LED driver-may have hit the jackpot.

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poynt99

Re: LED driver-may have hit the jackpot. « Reply #75 on: 2014-09-20, 22:24:16 »

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

Here is the average power in each LED.

LED1 = 5.2mW
LED2 = 5.65mW

LED driver-may have hit the jackpot.

PLEDs.png (13.24 kB, 717x587 - viewed 58 times.)

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TinMan

Re: LED driver-may have hit the jackpot. « Reply #76 on: 2014-09-21, 02:55:56 »

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Quote from: poynt99 on 2014-09-20, 16:53:38

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.

LED driver-may have hit the jackpot.

P-in scope shot..JPG (49.27 kB, 720x351 - viewed 60 times.)

« Last Edit: 2014-09-21, 04:22:03 by TinMan »

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poynt99

Re: LED driver-may have hit the jackpot. « Reply #77 on: 2014-09-21, 14:28:44 »

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

Brad,

I explained the process (perhaps poorly) in post #73.

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

LED driver-may have hit the jackpot.

current_trace_pulse.jpg (178.96 kB, 720x351 - viewed 61 times.)

« Last Edit: 2014-09-21, 15:53:59 by poynt99 »

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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

poynt99

Re: LED driver-may have hit the jackpot. « Reply #78 on: 2014-09-21, 14:44:10 »

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

Quote from: TinMan on 2014-09-16, 14:56:33

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] = 4.16mA

Note that if the current trace was already a pulse with 1.5Vp, we would have this:
(1.5Vp / 216.1) = 6.94mA

Note 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.

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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

Itsu

Re: LED driver-may have hit the jackpot. « Reply #79 on: 2014-09-21, 16:03:35 »

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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

LED driver-may have hit the jackpot.

IP 10mmled.png (118.4 kB, 875x712 - viewed 55 times.)

LED driver-may have hit the jackpot.

PhotoCell output 1.png (12.23 kB, 818x527 - viewed 53 times.)

« Last Edit: 2014-09-21, 22:57:26 by Itsu »

TinMan

Re: LED driver-may have hit the jackpot. « Reply #80 on: 2014-09-22, 01:45:56 »

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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

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poynt99	Re: LED driver-may have hit the jackpot. « Reply #81 on: 2014-09-22, 04:11:51 »
Group: Administrator Hero Member Posts: 3538 It's not as complicated as it may seem...	Why remove D1 and D3? --------------------------- "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

TinMan	Re: LED driver-may have hit the jackpot. « Reply #82 on: 2014-09-22, 06:07:17 »
Group: Elite Experimentalist Hero Member Posts: 5677	Quote from: poynt99 on 2014-09-22, 04:11:51 Why remove D1 and D3? So as we dont have to calculate the power disipation for these two components. --------------------------- Never let your schooling get in the way of your education.

TinMan

Re: LED driver-may have hit the jackpot. « Reply #83 on: 2014-09-22, 11:43:42 »

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Quote from: poynt99 on 2014-09-21, 14:28:44

Brad,

I explained the process (perhaps poorly) in post #73.

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.

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poynt99

Re: LED driver-may have hit the jackpot. « Reply #84 on: 2014-09-22, 13:40:34 »

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

Quote from: TinMan on 2014-09-22, 06:07:17

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?

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TinMan	Re: LED driver-may have hit the jackpot. « Reply #85 on: 2014-09-22, 13:46:47 »
Group: Elite Experimentalist Hero Member Posts: 5677	Quote from: poynt99 on 2014-09-22, 13:40:34 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. --------------------------- Never let your schooling get in the way of your education.

poynt99

Re: LED driver-may have hit the jackpot. « Reply #86 on: 2014-09-22, 13:55:47 »

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

Quote from: TinMan on 2014-09-22, 11:43:42

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:

Quote

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.

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.

---------------------------

"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

poynt99

Re: LED driver-may have hit the jackpot. « Reply #87 on: 2014-09-22, 14:13:38 »

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

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

Note: this only works when the current and voltage are unipolar, i.e. above or below 0V.

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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

ion

Re: LED driver-may have hit the jackpot. « Reply #88 on: 2014-09-22, 14:20:06 »

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It's turtles all the way down

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?

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poynt99

Re: LED driver-may have hit the jackpot. « Reply #89 on: 2014-09-22, 14:32:58 »

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

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.

---------------------------

"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

TinMan

Re: LED driver-may have hit the jackpot. « Reply #90 on: 2014-09-22, 14:33:42 »

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Quote from: ION on 2014-09-22, 14:20:06

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.

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TinMan

Re: LED driver-may have hit the jackpot. « Reply #91 on: 2014-09-22, 14:39:10 »

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Quote from: poynt99 on 2014-09-22, 14:32:58

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

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ion

Re: LED driver-may have hit the jackpot. « Reply #92 on: 2014-09-22, 14:49:42 »

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Quote from: TinMan on 2014-09-22, 14:33:42

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 .

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TinMan

Re: LED driver-may have hit the jackpot. « Reply #93 on: 2014-09-22, 14:58:08 »

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Quote from: ION on 2014-09-22, 14:49:42

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.

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poynt99

Re: LED driver-may have hit the jackpot. « Reply #94 on: 2014-09-22, 15:01:21 »

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Quote from: TinMan on 2014-09-22, 14:39:10

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.

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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

TinMan

Re: LED driver-may have hit the jackpot. « Reply #95 on: 2014-09-22, 22:33:57 »

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Quote from: poynt99 on 2014-09-22, 15:01:21

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.

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poynt99

Re: LED driver-may have hit the jackpot. « Reply #96 on: 2014-09-23, 04:39:53 »

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Quote from: TinMan on 2014-09-22, 22:33:57

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 ?

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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

TinMan

Re: LED driver-may have hit the jackpot. « Reply #97 on: 2014-09-23, 08:04:41 »

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Quote from: poynt99 on 2014-09-23, 04:39:53

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

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poynt99

Re: LED driver-may have hit the jackpot. « Reply #98 on: 2014-09-23, 14:46:05 »

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Quote from: TinMan on 2014-09-23, 08:04:41

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.

So the above becomes this:

VSG(avg peak) ~~x 0.4~~ x VDMM/CSR OR

VSG(avg peak) x VDMM/CSR

---------------------------

"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

TinMan

Re: LED driver-may have hit the jackpot. « Reply #99 on: 2014-09-23, 15:07:40 »

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Quote from: poynt99 on 2014-09-23, 14:46:05

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.

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.

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