Joule Thief 101

[MileHigh]

PW:

Just to be sure we are on the same page, the attached "death spike" screen cap from Brad's clip is indeed a current waveform.  I did not consider the loading of the scope ground on the transistor base input.

It's nice to see that you are willing to consider base reverse breakdown voltage as a possible explanation.  I am somewhat embarrassed for calling it "punch through."  I know the proper technical terms but they are not on the tip of my tongue anymore, so I use "slang" to get my points across.

MileHigh

[MileHigh]

MH
The v/CBO for the 2n3055 is 100 volt's.
I do not see 100volts anywhere in my scope shot's?.
I do not see anyway that this !punch through! could be taking place.

Brad

Yes that doesn't surprise me.  However, I believe there is a possibility that you can get avalanche effects, so perhaps a few nanoseconds could get the ball rolling.  Or the transistor is old and beaten up and not meeting spec any more.

My first impression was that there was a breakdown in the transistor itself, so perhaps you will get to the bottom of it and find out one way or another.  It looks very jarring to see that negative pulse of current.

MileHigh

[picowatt]

MH
The v/CBO for the 2n3055 is 100 volt's.
I do not see 100volts anywhere in my scope shot's?.
I do not see anyway that this !punch through! could be taking place.


Brad

I just looked at the data sheet, max V_EB is given as 7VDC.  Moreover, I just applied a current limited negative voltage between the base and emitter of a 2n3055 here on the bench and it broke down at a Vbe of -8.7VDC.  So hats off to MH!!

However, if the only current flowing thru the base circuit during the off time was related to V_EB breakdown, the current flow should not begin until Vbe exceeds -7VDC (probably closer to the -8,7VDC I observed).

I would think capacitive loading would create immediate current flow/loading of the base waveform as soon as the negative going transition begins and then be clamped at the base emitter breakdown voltage (ca -8.7VDC) once that junction breaks down.  If you look closely at Tinman's noisey diff measurement attempt, or the capture made with the scope ground on the base, there does appear to be a fairly flat period of current flow during the off time following the transitional spike.  I had thought that to be noise but it likely is base/emitter breakdown clamping current.

Perhaps as MH suggests, a bit more scrutiny of the turn off period at a faster sweep rate is in order.

PW

[tinman]

Brad:

Just a little reality check:



  If the transistor is switching properly for the high base resistance, then for sure it will be switching properly for the low base resistance and you will barely, if at all, see any difference in the collector-emitter voltage for the two cases.  Assuming the collector-emitter voltage is the same in both cases, then that means for a high base resistance or a low base resistance the build-up of current flowing through L1 will be identical.  That means that decreasing the base resistance does not increase the build up of the current flowing through L1, but it does increase the current flowing through L2.

However, when you decrease the base resistance, the timing changes, and the energizing cycle for the L1 coil gets longer, and that results in slightly more current flowing through L1 when the transistor switches off, and hence a very slightly brighter LED as observed in your last clip.

  The L1 coil and L2 coil in the Joule Thief transformer produce opposite and cancelling flux when the current flows from top to bottom for each coil.  Therefore increased current flow in L2 will reduce the magnetic energy in the core that was originally put there by current flow through L1.  This is the third time I am stating this.

MileHigh

In your last clip when you decrease the base resistance the brightness increase is not significant to the human eye, and that's what counts.  This is for the case when the battery voltage is about 1.1 volts.  It may be different for lower battery voltages.

Do you not remember what the whole idea was in regards to having the VR MH?
It was so as we could reduce the base resistance as the battery voltage dropped-and bit by bit. Not start swinging the resistance about while the battery voltage is still quite high.

Of course decreasing the base resistance will increase the current flow through L2.  However, PW is working with you to try to understand this in more detail.  What happens to the collector-emitter voltage drop when the transistor is ON for high base resistance vs. low base resistance?  You are supposed to make measurements of that.

Well lucky for me,i do things when i want,and get time to do them--not when MH thinks i should be doing them,and within his small time frame

With respect to increased current flow in L2 putting more magnetic energy into the core, and that makes the LED brighter, you are dead wrong.  Myself, TK, and PW have stated this.

Please repost these post where TK and PW have said this is incorrect.

Here is the conversation myself and PW had

PW-

In your video, we might "assume" that the base current increases as you decrease the value of the potentiometer connected to the transistor's base.  However, because the base voltage remains constant, there is no way to see (or know) what change is actually occurring to the base current as you adjust the pot with the tests you made in the video.

We now know the base current dose indeed increase.

My reply-Not entirely correct.
As the current flow in L1 is set once the transistor is fully switched on,then the only way the LED can receive more current from the inductive kickback is by way of a stronger magnetic field being produced during each on time pulse,and as L1's current is set,then the only way to increase that magnetic field is by way of L2. For this to happen,then L2 must be receiving more current flowing through it,and we know this would be the case if we reduce the base resistance value..-->and we do now know for sure that the base/L2 current dose increase when we reduce the base resistance.

PWs

reply-Yes, one can "assume" that is what is happening and use what is observed as a proxy for base current.  My point was that nowhere is base current itself being directly measured or observed in the video

.

But shortly after this post,we confirmed an increase in base current as we reduced the base resistance.
I have not seen a post from PW that states i am wrong
And the only two post i have seen from TK,said nothing about it.
So where are these post that PW and TK state i am wrong?-->you up to your old tricks again MH?.

Now,time for you to think about things a little here,as either way you are wrong.
What is needed in order for the LED to emit more light from the inductive kickback spike?-a larger/stronger magnetic field collapse?
How do we build a larger/stronger magnetic field?--a higher value of overall current flow through the inductor during the on time perhaps?
How do we get this higher value of current flow through the inductor during the on time?-the transistor switching on for a longer period of time perhaps ?.
What needs to happen in order to get the transistor to switch on for a longer period of time?-a longer period and higher value of current flow through L2 perhaps?.
How is this achieved?--well as clearly seen,we reduce the base resistance.

I am lost as to how you can say that the larger amount of current flow through L2 is not responsible for the building of a larger/stronger magnetic field in the inductor

I am also at a loss as to how you can say that the two fields built by L1 and L2 will cancel each other out. First up,the current traveling in the two coils is in opposite directions,and there for the two fields will buck,creating a larger field. Second,if the two fields canceled each other out,then when we raise the current flow in L2,the current should actually drop--not rise. But as we know,there is no decrease in current flow,there is only an increase in current flow,and there for there is no cancellation of magnetic fields,as there is no decrease in current flow.
If there was a field cancellation going on,then the circuit would not work as it dose.
You posted the working your self,and now you disagree with how the circuit work's.
Start of cycle.
current starts to flow through L2 as the transistor starts to conduct.
At this point current starts to flow through L1.
The current flowing through L1 increases the current flowing through L2.
This cascade effect/transformer effect keeps going on between the two coils until the transformer is fully switched on,and remains on until such time as the magnetic field in the inductor has reached it's peak-no longer varying in time. At this point,the current flowing through L2 drop's right down,and the transistor opens,and the inductive spike current from L1 is sent through the LED.

How on earth can any of that happen if the two coils fields cancel each other out?

Brad

[tinman]

I am confused about the math trace business because the purple traces in your attached scope captures look like perfectly clean math traces except for the fact that it looks like they are upside-down, and I can't be sure of your "virtual ground reference" for the purple traces.

If the purple traces are indeed upside-down, then it would be trivial to set them right-side-up and dim out the other two traces and then you are good to go.  Am I missing something?

MileHigh

Because MH-as i stated to PW,that is a variation circuit,where the LED was place across the collector/base junction. That is why the math trace is very clean there,as the LED filtered out all the noise that you see in the later scope shot,where the original test circuit is being used,and the math trace is very noisy.

Brad