Author Topic: Joule Thief 101 (Read 938824 times)

partzman · « **Reply #2625 on:** May 20, 2016, 08:50:13 PM »

webby1

You were correct regarding the instantaneous voltages across the resistors at T=5

For the question regarding the general shape of the voltages and currents when T>5 secs, I attached a sim showing the results which will speak more clearly than an explanation with words.

The Vin and Vout traces are superimposed until T=5 when Vin goes to zero and Vout shows the voltage across the 1.6666 ohm and 10 ohm resistors respectively. The current from L1 for each resistance is shown and marked as voltage/current pairs. To be clear, R1 has been stepped with 1.6666 ohms and 10 ohms to produce the overall plot with two pairs of voltage and current from T5 to the end.

There are also two cursors on the plot indicating the time and current magnitude for each pair at one time constant or T=L/R which indicates the currents have dropped ~63% from the starting value of 2.4 amps.

Several things to note and that is, the polarity of the voltage(s) across R1 and the shape of the voltage and current waveforms.

Also, we can conclude from this that there is no value of "fixed" resistance that can reproduce the same current waveform as MH's from T5 to T8.

partzman

partzman · « **Reply #2626 on:** May 21, 2016, 05:23:38 PM »

webby1,

I've attached a sim that uses a non-linear resistor R1 from T5 to T8 that creates a complimentary negative going current ramp to the voltage generated positive going current ramp from T0 to T3. This is shown for comparison to the fixed resistor R1s in the previous sim.

Since dI = E*t/L and L is fixed at 5H and we wish to start with I = 2.4 amps at T5 and ramp to 0 amps at T8, we can solve for R using either t or E. I chose to solve using t as follows. With R=E/I and dI = E*t/L, through substitution and simplification we arrive at R = L/t.

The denominator t is actually T-dt with T equal to the overall time from T5 to T8 to ramp from 2.4 amps to zero amps and dt is the incremental change in time during this period. However, R1 is calculated during the entire plot as is seen in the red "5/(8-time)" trace and at t5, the resistance is 1.66666 ohms which is the starting value needed to produce 4 volts at 2.4 amps. Also at T5, S1 is opened to disconnected Vin from L1 and S2 then connects R1 to L1.

To clarify, the label at the far left of the plot is the resistance of R1 so "100s-1" should read 100 ohms, etc.

As we approach T8, "T-dt" approaches zero and R1 approaches infinite resistance.

partzman

MileHigh · « **Reply #2627 on:** May 21, 2016, 07:17:01 PM »

Quote from: partzman on May 21, 2016, 05:23:38 PM

webby1,

I've attached a sim that uses a non-linear resistor R1 from T5 to T8 that creates a complimentary negative going current ramp to the voltage generated positive going current ramp from T0 to T3. This is shown for comparison to the fixed resistor R1s in the previous sim.

Since dI = E*t/L and L is fixed at 5H and we wish to start with I = 2.4 amps at T5 and ramp to 0 amps at T8, we can solve for R using either t or E. I chose to solve using t as follows. With R=E/I and dI = E*t/L, through substitution and simplification we arrive at R = L/t.

The denominator t is actually T-dt with T equal to the overall time from T5 to T8 to ramp from 2.4 amps to zero amps and dt is the incremental change in time during this period. However, R1 is calculated during the entire plot as is seen in the red "5/(8-time)" trace and at t5, the resistance is 1.66666 ohms which is the starting value needed to produce 4 volts at 2.4 amps. Also at T5, S1 is opened to disconnected Vin from L1 and S2 then connects R1 to L1.

To clarify, the label at the far left of the plot is the resistance of R1 so "100s-1" should read 100 ohms, etc.

As we approach T8, "T-dt" approaches zero and R1 approaches infinite resistance.

partzman

Awesome work Partzman. Getting the current curve to ramp down in a linear fashion with a variable resistor is a not-too-difficult thought experiment but it is really cool, even beyond cool, to see you work the variables to define the variable resistor as a function of time for the simulation and then run it and show the fruits of your labour.

I can imagine the usual grumbling, "That's not ever going to happen in real life," and "That's useless and I can't see any reason for doing that."

Well, what about a rocket? When a rocket launches it obeys the usual f = ma to accelerate and get off the ground. However, the mass of the rocket 'm' is actually 'm(t).' In other words the mass of the rocket is a function of time t and always decreasing because it is burning fuel, i.e.; f = m(t)a. But then there is another thing to consider, let's assume that the force from some rocket engines is proportional to the amount of fuel in the engine, and so we can say that the force is also a function of time. So you get f(t) = m(t)a. So rewriting it, you get a = f(t)/m(t). Now your variable resistor that is a function of time is starting to look a lot more interesting.

Finally, when it comes to launching a rocket, you can't forget the air resistance that is slowing the rocket down. Let's say that we say that the air resistance is a function of both the height h and the velocity v. i.e; The force of the air resistance fair = fair(h,v)

So now you have a new formula for the acceleration of the rocket: a = [f(t) - fair(h,v)]/m(t).

Now all of a sudden that resistance that's a function of time starts to look pretty sweet.

partzman · « **Reply #2628 on:** May 21, 2016, 07:43:39 PM »

Quote from: MileHigh on May 21, 2016, 07:17:01 PM

Awesome work Partzman. Getting the current curve to ramp down in a linear fashion with a variable resistor is a not-too-difficult thought experiment but it is really cool, even beyond cool, to see you work the variables to define the variable resistor as a function of time for the simulation and then run it and show the fruits of your labour.

I can imagine the usual grumbling, "That's not ever going to happen in real life," and "That's useless and I can't see any reason for doing that."

Well, what about a rocket? When a rocket launches it obeys the usual f = ma to accelerate and get off the ground. However, the mass of the rocket 'm' is actually 'm(t).' In other words the mass of the rocket is a function of time t and always decreasing because it is burning fuel, i.e.; f = m(t)a. But then there is another thing to consider, let's assume that the force from some rocket engines is proportional to the amount of fuel in the engine, and so we can say that the force is also a function of time. So you get f(t) = m(t)a. So rewriting it, you get a = f(t)/m(t). Now your variable resistor that is a function of time is starting to look a lot more interesting.

Finally, when it comes to launching a rocket, you can't forget the air resistance that is slowing the rocket down. Let's say that we say that the air resistance is a function of both the height h and the velocity v. i.e; The force of the air resistance fair = fair(h,v)

So now you have a new formula for the acceleration of the rocket: a = [f(t) - fair(h,v)]/m(t).

Now all of a sudden that resistance that's a function of time starts to look pretty sweet.

Thanks MH,

partzman

tinman · « **Reply #2629 on:** July 10, 2016, 08:49:01 AM »

Going back to the simple JT,which circuit would be more efficient?

We shall use MH's term for efficiency as stated on this thread in reply 166

For purposes of a fair discussion let's put aside the "Thief" part of the Joule Thief that can extract energy from nearly dead batteries. In other words, let's just look at lumens per watt of supplied power.

So what is a JT circuit?
Well to also quote MH on post 166

When is a circuit a Joule Thief or not? I think that there is a simple answer to that one. If the circuit can power a LED with a battery whose output voltage is lower than the normal drive voltage for the LED, and the LED is driven using the technique of a discharging inductor acting as a current source, then you have a Joule Thief.

As some would remember,MH later on decided that the first circuit-circuit 1 is the JT circuit,although it would seem that this go's against his first post on what a JT circuit is.

But for the sake of the discussion,lets just stick to the two circuits below.
Which do you believe to be the more efficient circuit out of the two?

Brad

TinselKoala · « **Reply #2630 on:** July 10, 2016, 09:07:00 AM »

We need a little more information.

Please specify:
-Number of turns of L1 and L2
-Transistor part number (choose one of MPSA18, BC337-25, 2n2222a, 2n3904)

tinman · « **Reply #2631 on:** July 10, 2016, 09:39:18 AM »

As we go along in this thread,i will try and clean up some of the rubbish that has been dumped in the thread as it went on.

The hope is ,that at the end of the day,we will have the !correct! knowledge to build the most efficient JT we can.

The first thing that needs to be cleared up is the below.

Quote MH post 207
Also, the Joule Thief will not work as a Joule Thief, if it woks at all, without the inductive coupling between the two windings. Saying it works because of "the C value of the transistor" is just more word salad.
The fundamental timing and operation of a Joule Thief is based on L/R time constants and there is no resonance at play at all - the Joule Thief timing and operation is governed by the interaction between inductance and resistance and not capacitance.

If we are to go on MHs description of a JT,( If the circuit can power a LED with a battery whose output voltage is lower than the normal drive voltage for the LED, and the LED is driven using the technique of a discharging inductor acting as a current source, then you have a Joule Thief. If the circuit does not meet these two conditions then it is not a Joule Thief.)then the above is not true at all,and the cool joule circuit dose indeed work due to the miller capacitance effect,which is the junction C value of the transistor,and also meets the two conditions set out by MH to qualify it as a JT circuit.

https://www.youtube.com/watch?v=5Mbp1iuB7as

MHs answer to the cool joule was this
Quote post 234: So it's not a Joule Thief because it does not do anything special to extract energy from a very-low-voltage battery.You are not going there Brad. You have been fully aware of what the standard Joule Thief circuit is for years
But then we had this a couple of pages before
Quote: If the circuit can power a LED with a battery whose output voltage is lower than the normal drive voltage for the LED, and the LED is driven using the technique of a discharging inductor acting as a current source, then you have a Joule Thief.

Seems to be some conflict going on here

Anyway,the aim of the game here,is to decide which of the two circuits is more efficient,in way of MHs example of efficiency=let's just look at lumens per watt of supplied power.

TK has already carried out a few tests to this regard,but i suspect he will be back with some different findings soon.

Brad

tinman · « **Reply #2632 on:** July 10, 2016, 09:42:06 AM »

Quote from: TinselKoala on July 10, 2016, 09:07:00 AM

We need a little more information.

Please specify:
-Number of turns of L1 and L2
-Transistor part number (choose one of MPSA18, BC337-25, 2n2222a, 2n3904)

TK
It dose not matter as to turns or transistor used,as long as you use the very same DUT when testing both LED positions.
I will be using a 2n3055 for most of my experiments on the JT circuits.

Brad

tinman · « **Reply #2633 on:** July 10, 2016, 09:51:35 AM »

To quote MH from other thread.

Quote

I think every single time I have asked you to explain one of your procedures you play the bullshit "I won't bow to your demands" card. You clam up and freeze up.
Right now I am operating under the assumption that either you don't know how to measure the output impedance of a battery or you think that you do and whatever you do has some hapless tragic mistake in it. I have seen things like this before.
Or, you can wipe the ridiculous attitude away, and simply explain how you measure the output impedance so we can check if if makes sense.
Why are you freezing up?

The internal resistance value of the battery can be calculated on the fly MH--while the JT circuit is operating.
I wonder if you actually know how to do this?
My bet is-like always,you wait for some one else to give the answer,and then you say--oh yes,that is how it is done.
So not this time MH-this time you will have to work that out for your self.
We could just do the internal resistance test of the battery,while it is not in use in the circuit,but i wanted to be more accurate,and know what it was during use in the JT circuit.

Brad

tinman · « **Reply #2634 on:** July 10, 2016, 09:55:52 AM »

Quote TK

Quote

The measurement points are indicated on one of the photos of the apparatus. I am measuring input current by the voltage drop across a 0.1 ohm non-inductive resistor on the negative side of the circuit, and input voltage simply across the input terminals. I have done both measurements with oscilloscope and DMMs. We have seen from Poynt99's work that the DMMs do a very good job of averaging pulsed inputs, and I also used the oscilloscope's "average" measurements of each channel's raw readings to confirm the DMM readings. I used the scope's Math function to multiply the raw (not averaged) instantaneous voltage and current inputs, then had the scope compute the "average" of this power measurement,which I then used in the Lux per Watt calculation. The current traces are very different between the two circuits, as you probably know yourself.

TK

Did you work out as to why when using the battery,the efficiency between the two circuits swapped over from that of using your PSU ?

I think you need to look a little closer as to what happens during the off time(transistor open) of circuit 1--where the battery is included.
If you think about it a little,you will see what changed when switching to the battery,and how all of a sudden,circuit 1 became more efficient--or appeared to

Brad

MileHigh · « **Reply #2635 on:** July 10, 2016, 10:38:32 AM »

Quote from: tinman on July 10, 2016, 09:51:35 AM

To quote MH from other thread.

The internal resistance value of the battery can be calculated on the fly MH--while the JT circuit is operating.
I wonder if you actually know how to do this?
My bet is-like always,you wait for some one else to give the answer,and then you say--oh yes,that is how it is done.
So not this time MH-this time you will have to work that out for your self.
We could just do the internal resistance test of the battery,while it is not in use in the circuit,but i wanted to be more accurate,and know what it was during use in the JT circuit.

Brad

And now you are back in full-blown stupid-ass trash talk mode. Look at you! Your brain is frying.

Quote

My bet is-like always,you wait for some one else to give the answer,and then you say--oh yes,that is how it is done.

That is complete and total BS because your brain is frying and you will say any stupid negative thing about me, even if you have to shamelessly lie like a jackass.

Quote

So not this time MH-this time you will have to work that out for your self.

Kiss my butt! O hapless one!

Quote

We could just do the internal resistance test of the battery,while it is not in use in the circuit,but i wanted to be more accurate,and know what it was during use in the JT circuit.

Then just state the two procedures and stop the ridiculous soap opera drama, for crying out loud.

MileHigh · « **Reply #2636 on:** July 10, 2016, 10:54:13 AM »

I am going to give you a little reality check Brad.

I argued the first circuit was "more efficient" because it looked like it offered a longer run time by going down to a lower battery voltage and therefore could extract more energy from a dead battery. That is the basic Joule Thief premise.

Then you discussed the first circuit and the second circuit with respect to looses due to the internal impedance of the battery. Your argument was that since in the second circuit some of the time the battery was OFF and not supplying any current, that there would be less internal battery impedance losses. I proved to you that that notion is false if you assume the exactly the same power draw for both circuits. In fact, it turns out the first circuit will be more efficient using that efficiency parameter which is the opposite of what you thought.

I have no idea where you want to go with this. I proved that your assumption for the second circuit was wrong in terms of internal energy lost in the battery. If you can't understand what I said or how I arrived at that conclusion, then have a good brain fry, bring along some marshmallows.

Beyond that, the issue of most light per unit of input power is interesting. I already indicated in my first long posting that TK responded to that it's a quite complicated matter with many parameters to consider. I am not really that interested in getting into a long discussion about it.

MileHigh · « **Reply #2637 on:** July 10, 2016, 11:42:59 AM »

Brad:

Quote

And so that is the difference between bench workers and pen pushers MH--the bench man will go searching as to why the two efficiency differences between PSU and battery exist,and will find that reason,while you(the pen pusher) will just keep on wallowing in your own self pity,and continue to peddle rubbish-like you are here and now.

For stooping to that low a level of trash talk, go foof yourself you semi-literate jackass. I can assure you like I said before, I can spin circles around you on a bench and its been 25 year now. As of a few months ago, you didn't even understand how a Joule Thief works, you did not understand how resonance works, and you did not understand how an inductor works. You could barely punch your way out of a wet paper bag. Look at what you posted to RMS seven months ago. You tried to make two points to him, and on both of your points you fell flat on your face, wrong and wrong again. That's really impressive "six years under your belt" bench work noggin shining there Brad.

Your bullshit pitch that I am lost and you are the one with all the answers when it comes to a Joule Thief is a joke. It's just you being steaming mad. This is your cue to mention a JFET fool.

Quote

So i stand by my answer-->you cannot place an ideal voltage across an ideal inductor.
If you did(theoretically),the current would rise instantly to an infinite value.

But you were saying just last week that no current would flow because of the CEMF!

Time to crack another egg and put it on that piping hot skillet.

tinman · « **Reply #2638 on:** July 10, 2016, 12:14:36 PM »

author=MileHigh link=topic=8341.msg488201#msg488201 date=1468143779]

Quote

For stooping to that low a level of trash talk, go foof yourself you semi-literate jackass.

Foof?

I am not sure how dioxygen difluoride fits into it,but anyway

But i see it did not take long for those profanities to make an appearance MH

But if it makes you feel better,then you go right ahead,as us Aussies can take a whole lot more than you can give--were all convicts remember lol

Quote

I can assure like I said before, I can spin circles around you on a bench and its been 25 year now.

But alas -cannot prove it--even when some one else well gifted in the art of EE was going to do all the bench work for you.

Quote

As of a few months ago, you didn't even understand how a Joule Thief works, you did not understand how resonance works, and you did not understand how an inductor works.

It would seem that you have made that assumption based on your own misunderstandings,and lack of knowledge. Kick back a while,you might just learn something

Quote

You could barely punch your way out of a wet paper bag

.

I think i could manage that MH

Quote

Your bullshit pitch that I am lost and you are the one with all the answers when it comes to a Joule Thief is a joke. It's just you being steaming mad. This is your cue to mention a JFET fool.

Why would you think i am mad MH?
And i was not going to mention anything about you not knowing what a J/FET was.

Hang in there MH--we'll learn ya something soon enough-->hows that for some spot on English

But anyway,if you cant keep up,just do the best you can

Brad

MileHigh · « **Reply #2639 on:** July 10, 2016, 12:25:56 PM »

Everything you just posted was completely FAKE Brad and you are not fooling anybody.

When push comes to shove you degenerate into this weak little fake character and I have no problem being completely real. It must feel miserable inside having to put up this fake front when you know that most people know that you are faking it. Go build a Joule Thief, it will be good therapy for you.

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Author Topic: Joule Thief 101 (Read 938824 times)

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