LED driver-may have hit the jackpot.

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TinMan

Re: LED driver-may have hit the jackpot. « Reply #100 on: 2014-09-23, 15:23:56 »

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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.

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TinMan

Re: LED driver-may have hit the jackpot. « Reply #101 on: 2014-09-23, 15:35:39 »

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Quote from: poynt99 on 2014-09-20, 22:24:16

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

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poynt99

Re: LED driver-may have hit the jackpot. « Reply #102 on: 2014-09-23, 16:28:42 »

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

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.

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poynt99

Re: LED driver-may have hit the jackpot. « Reply #103 on: 2014-09-23, 18:02:17 »

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

Quote from: TinMan on 2014-09-23, 15:35:39

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.

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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 #104 on: 2014-09-24, 00:16:56 »

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It's turtles all the way down

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.

LED driver-may have hit the jackpot.

LED Power.jpg (17.86 kB, 583x120 - viewed 75 times.)

« Last Edit: 2014-09-24, 02:25:02 by ION »

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poynt99	Re: LED driver-may have hit the jackpot. « Reply #105 on: 2014-09-24, 03:18:53 »
Group: Administrator Hero Member Posts: 3538 It's not as complicated as it may seem...	Never thought LEDs would be that inefficient. --------------------------- "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 #106 on: 2014-09-24, 14:34:32 »
Group: Administrator Hero Member Posts: 3538 It's not as complicated as it may seem...	Here is a document which talks about LED efficiency and efficacy. led_energy_efficiency.pdf (961.21 kB - downloaded 14 times.) --------------------------- "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 #107 on: 2014-09-24, 15:00:05 »

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Quote from: poynt99 on 2014-09-24, 14:34:32

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.

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poynt99	Re: LED driver-may have hit the jackpot. « Reply #108 on: 2014-09-24, 15:02:12 »
Group: Administrator Hero Member Posts: 3538 It's not as complicated as it may seem...	Yes. --------------------------- "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 #109 on: 2014-09-24, 15:06:45 »
Group: Elite Experimentalist Hero Member Posts: 5677	Quote from: poynt99 on 2014-09-24, 15:02:12 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 ? --------------------------- Never let your schooling get in the way of your education.

ion

Re: LED driver-may have hit the jackpot. « Reply #110 on: 2014-09-24, 15:17:41 »

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It's turtles all the way down

Quote from: poynt99 on 2014-09-24, 14:34:32

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

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TinMan

Re: LED driver-may have hit the jackpot. « Reply #111 on: 2014-09-24, 15:38:35 »

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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 ?

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poynt99

Re: LED driver-may have hit the jackpot. « Reply #112 on: 2014-09-24, 19:08:22 »

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

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.

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TinMan

Re: LED driver-may have hit the jackpot. « Reply #113 on: 2014-09-24, 23:35:45 »

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Quote from: poynt99 on 2014-09-24, 19:08:22

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 ?.

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poynt99

Re: LED driver-may have hit the jackpot. « Reply #114 on: 2014-09-25, 00:16:28 »

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

Quote from: TinMan on 2014-09-24, 15:38:35

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.

LED driver-may have hit the jackpot.

eyeball_vs_actual_LED1.png (14.83 kB, 713x588 - viewed 50 times.)

« Last Edit: 2014-09-25, 01:43:59 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

TinMan

Re: LED driver-may have hit the jackpot. « Reply #115 on: 2014-09-25, 06:08:39 »

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Quote from: poynt99 on 2014-09-25, 00:16:28

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

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poynt99

Re: LED driver-may have hit the jackpot. « Reply #116 on: 2014-09-25, 14:00:19 »

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

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?

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TinMan

Re: LED driver-may have hit the jackpot. « Reply #117 on: 2014-09-25, 14:29:53 »

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Quote from: poynt99 on 2014-09-25, 14:00:19

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.

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poynt99	Re: LED driver-may have hit the jackpot. « Reply #118 on: 2014-09-25, 14:44:40 »
Group: Administrator Hero Member Posts: 3538 It's not as complicated as it may seem...	Agreed. So in my sim it would be the power dissipated by the 9.2 Ohm resistor. --------------------------- "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 #119 on: 2014-09-25, 15:06:12 »

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Quote from: poynt99 on 2014-09-25, 14:44:40

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?.

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poynt99

Re: LED driver-may have hit the jackpot. « Reply #120 on: 2014-09-25, 15:40:13 »

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

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.

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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 #121 on: 2014-09-25, 23:18:58 »

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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?.

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poynt99

Re: LED driver-may have hit the jackpot. « Reply #122 on: 2014-09-25, 23:53:43 »

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

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.

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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 #123 on: 2014-09-26, 01:15:14 »
Group: Elite Experimentalist Hero Member Posts: 5677	Quote from: poynt99 on 2014-09-25, 23:53:43 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 --------------------------- Never let your schooling get in the way of your education.

poynt99

Re: LED driver-may have hit the jackpot. « Reply #124 on: 2014-09-27, 00:19:20 »

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

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)

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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