seeing an increase or decrease of light output power.
Quote: As the DMM is already accounting for the duty cycle, it becomes not only unnecessary, but incorrect to again multiply by the duty cycle with this method.
This i know,and the DMM's amp reading was left as the DMM read. This amount was multiplied by the average voltage of the complete cycle.--> Average voltage as shown on scope for 40% duty cycle x 40% x DMM amp reading.The DMM's amp reading was never x 40%,but left as is.
Quote from: TinMan on 2014.09.22, 21:33:57
No Darren,the P/out from the PVC are no where near my P/in.
Yes of course, I forgot. You were comparing to a DC supply at some voltage. Well, same point applies. You ought to revisit those pulse measurements and conclusions.
Quote
Quote: As the DMM is already accounting for the duty cycle, it becomes not only unnecessary, but incorrect to again multiply by the duty cycle with this method.
This i know,and the DMM's amp reading was left as the DMM read. This amount was multiplied by the average voltage of the complete cycle.--> Average voltage as shown on scope for 40% duty cycle x 40% x DMM amp reading.The DMM's amp reading was never x 40%,but left as is.
So to restate what you wrote in the last part of your quote, do you mean the following?:
VSG(avg peak) x 0.4 x VDMM/CSR ?
Quote from: poynt99 on 2014.09.23, 03:39:53
Yes of course, I forgot. You were comparing to a DC supply at some voltage. Well, same point applies. You ought to revisit those pulse measurements and conclusions.
So to restate what you wrote in the last part of your quote, do you mean the following?:
VSG(avg peak) x 0.4 x VDMM/CSR ?
Yes, that is 1of the methods I used to verify the accuracy of my DMM amp meter reading
Quote from: TinMan on 2014.09.23, 07:04:41
Yes, that is 1of the methods I used to verify the accuracy of my DMM amp meter reading
I see.
Brad, I wrote this:
VSG(avg peak) x 0.4 x VDMM/CSR
and you just said that this is correct and what you have been doing to calculate your Pin.
Therein lies the problem. I've reiterated a few times now that you DO NOT multiply by 0.4. :o
So the above becomes this:
VSG(avg peak)
x 0.4 x VDMM/CSR OR
VSG(avg peak) x VDMM/CSR
Quote from: poynt99 on 2014.09.23, 13:46:05
I see.
Brad, I wrote this:
VSG(avg peak) x 0.4 x VDMM/CSR
and you just said that this is correct and what you have been doing to calculate your Pin.
Therein lies the problem. I've reiterated a few times now that you DO NOT multiply by 0.4. :o
So the above becomes this:
VSG(avg peak) x 0.4 x VDMM/CSR OR
VSG(avg peak) x VDMM/CSR
Well yes and no to the confusion.
1st method-I multiply the VSG by .4 to get my average voltage over a full cycle. As my DMM set on amp's reads an average current over the full cycle already,i then multiply the average voltage by the DMM's current reading to get my P/in
2nd method-As my DMM set on VDC reads an average voltage over a full cycle,i can see what voltage i have over the CSR. This is then VDMM/CSR to get my I/in. This is then multiplied by the average voltage of the SG to get my P/in.
VSG average is VSG avg peak during the on time(40%) x .4 Then I average is VDMM/CSR-->V average x I average is P/in average.
OK-to sum up how i get my P/in.
First i work out the average V/in over the 40% on time using my scope. I then multiply this by 40% or x .4 to get my average voltage over a full cycle-or continuous voltage in. I then multiply this by the average current that my DMM show's.
Second method. I use my VDMM to get the average continuous voltage in,and multiply this by the IDMM to get my P/in.
Third method-i use my VDMM to get average V/in and use the second VDMM/CSR to get my current in.
Last method was using the scope to get both current average in,and voltage average in.
All methods are within 5% of each other,so must be close to right.
Quote from: poynt99 on 2014.09.20, 21:24:16
Here is the average power in each LED.
LED1 = 5.2mW
LED2 = 5.65mW
Looking at the scope shot associated with this post Darren, i think you have the two P/out's around the wrong way?.Should it read LED1= 5.65mW
LED2 = 5.2mW.
Anyway.looking forward to those measurements as soon as you get time.
just a recap.
First up, please remove D1 and D3 from the circuit.
1-what is the average P/in over the full cycles
1A-what is the average V/in over the 40% on time
1B-what is the average I/in over the 40% on time
2-what is the average voltage across LED1 for the 40% on time
3-what is the average current flowing into LED1 for that 40% on time
4-what is the average voltage across LED1 for the 60% off time
5-what is the average current across LED 1 for the 60% off time
6-same as q4 and q5 , but for LED2
Cheers
Brad
Again Brad,
You are still multiplying by an extra duty factor of 0.4 when it is unnecessary. The DMM is already applying that duty factor for you.
It doesn't matter which parameter the duty factor is applied to, V or I, as all three are multiplied together anyway. So:
V x I x 0.4 is the same as 0.4 x V x I is the same as I x V x 0.4. They all account for the duty cycle ONCE, which is all that we want anyway.
Quote from: TinMan on 2014.09.23, 14:35:39
First up, please remove D1 and D3 from the circuit.
OK, but two things, the circuit operates different now, some current below 0V. Also D2 is still there so still not accounting for all power dissipation.
Quote
1-what is the average P/in over the full cycles
About 16.5mW
Quote
1A-what is the average V/in over the 40% on time
Vin (Vsg) is clean so it is 9Vp.
Quote
1B-what is the average I/in over the 40% on time
About 4.8mA
Quote
2-what is the average voltage across LED1 for the 40% on time
About 3.3V for my LED model.
Quote
3-what is the average current flowing into LED1 for that 40% on time
Same as the input current, about 4.8mA.
Quote
4-what is the average voltage across LED1 for the 60% off time
About 3.1V for my LED model.
Quote
5-what is the average current across LED 1 for the 60% off time
About 0.75mA
Quote
6-same as q4 and q5 , but for LED2
q4: About 3.25V
Quote
6-same as q4 and q5 , but for LED2
q5: About 3.2mA
I'm not sure what all these numbers mean to you, but there you are. Keep it real man. ;)
Using a Photodyne Chopped Light Multimeter and grabbing a bright white led off the bench of unknown vintage, I ran these numbers using a HP constant current source over three decades of current.
Voltage across the Led is not the best way to characterize leds, better to use current through the LED, as the voltage drifts with heating. When voltage finally stabilizes, power may then be computed.
If these numbers are correct, they show a rather low efficiency, which can possibly be improved a bit with pulse mode operation. There is an interesting decline in efficiency when operated at higher steady state power.
We are still a very far cry from OU, even with the very best PV cells, since we lose almost 98% of the LED input power to heat and get less than 2% light power.
The Led was held right up to the sensing silicon receptor, positioned as close as possible. Of course I don't have full specs on the meter and it may not be seeing the LED at it's optimum power point or wavelength, nor do I have specs on the LED.
I also suspect the LED I tested may be partially damaged although it seems to light very brightly to the eye with 0.1 mA current. Until i get my hands on a new supply with data sheets and curves, I will consider my numbers suspect.
Never thought LEDs would be that inefficient. :o
Here is a document which talks about LED efficiency and efficacy.
Quote from: poynt99 on 2014.09.24, 13:34:32
Here is a document which talks about LED efficiency and efficacy.
This may sound like a stupid question,but how do you calculate power disipated by the LED?
If we look at your measurements for the 40% on time Darren,is it VxA=Watt's?
So LED 1 during the 40% on time is 3.3v x 4.8mA=15.84mW.
Yes.
Quote from: poynt99 on 2014.09.24, 14:02:12
Yes.
OK,so for the 60% off time it is 3.1v x .75mA=2.325mW.
So do we add these two amounts together to get the total disipated power by LED 1 for a full 100% cycle?
15.84mW + 2.325mW= 18.165mW ?
Quote from: poynt99 on 2014.09.24, 13:34:32
Here is a document which talks about LED efficiency and efficacy.
Turns out I just downloaded and read that document and a pile of others on the subject yesterday. Yes, there is a lot of confusion between the terms as applied to light sources. Wikipedia has a good entry on the subject also.
The bright white LED I used for the tests was probably eight to ten years old, I'm sure a lot of improvements have been made since then.
I ran a few lower level tests this morning. At 0.001 mA (1uA) I can just detect with my eye the light of the LED. It is dropping 2.110 volts at that point. So that is 2.110 uWatts input. My photo meter does not register this low. I would probably have to set up a sensitive phototube detector.
At 10 uA input, Vdrop is 2.511 volts for 25.11 uWatts input. This is definitely visible, and is why some of the LED long run devices (Lasersaber etc.) fascinate some people.
Using pulse mode and concentrating that 25.11 uW into a 10% duty cycle 251.1 uW pulse, at a 20 pps rate, the eye will see what appears to be a very bright LED running continuously.
BTW, the supposed 230% efficient LED's light cannot be seen with the unaided eye as it is in the range of 60 pW.
Still they are interesting if scavenging energy like a heat pump and turning it into light.
Anyone have a manual for a Photodyne 33XLC with a #550 measuring head?
Regards, ION
OK-summery here--> if correct.
Using your V/in and I/in Darren(as posted) we have for the 40% on time-Vsg 9v.
Current is 4.8mA.
Total P/in for 40% on time is 43.2mW. We then multiply this by .4 to get P/in average for full 100% cycle.
17.28mW.
LED 1 power disipation for 40% on time is-->3.3v x 4.8mA=15.84mW.-->This you confirmed in post 108
LED 1 power disipation for 60% off time is-->3.1v x .75mA=2.325mW
So LED 1 disipated 15.84mW of power in the first 40% of the cycle(on time),and another 2.325mW in the other 60% of the cycle.
So LED 1 has in total disipated 18.165mW over 1 full cycle.
So-why is this more than your average P/in of 17.28mW ?
I eye-balled all those figures, as they are quite difficult to obtain from any kind of measurement. So it could be that I was off enough that the numbers don't quite add up to zero.
It would be far easier, and make more sense to get the average power over several cycles, then compare. This is how I always do it. It is not standard practice to do power measurements they way you are doing them here.
Quote from: poynt99 on 2014.09.24, 18:08:22
I eye-balled all those figures, as they are quite difficult to obtain from any kind of measurement. So it could be that I was off enough that the numbers don't quite add up to zero.
It would be far easier, and make more sense to get the average power over several cycles, then compare. This is how I always do it. It is not standard practice to do power measurements they way you are doing them here.
Darren
We spent how much time in our last chat you trying to explain to me how we get the average P/in-and finally i seen my mistake. Now after all that,you are saying -this is not standard practice to do it this way-->this is after you have shown me that your P/in measurement method was correct,and gave us our average P/in. You also say that the 9vsg is very clean and accurate,so that only leaves a slight +/- for the I/in-->but it must be close !right?! . Then above,you say Quote: I eye-balled all those figures, as they are quite difficult to obtain from any kind of measurement. So it could be that I was off enough that the numbers don't quite add up to zero.
This is fair enough,but we are only looking at the power disipated by LED 1--> we havnt even taken into account LED 2,the inductor/resistor and D2 losses yet. I am also sure that if we take say 10 complete cycles,and average them all out,the result will be the same. You have always stated that your sim will give very close and accurate measurements to that of a real device.
Quote post 103--I'm not sure what all these numbers mean to you, but there you are. Keep it real man.
This is the reason i wanted these numbers,and they have a whole lot of meaning-as you can see. From these numbers you have given,there comes answers. Cycle by cycle we can look at what is happening within the circuit,so either your P/in average is way off,or your disipated power over the components within the circuit is way off,or we are doing the math wrong-the answer has to be one of these, because as i stated,we are only looking at LED 1 ATM,and no account for power disipated in LED 2,the inductor/resistor and D1 have been taken into account yet.
Do you think it deserves a closer look now ?.
Quote from: TinMan on 2014.09.24, 14:38:35
OK-summery here--> if correct.
Using your V/in and I/in Darren(as posted) we have for the 40% on time-Vsg 9v.
Current is 4.8mA.
Total P/in for 40% on time is 43.2mW. We then multiply this by .4 to get P/in average for full 100% cycle.
17.28mW.
LED 1 power disipation for 40% on time is-->3.3v x 4.8mA=15.84mW.-->This you confirmed in post 108
LED 1 power disipation for 60% off time is-->3.1v x .75mA=2.325mW
So LED 1 disipated 15.84mW of power in the first 40% of the cycle(on time),and another 2.325mW in the other 60% of the cycle.
So LED 1 has in total disipated 18.165mW over 1 full cycle.
So-why is this more than your average P/in of 17.28mW ?
Because you are not performing the computations correctly. For your LED power dissipation, you still have to multiply the VxI result by it's associated duty cycle. Remember, you asked for the average current and voltage during the 40% & 60% duty cycles separately (not the entire cycle), and multiplied together, yes you get around 15.84mW (for the 40%). But now you have to multiply that by 0.4 because you still have to account for what portion this represents over the entire cycle. So, your LED1 power computation should look more like this:
3.3v x 4.8mA x 0.4 = 6.34mW
and
3.1v x .75mA x 0.6 = 1.4mW
For a total of about
7.74mW. The actual power measured in the circuit (see attached) is about
7.42mW, so my eyeball approximations were not too far off actually.
Quote from: poynt99 on 2014.09.24, 23:16:28
Because you are not performing the computations correctly. For your LED power dissipation, you still have to multiply the VxI result by it's associated duty cycle. Remember, you asked for the average current and voltage during the 40% & 60% duty cycles separately (not the entire cycle), and multiplied together, yes you get around 15.84mW (for the 40%). But now you have to multiply that by 0.4 because you still have to account for what portion this represents over the entire cycle. So, your LED1 power computation should look more like this:
3.3v x 4.8mA x 0.4 = 6.34mW
and
3.1v x .75mA x 0.6 = 1.4mW
For a total of about 7.74mW. The actual power measured in the circuit (see attached) is about 7.42mW, so my eyeball approximations were not too far off actually.
I dont think this is right, as LED1 is on 100% of the cycle-so why the need to multiply by the input duty cycle.
The sum of the 40% cycle and the 60% cycle of LED1 should be the average of the complete cycle
Well, it IS right Brad. Think about it.
And the measurement that I posted is in close agreement. Are you really going to argue against actual measurements? ???
Quote from: poynt99 on 2014.09.25, 13:00:19
Well, it IS right Brad. Think about it.
And the measurement that I posted is in close agreement. Are you really going to argue against actual measurements? ???
OK-lets say i agree for the time being.
Next question-during the 40% on time,what is the power being disipated by the inductor. We know LED 1 is disipating 15.84mW during that 40% of the cycle,and averages out over the complete cycle to 6.336mW. But the inductor is a wire wound resistor,and anything with resistance disipates heat when a current flows through it. I understand that there is also losses due to inductive heating of the core,but lets just look at the resistive side of it for now.
From this 40% measurement,we then of course multiply by .4 to get the over all disipation of the inductor over a full cycle.
Agreed.
So in my sim it would be the power dissipated by the 9.2 Ohm resistor.
Quote from: poynt99 on 2014.09.25, 13:44:40
Agreed.
So in my sim it would be the power dissipated by the 9.2 Ohm resistor.
So now i want to make sure this is correct.
The inductors resistance will be higher than the 9.2 ohm,s due to the skin effect-this we can work out,but not nessasary.
The P/in for the 40% is 9v x 4.8mA = 43.2mW. Averaged out over the complete cycle,it is 17.28mW
LED 1 power disipation during the 40% on time is 3.3 x 4.8 =15.84mW-the average over a complete cycle is 6.336mW
The inductor/resistor power disipation would there for be the difference between P/in and LED 1,as it is a series circuit during the 40% on time.
So inductor/resistor is 9v-3.3v=5.7v x 4.8mA=27.36mW. So the average would be 10.944mW.
10.944mW + 6.336mW=our P/in of 17.28mW
Is there an error in the above?.
If there was no CSR resistor of 200 Ohms, then there would be no error. The CSR is also in series and will have dissipation. So that remaining 5.7V will be split between the inductor and the CSR.
OK-yes,i forgot about the 200ohm CSR. Lets remove that for now.
So you are sure the below is correct.
Quote Brad: The P/in for the 40% is 9v x 4.8mA = 43.2mW. Averaged out over the complete cycle,it is 17.28mW
LED 1 power disipation during the 40% on time is 3.3 x 4.8 =15.84mW-the average over a complete cycle is 6.336mW
The inductor/resistor power disipation would there for be the difference between P/in and LED 1,as it is a series circuit during the 40% on time.
So inductor/resistor is 9v-3.3v=5.7v x 4.8mA=27.36mW. So the average would be 10.944mW.
10.944mW + 6.336mW=our P/in of 17.28mW
Is there an error in the above?.
Quote Darren: If there was no CSR resistor of 200 Ohms, then there would be no error.
This would mean that at this time,the inductors resistance is 1187.5 ohms
We could check this simply by replacing the inductor on your sim with the above value resistor(-the CSR),and we still should get the same power disipated by LED 1 during the 40% on time-correct?.
Brad,
If you short out the 200 Ohm resistor and then measure the LED1 power with the inductor resistance first at 9.2 Ohms, then at 1190 Ohm, there is a huge difference in power, like 40mW vs. 6mW.
This is a dynamic circuit, and things are more complicated than just replacing the inductor with a big resistor. Where are you headed with this? I have to admit this is becoming quite tedious.
Quote from: poynt99 on 2014.09.25, 22:53:43
Brad,
Where are you headed with this? I have to admit this is becoming quite tedious.
All good Darren
Wont take up any more of your time
Thanks for your help
Brad
To round off this discussion in case you're interested, here is a tally of all the average power dissipations in the simulated circuit. All components included:
Pin (PSG) = 15.2mW
D3 = 1.15mW
D2 = 0W
D1 = 1.15mW
LED2 = 5.1mW
LED1 = 5.58mW
Rcoil = 0.15mW
CSR (216.1) = 1.9mW
Sum total (not including Pin) = 15.03mW (rounding and estimating errors account for the small difference with Pin)
Quote from: poynt99 on 2014.09.26, 23:19:20
To round off this discussion in case you're interested, here is a tally of all the average power dissipation's in the simulated circuit. All components included:
Pin (PSG) = 15.2mW
D3 = 1.15mW
D2 = 0W
D1 = 1.15mW
LED2 = 5.1mW
LED1 = 5.58mW
Rcoil = 0.15mW
CSR (216.1) = 1.9mW
Sum total (not including Pin) = 15.03mW (rounding and estimating errors account for the small difference with Pin)
Thanks Darren.
I will have to look at this myself with my circuit,as the number above make no sense in regards to our previous number crunching.
Quote: So inductor/resistor is 9v-3.3v=5.7v x 4.8mA=27.36mW. So the average would be 10.944mW.
10.944mW + 6.336mW=our P/in of 17.28mW
Is there an error in the above?.
Quote Darren: If there was no CSR resistor of 200 Ohms, then there would be no error.
So if we take away the power dissipation of the 200 ohm CSR(1.9mw),how is it that your above R/coil is dissipating only .15mW?
Where the numbers you presented in post 103 with the 200 OHM CSR in place?,or without it-this may be where I'm messing up here,as i assumed it was still in play,and that is how you were getting your current measurement.
What I'm trying to work out is-dose the inductor(being a resistor as well) dissipate as much power(in the form of heat)as a normal resistor would with the same amount of current passing through it?-in this pulsed state of course.
It is this that has me confused -Quote: If there was no CSR resistor of 200 Ohms, then there would be no error.
This was in reply to my statement Quote: So inductor/resistor is 9v-3.3v=5.7v x 4.8mA=27.36mW. So the average would be 10.944mW.
10.944mW + 6.336mW=our P/in of 17.28mW
So all our P/in is accounted for during the 40% on time-->Which brings the question-where dose the power come from that is dissipated in LED1, LED2 and the inductor during the 60% off time?
You ask where I'm going with this-well that is dependent on wether the inductor/resistor dissipates as much power(in way of heat) as a standard resistor would during only the 40% on time. If it dose,then the magnetic field built up around the inductor during the 40% on time took no power to create,and thus,the power delivered back to the system during the 60% off time is excess power.This would also explain as to why i can get more actual light power output than that of straight DC of the same P/in.
I will put together a circuit,where as i can swap the inductor out for a resistor of the same value(calculated while circuit is running),and measure the temp of both the inductor and resistor over a set run time. This will tell me if both give off the same amount of heat in the same situation.I would expect that the inductor should give off more heat,as it also has current flowing through it during the 60% off time-is this something you may have missed?.
Anyway,don't spend any more time on it Darren,as you have done enough already,and i thank you for that.
Cheers
Brad
Quote from: TinMan on 2014.09.27, 03:01:43
Thanks Darren.
I will have to look at this myself with my circuit,as the number above make no sense in regards to our previous number crunching.
So if we take away the power dissipation of the 200 ohm CSR(1.9mw),how is it that your above R/coil is dissipating only .15mW?
Where the numbers you presented in post 103 with the 200 OHM CSR in place?,or without it-this may be where I'm messing up here,as i assumed it was still in play,and that is how you were getting your current measurement.
If you remove the CSR, then the circuit needs to be remeasured because a lot has now changed. We can't just remove its power dissipation from the present measurements. Yes, the CSR was in place, and D1 and D3 were shorted. With the CSR in place, the CSR should dissipate quite a bit more power than the inductor.
Quote
So all our P/in is accounted for during the 40% on time-->
No, and not sure how you reached that conclusion (and if I understand correctly).
Quote
This would also explain as to why i can get more actual light power output than that of straight DC of the same P/in.
I won't be surprised if the difference you speak of is the result of efficiencies associated with each drive method.
Quote
I will put together a circuit,where as i can swap the inductor out for a resistor of the same value(calculated while circuit is running),and measure the temp of both the inductor and resistor over a set run time. This will tell me if both give off the same amount of heat in the same situation.I would expect that the inductor should give off more heat,as it also has current flowing through it during the 60% off time-is this something you may have missed?.
I'm fairly certain I haven't missed anything. I predict your resistor (of equal mass) will get about 20x hotter than the inductor. So if the inductor raises by 1ºC, the resistor will get about 20ºC above ambient.
:P
Ok-here is why i find our discusions on this confusing.Below are the last numbers you posted.
Pin (PSG) = 15.2mW
D3 = 1.15mW
D2 = 0W
D1 = 1.15mW
LED2 = 5.1mW
LED1 = 5.58mW
Rcoil = 0.15mW
CSR (216.1) = 1.9mW
Post 103- Also D2 is still there so still not accounting for all power dissipation.
And the above says D2=0mW ?
Post 117 i quoted: Next question-during the 40% on time,what is the power being disipated by the inductor. We know LED 1 is disipating 15.84mW during that 40% of the cycle,and averages out over the complete cycle to 6.336mW. But the inductor is a wire wound resistor,and anything with resistance disipates heat when a current flows through it. I understand that there is also losses due to inductive heating of the core,but lets just look at the resistive side of it for now.
From this 40% measurement,we then of course multiply by .4 to get the over all disipation of the inductor over a full cycle.
Your answer on post 118 was quote: Agreed.So in my sim it would be the power dissipated by the 9.2 Ohm resistor.
So using the numbers you posted on post 103 with D1 and D3 omited,we have during the 40% on time
VSG 9v
I/in 4.8mA
This is 43.2mW in during the 40% on time.
LED1 during this 40% on time is
V=3.3
I=4.8
This is 15.84mW
The 216.1 ohm CSR would be disipating 4.978mW during this 40% on time.Our series circuit at this 40% on time is P/in-CSR-inductor/resistor-LED1-ground
P/in43.2mW
Power disipated is LED1 + CSR + Inductor/resistor-the only things in series during the 40% on time.
15.84 + 4.978=20.818mW.This means that the inductor/resistor must be disipating 22.382mW at this time. This averaged out is 18.382mW over the full cycle.
Your reply to my post 119 was-Quote:If there was no CSR resistor of 200 Ohms, then there would be no error. The CSR is also in series and will have dissipation. So that remaining 5.7V will be split between the inductor and the CSR.
So the CSR being 216.1 ohms and has a current of 4.8mA going through it,will have a voltage of 1.037 across it. This is then subtracted from the 5.7v as you stated,giving us 4.663v across the inductor/resistor. 4.663 x 4.8mA is 22.382mW-the exact same amount i stated above.
All these amounts are calculated from your amounts given in post 103.
Also your statement-Quote:So in my sim it would be the power dissipated by the 9.2 Ohm resistor.
This is not correct,as the inductors resistance would not be 9.2 ohms at these frequencies.In post 121 i stated-Quote: This would mean that at this time,the inductors resistance is 1187.5 ohms
Turns out this is correct,as the 216.1 ohms is also included in that resistance value.-the one you thought i left out
4.663v across the inductor/resistor with 4.8mA flowing through it,give us a resistance of 971.44 ohms. If we add the 216.1 ohm CSR to this,we get 1187.54 ohms-the very same amount i posted in post 121.
If the inductors resistance was what you quoted(9.2 ohms),then the voltage across it at 4.8mA would only be .044v,and be disipating only .211mW.
If we add all the voltages up using your 9.2 ohms,we are missing 4.619volts.
If we add al the voltages up that i calculated,you will see that it is exactly the same as the V/in
LED1 3.3 + 216.1 ohm CSR 1.037 + inductor/resistor 4.663=9v
9v x 4.8=43.2-the very P/in we started with.
Darren-all my numbers are correct,and account for all the P/in in that 40% on time. They also average out to be exactly the same as the P/in average.The inductor has a resistance of 971.4 ohms,and4.8mA flowing through it(all the numbers tell us this)The reason this resistance has rissen to this amount from the 9.2 ohms you posted(and i gave)is of course because it is an inductor aswell,and the skin effect due to frequency has caused the resistance to rise. So in acordence with ohms law,the power disipated by the inductor during tho 40% on time is indeed 22.382 milliwatts. This is also proven by dropping the frequency from the SG. As the frequency drops,so dose the resistance-the skin effect is becoming less. this also explains why LED1 gets brighter and a higher voltage is measured across it when i drop the frequency down-the resistance of the inductor drops aswell,so there is less power disipated by the inductor,and our I/in go's up due to the lower resistance.
So your R/coil disipation of .15mW is incorrect because your resistance value of 9.2 ohms is incorrect.
Quote from: TinMan on 2014.09.27, 07:47:10
Ok-here is why i find our discusions on this confusing.Below are the last numbers you posted.
Pin (PSG) = 15.2mW
D3 = 1.15mW
D2 = 0W
D1 = 1.15mW
LED2 = 5.1mW
LED1 = 5.58mW
Rcoil = 0.15mW
CSR (216.1) = 1.9mW
Post 103- Also D2 is still there so still not accounting for all power dissipation.
And the above says D2=0mW ?
Yes, that's right. It turns out that D2 (the middle diode) plays virtually no role at all in this circuit (at least in my sim). It has only uA of current and less than 5uW of power dissipation. It can be removed with virtually no change in circuit operation (try it to confirm).
Quote
Your answer on post 118 was quote: Agreed.So in my sim it would be the power dissipated by the 9.2 Ohm resistor.
So using the numbers you posted on post 103 with D1 and D3 omited,we have during the 40% on time
VSG 9v
I/in 4.8mA
This is 43.2mW in during the 40% on time.
LED1 during this 40% on time is
V=3.3
I=4.8
This is 15.84mW
The 216.1 ohm CSR would be disipating 4.978mW during this 40% on time.Our series circuit at this 40% on time is P/in-CSR-inductor/resistor-LED1-ground
P/in43.2mW
Power disipated is LED1 + CSR + Inductor/resistor-the only things in series during the 40% on time.
15.84 + 4.978=20.818mW.This means that the inductor/resistor must be disipating 22.382mW at this time. This averaged out is 18.382mW over the full cycle.
You sure 18.382mW is correct? Did you apply a factor of 0.4 to 22.382mW to average it out over the full cycle as you stated?
Anyway, I understand your present confusion. You are correct, the power dissipations for the 40% don't add up. It looks like we are putting more power into the circuit than what can be accounted for in the LED1, CSR and inductive resistor.
I think the first step to understanding what is going on in this circuit is to realize, or at least remember, that pure inductors and capacitors don't dissipate any real power. They exhibit
reactance, and while they appear to dissipate power in one part of the cycle, they return all of that power in the next part of the cycle, for a net power of 0W. In the case of an inductor (with a perfect lossless core) only the resistive series resistance Rs actually dissipates power, in other words only the resistance in the inductor's wiring dissipates power and heats up. In our case, the 9.2 Ohms resistance is the only part of the coil that dissipates any
real power. The inductance portion of the coil only exhibits reactive or apparent power. Real power results in heating, while reactive power does not.
Now we know that the 9.2 Ohm resistor "Rs" is only dissipating about 0.22mW (with D1 and D3 shorted), the LED1 about 15.84mW, and the CSR about 4.98mW for a total of about 21.04mW. With a Pin of 43.2mW, where is the remaining 22.16mW during this 40% part of the cycle?
Well, we have to go back to the discussion about inductors and how they work. As you pulse it, the inductive reactance of the coil begins to energize, and during this process it uses energy from the supply SG. In our case an average of about 22.16mW of power over that 40% portion of the cycle. But over the remaining 60% portion of the cycle, it then returns most of that "energy" back to the circuit (see scope shot below). If we look at the average power dissipated only in the inductance (reactance) of the coil, we will see that it is rapidly heading towards 0mW of power, but it takes some time. Eventually if the sim was run long enough the average goes down to 0mW (see second scope shot below). Understand that it is well known that inductive reactances don't dissipate any net power. All they can do is store and release energy, minus any losses due to core heating and resistance in the wire.
Quote from: poynt99 on 2014.09.27, 14:21:11
Yes, that's right. It turns out that D2 (the middle diode) plays virtually no role at all in this circuit (at least in my sim). It has only uA of current and less than 5uW of power dissipation. It can be removed with virtually no change in circuit operation (try it to confirm).
You sure 18.382mW is correct? Did you apply a factor of 0.4 to 22.382mW to average it out over the full cycle as you stated?
Anyway, I understand your present confusion. You are correct, the power dissipation's for the 40% don't add up. It looks like we are putting more power into the circuit than what can be accounted for in the LED1, CSR and inductive resistor.
I think the first step to understanding what is going on in this circuit is to realize, or at least remember, that pure inductors and capacitors don't dissipate any real power. They exhibit reactance, and while they appear to dissipate power in one part of the cycle, they return all of that power in the next part of the cycle, for a net power of 0W. In the case of an inductor (with a perfect lossless core) only the resistive series resistance Rs actually dissipates power, in other words only the resistance in the inductor's wiring dissipates power and heats up. In our case, the 9.2 Ohms resistance is the only part of the coil that dissipates any real power. The inductance portion of the coil only exhibits reactive or apparent power. Real power results in heating, while reactive power does not.
Now we know that the 9.2 Ohm resistor "Rs" is only dissipating about 0.22mW (with D1 and D3 shorted), the LED1 about 15.84mW, and the CSR about 4.98mW for a total of about 21.04mW. With a Pin of 43.2mW, where is the remaining 22.16mW during this 40% part of the cycle?
Well, we have to go back to the discussion about inductors and how they work. As you pulse it, the inductive reactance of the coil begins to energize, and during this process it uses energy from the supply SG. In our case an average of about 22.16mW of power over that 40% portion of the cycle. But over the remaining 60% portion of the cycle, it then returns most of that "energy" back to the circuit (see scope shot below). If we look at the average power dissipated only in the inductance (reactance) of the coil, we will see that it is rapidly heading towards 0mW of power, but it takes some time. Eventually if the sim was run long enough the average goes down to 0mW (see second scope shot below). Understand that it is well known that inductive reactances don't dissipate any net power. All they can do is store and release energy, minus any losses due to core heating and resistance in the wire.
That is my point Darren,the resistance is not 9.2 ohms at these frequencies. the resistance increases as the skin effect dose. As you know,the skin effect reduces the amount of current that can flow through the inductors winding,as the cross section of the conducting wire is reduced in effect-and of course,the less current that can flow through this conducting wire means a higher resistance is present. The other mistake here(and by most)is the thinking that it takes power to form the magnetic field around the inductor-which is what delivers the power to the LED's when the 40% on cycle switches off.This is not correct. It takes no power to form the magnetic field,as it is only the bi product of current flowing through the conducting winding's. This magnetic field can be increased in strength without any additional power input to the inductor coil. The power given by the magnetic fields collaps around the inductor can be taken and used without any effect at all to the power input to that inductor. When the inductor is being pulsed at a set frequency and P/in,it will use x amount of power. If we unwind the wire from the inductor,and ran it in a straight line,it will still consume x amount of power and dissipate the exact same amount of power as it did formed as a coil-with or without the core.
I know all this to be true,as i carried out all these test today.
Quote from: TinMan on 2014.09.27, 15:03:10
That is my point Darren,the resistance is not 9.2 ohms at these frequencies. the resistance increases as the skin effect dose. As you know,the skin effect reduces the amount of current that can flow through the inductors winding,as the cross section of the conducting wire is reduced in effect-and of course,the less current that can flow through this conducting wire means a higher resistance is present.
The resistance may be slightly higher due to the skin effect, but not much. We should consider the series resistance of the inductor as 9.2 Ohms.
Quote
The other mistake here(and by most)is the thinking that it takes power to form the magnetic field around the inductor-which is what delivers the power to the LED's when the 40% on cycle switches off.This is not correct.
It does indeed require energy from the supply to energize an inductor, and this is clearly shown in the scope shot above. In fact, this is the mysterious missing energy previously not accounted for, as per my post above.
Quote
It takes no power to form the magnetic field,as it is only the bi product of current flowing through the conducting winding's.
This is not correct. It would seem you are making assumptions, and/or misinterpreting your results, and this is skewing your understanding. If you replace the windings with a straight piece of wire, where does your stored energy (built up magnetic field) go? It's gone. But at the same time, the entire circuit operation has dramatically changed as well.
The magnetic field is built up BECAUSE the inductive windings are present, and as this winding represents an inductive reactance (an imaginary resistance if you will), it draws power/energy from the supply and stores it as a field around those windings. Inductance is a "dynamic resistance" (reactance) to a changing current and as you know resistance times current
2 equals power. So the coil does indeed draw power from and store energy from the SG supply. You are only fooling yourself if you believe otherwise.
When the power is cut off, the inductor returns that stored energy back to the circuit, wherever there is a current path, in this case through LED2. So for the 40%, all the components, excluding LED2 are dissipating some x power, AND at the same time the inductor is storing additional power from the supply, only to be used later in LED2.
Brad,
When you charge up a capacitor with a battery, is there not a transfer of energy from the battery to the capacitor? If your answer is "yes", then it is no different with an inductor, other than a capacitor is a voltage/parallel device, and an inductor is a current/series device. Current (i.e. series) is required to energize an inductor, whereas voltage (i.e. parallel) is required to charge a capacitor.
Quote from: poynt99 on 2014.09.27, 16:00:19
The resistance may be slightly higher due to the skin effect, but not much. We should consider the series resistance of the inductor as 9.2 Ohms.
The magnetic field is built up BECAUSE the inductive windings are present, and as this winding represents an inductive reactance (an imaginary resistance if you will), it draws power/energy from the supply and stores it as a field around those windings. Inductance is a "dynamic resistance" (reactance) to a changing current and as you know resistance times current2 equals power. So the coil does indeed draw power from and store energy from the SG supply. You are only fooling yourself if you believe otherwise.
When the power is cut off, the inductor returns that stored energy back to the circuit, wherever there is a current path, in this case through LED2. So for the 40%, all the components, excluding LED2 are dissipating some x power, AND at the same time the inductor is storing additional power from the supply, only to be used later in LED2.
I understand and believe that it is the magnetic field collapse around the inductor/coil that returns power back to the system,but i dont believe that this magnetic field requires any additional power above the resistance value of the coil to create it(still could be wrong on this one). If we take a long piece of wire that has a resistance of say 8 ohms,and we provide enough DC power to create say 8 volts across that wire,we would be using 8 watts of power. If we now wrap that wire to form multiple loops,we would create a magnetic field. Our power draw at this time would still be 8 watts,so it took no additinal power to create that magnetic field. I now can slide a core into that coil,and increase the strength of that magnetic field without anything changing on our P/in side. Power is only consumed when work is being done,and for work to be done there must be a resistance acting against a force. As there is no resistance to act against the magnetic field being built,then there is no work being done.The resistance only comes when that magnetic field collapses--> in the case of our circuit.
Today i will carry out some more test on pulsed inductors,and see where i end up.
Thanks once again Darren for your help and input.
Brad
Quote from: TinMan on 2014.09.27, 23:44:09
I understand and believe that it is the magnetic field collapse around the inductor/coil that returns power back to the system,
At least we agree on this part! 8)
Quote
but i dont believe that this magnetic field requires any additional power above the resistance value of the coil to create it(still could be wrong on this one).
This statement is errant, but I think I understand what you are trying to say. My answer: No, not true. Would a coil made from superconducting wire require energy from a source to build up a magnetic field?
Quote
If we take a long piece of wire that has a resistance of say 8 ohms,and we provide enough DC power to create say 8 volts across that wire,we would be using 8 watts of power.
How much
energy are you using though? See, now you are changing the playing field. You can't do that! We are dealing with a
dynamic circuit here. You are now equating this dynamic circuit with a static circuit, but they are far from equal.
Study my last scope shots (http://www.overunityresearch.com/index.php?action=dlattach;topic=2603.0;attach=15076) until they really sink in.
I mean really study it and think about it hard. Understanding that will save you a lot of time. If you don't understand it, ask; I'd be happy to explain it further.
Quiz: In my scope shot, has the energy buildup in the coil maxed out?
Quote: Quiz: In my scope shot, has the energy buildup in the coil maxed out?
As the blue current trace seems to be still climbing,and not yet leveled out,then i would say no. For it to reach its maximum energy level,i would think that would be at the point when the current levels out,but of course,by then the voltage has also began to drop accross the coil.
So i now need to see if what i have learned so far is correct. I have made a simple circuit (pictured below). Looking at that circuit,can you tell me what the average I/in is, and P/in average. Also,what amount of the P/in average will be disipated through the CSR and inductor?.
Cheers.
Brad
Any idea of the inductance?
Confirm the frequency; 70 Hz?
Quote from: poynt99 on 2014.09.28, 04:48:15
Any idea of the inductance?
Confirm the frequency; 70 Hz?
No idea of inductance-steel laminated core. Coil from shaded pole motor from microwave oven fan-240volt AC.
frequency is 70 Hz on SG.
I just happen to have a 220v coil of that type :)
If it helps without core 158mH
with core 630mH I'm sure that is going to be in the ball park of yours
regards
Mike 8)
Quote from: Centraflow on 2014.09.28, 10:17:52
I just happen to have a 220v coil of that type :)
If it helps without core 158mH
with core 630mH I'm sure that is going to be in the ball park of yours
regards
Mike 8)
Cheers Mike.
I must get me one of those flash meters that measure things like this on inductors.
Darren
Below is a scope shot on the last little circuit i posted. The yellow trace is across the 100 ohm CSR,and the blue trace is across the circuit/SG. Are my quotes on the shot correct? i believe so. I also have this funny feeling that my SG cannot supply enough current to keep the voltage up,and that is why the rapid drop in voltage across the circuit. I know there will be CEMF,but i didnt think it would be as much as seen on the scope shot?. I also find it odd that the current rises as the voltage drop's at these low frequencies-is this normal?
P.S - When i place the LED in the circuit,it has no effect what so ever on the voltage or current trace (P/in) cycle.The traces during the on time stay the same with or without the LED,as dose the P/in on the meters.
Cheers
Brad
Quote from: Centraflow on 2014.09.28, 10:17:52
I just happen to have a 220v coil of that type :)
If it helps without core 158mH
with core 630mH I'm sure that is going to be in the ball park of yours
regards
Mike 8)
Thanks Mike. Is yours about 100 Ohms DC as well?
Here is my sim to closely match your circuit.
Indeed your SG seems to have a relatively high output resistance. I added a 240 Ohm resistor to the SG output to match your wave forms.
Quote from: TinMan on 2014.09.28, 10:50:20
I also find it odd that the current rises as the voltage drop's at these low frequencies-is this normal?
You are getting a big voltage drop across the output resistance of the SG. So the wave forms do make sense from that perspective.
Quote
P.S - When i place the LED in the circuit,it has no effect what so ever on the voltage or current trace (P/in) cycle.The traces during the on time stay the same with or without the LED,as dose the P/in on the meters.
Cheers
Brad
Well, on my sim the Pin is 150mW, and the PLED is only 4.6mW, so you may not notice much difference in the wave forms with and without the LED installed. btw, almost half of the Pin is being burned up in the SG's output resistance of 240 Ohms (74.5mW).
The scope shot below indicates when the LED is receiving power. It is a much shorter duration than what you indicated on your scope shot.
Yes, this LED power is from the stored energy in the inductor, but only a small portion of it. The rest of the stored coil energy is burned up in the coil's own 111 Ohm series resistance (the majority), the CSR, and the SG's output resistance.
Quote from: poynt99 on 2014.09.28, 13:26:10
You are getting a big voltage drop across the output resistance of the SG. So the wave forms do make sense from that perspective.
Well, on my sim the Pin is 150mW, and the PLED is only 4.6mW, so you may not notice much difference in the wave forms with and without the LED installed. btw, almost half of the Pin is being burned up in the SG's output resistance of 240 Ohms (74.5mW).
Im not sure about the voltage drop,as i have the output resistance on the SG set to 0. My SG is capable of putting out 120mA-no problem at all. So really not sure why this is happening.
My I/average in is 22.42mA,and a Pv of 13v.If i remember correctly,we don't have to average out the voltage,as the current is already averaged out?. Once again if we do the sum using these numbers,we get 291.46mW. This would mean that the resistance would have to be 579.8 ohm's-which it is not,my CSR and inductor together have a combined resistance of 211 ohms. At these low frequencies,i think we could eliminate the skin effect altogether.
My best bet would be to make a short video,and show you what i see. When i add the LED,or even charge a battery from the flyback,my input current actually go's down slightly. It seems that i can use the stored energy without it effecting the P/in. I don't see core saturation being an issue at these low power and frequency levels.
I have never really looked this closely into what is happening within these types of circuits before,and having the new scope is great for doing this. Pulse by pulse i can follow the voltage and current flow through the circuit O0. So now my goal is to get to understand correctly what my scope is showing me. My dad was in the process of showing and teaching me,but shortly after we began,he became ill and died. As i know no other people around my area that is into this sort of stuff,i rely on people like your self Darren to teach me a few thing's. I know that i become demanding some time's,and when you have had enough,just let me know. But there is a reason that i want to learn,as there is a problem left to me by my dad that needs solving.
Anyway,i will get that short video made tomorrow,so as you can see first hand what I'm seeing-as some times i may not explain things correctly in words.
Cheers
Brad
Quote from: poynt99 on 2014.09.28, 12:25:00
Thanks Mike. Is yours about 100 Ohms DC as well?
No, 278 ohm
regards
Mike 8)
No problem helping out Brad. Looking forward to your video.
Thanks Mike.