LED driver-may have hit the jackpot.

[TinMan]

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

[poynt99]

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.

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).

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.

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.

[TinMan]

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.

[poynt99]

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).

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.

[TinMan]

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.

[poynt99]

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.

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.

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 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.

[poynt99]

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.

[TinMan]

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

[poynt99]

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!  

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?

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 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?

[TinMan]

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

[poynt99]

Any idea of the inductance?

Confirm the frequency; 70 Hz?

[TinMan]

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.

[Centraflow]

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

[TinMan]

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

Cheers Mike.
I must get me one of those flash meters that measure things like this on inductors.

[TinMan]

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

[poynt99]

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

Thanks Mike. Is yours about 100 Ohms DC as well?

[poynt99]

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.

[poynt99]

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.

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).

[poynt99]

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.

[TinMan]

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 . 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

[Centraflow]

Thanks Mike. Is yours about 100 Ohms DC as well?

No, 278 ohm

regards

Mike

[poynt99]

No problem helping out Brad. Looking forward to your video.





Thanks Mike.