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Author Topic: LED driver-may have hit the jackpot.  (Read 1157 times)
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img  TinMan
LED driver-may have hit the jackpot. « on: 2014-09-12, 19:19:58 »

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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.


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Re: LED driver-may have hit the jackpot. « Reply #1 on: 2014-09-13, 00:42:43 »

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It's not as complicated as it may seem...
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?


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Re: LED driver-may have hit the jackpot. « Reply #2 on: 2014-09-13, 01:29:13 »

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Quote from: poynt99 on 2014-09-13, 00:42:43
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.


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Re: LED driver-may have hit the jackpot. « Reply #3 on: 2014-09-13, 04:41:22 »

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@ 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.


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Re: LED driver-may have hit the jackpot. « Reply #4 on: 2014-09-13, 05:21:12 »

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@ 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


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Re: LED driver-may have hit the jackpot. « Reply #5 on: 2014-09-13, 08:12:26 »

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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.



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Re: LED driver-may have hit the jackpot. « Reply #6 on: 2014-09-13, 10:21:27 »

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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.


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Re: LED driver-may have hit the jackpot. « Reply #7 on: 2014-09-13, 11:54:08 »

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@ 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


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Re: LED driver-may have hit the jackpot. « Reply #8 on: 2014-09-13, 13:20:59 »

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It's not as complicated as it may seem...
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


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Re: LED driver-may have hit the jackpot. « Reply #9 on: 2014-09-13, 13:32:55 »

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It's not as complicated as it may seem...
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.


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Re: LED driver-may have hit the jackpot. « Reply #10 on: 2014-09-13, 13:43:54 »

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It's not as complicated as it may seem...
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.


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Re: LED driver-may have hit the jackpot. « Reply #11 on: 2014-09-13, 13:50:03 »

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Quote from: poynt99 on 2014-09-13, 13:43:54
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?


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Re: LED driver-may have hit the jackpot. « Reply #12 on: 2014-09-13, 14:09:25 »

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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.


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Re: LED driver-may have hit the jackpot. « Reply #13 on: 2014-09-13, 14:23:08 »

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It's not as complicated as it may seem...
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.


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"Some scientists claim that hydrogen, because it is so plentiful, is the basic building block of the universe. I dispute that. I say there is more stupidity than hydrogen, and that is the basic building block of the universe." Frank Zappa
   
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Re: LED driver-may have hit the jackpot. « Reply #14 on: 2014-09-13, 14:26:31 »

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It's not as complicated as it may seem...
So,

26.4mV/15.2 x 6Vp x 0.38 = 3.96mW for Pin.


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Re: LED driver-may have hit the jackpot. « Reply #15 on: 2014-09-13, 14:41:14 »

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Quote from: poynt99 on 2014-09-13, 14:26:31
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.


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Re: LED driver-may have hit the jackpot. « Reply #16 on: 2014-09-13, 15:13:15 »

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It's not as complicated as it may seem...
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.


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Re: LED driver-may have hit the jackpot. « Reply #17 on: 2014-09-13, 15:20:59 »

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Quote from: poynt99 on 2014-09-13, 15:13:15
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.


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Re: LED driver-may have hit the jackpot. « Reply #18 on: 2014-09-13, 16:13:24 »
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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.


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Re: LED driver-may have hit the jackpot. « Reply #19 on: 2014-09-13, 17:33:51 »

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Quote from: ION on 2014-09-13, 16:13:24
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.


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Re: LED driver-may have hit the jackpot. « Reply #20 on: 2014-09-14, 07:29:48 »

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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%



« Last Edit: 2014-09-14, 09:04:07 by TinMan »


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Re: LED driver-may have hit the jackpot. « Reply #21 on: 2014-09-14, 07:55:42 »

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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%


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Re: LED driver-may have hit the jackpot. « Reply #22 on: 2014-09-14, 09:59:14 »

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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.


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Re: LED driver-may have hit the jackpot. « Reply #23 on: 2014-09-14, 12:55:18 »

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It's not as complicated as it may seem...
Brad,

How did you determine that Iin is 2mA?

Why is LED1 on 100%? Isn't it also in series with LED2?


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Re: LED driver-may have hit the jackpot. « Reply #24 on: 2014-09-14, 13:11:56 »

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Quote from: poynt99 on 2014-09-14, 12:55:18
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.


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