Joule Thief 101

[TinselKoala]

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]

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]

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]

To quote MH from other thread.

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]

Quote TK

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