Joule Thief 101

[tinman]

When Q1 turns off and L2 collapses, the current flow thru L2 and the LED also induces a voltage across L2, which is the negative voltage portion of the base waveform.  During that off time, whenever the base drive voltage is below the Vbe forward voltage, the base leg is basically an open circuit because Q1's base is not conducting.  Under DC conditions, there would be a bit of reverse leakage current, but this would only be a few microamps at most.

As we have seen in your first video, as the variable base resistor's value is decreased, the negative going portion of the base waveform increases in amplitude slightly. The base resistor was only changed about 500 ohms or so.  We know that the reverse leakage current of Q1's base is fairly low.  Under DC conditions, we not expect to see any significant change in the reverse biased voltage observed at the base of Q1 by changing the resistance in series with the base by only 500 ohms (think of the reversed bias base as being a several megohm resistor forming a voltage divider with the L2 and base resistor).

The fact that the reverse voltage at the base does vary noticeably when the resistance is changed indicates that there remains a significant load on the waveform observed at the base of Q1 during the Q1 off time while the base is reverse biased.  This can only mean that there is a significant load at the base with respect to AC conditions (far greater than the megohms expected under DC conditions).

Keeping in mind that the negative going portion of the base waveform is a rather fast transition (edge) containing significant high frequency components, the most likely explanation for the amplitude of that portion of the base waveform changing as the base resistor is changed is due to there being a significant capacitance loading the circuit at the base of Q1.

Some of this capacitance may be probe related, which is why it is very important to use your scope probes set to their 10X position (assuming they are 1X/10X switchable).  More so important than increasing the probe's DC resistance to 10meg is the reduction of probe capacitance (AC reactance) while in the 10X mode.



The base current trace became very spikey when you attached the scope ground to the base, as you were forcing the rest of the circuit to change potential with respect to the scope ground during that fast transition.  This greatly greatly increased the stray capacitance and current seen across the 10R CSR at the base during the fast negative going edge of the waveform.

When the CSR (base current) is properly viewed, I would expect to see a very narrow spike (as in your recent CH1-CH2 attempts) as any base capacitance is charged, followed by a period of little to no current flow, a smaller. slower (more rounded) spike at Q1 turn on, and then the actual base current of 2ma or so as Q1 remains on until turn off.



PW

That said, however, it is very likely that there is a significant amount of junction capacitance between the base and CE junctions of Q1 (a 2n3055 has a fairly large die area).  During the fast transition portion of the waveform, the base waveform is going negative while the collector waveform is going positive.  I would expect Miller capacitance alone (collector to base capacitance) to present a significant load to the otherwise open base during this time.

Thank you PW for clearing this up.

Have the scope owners that successfully changed their software seen the math issue fixed/resolved?

To those i have spoken to,yes,there math is fully functional now since the firmware update.
But it has to be done right the very first time. If the power happens to go out when doing the upgrade,then you throw your scope in the bin.
One guy said he now gets channel drift on his scope since the upgrade,and has been unable to fix it.
I cannot afford a new scope ATM,so i am not going to chance it,as it seems to be only about a 90% success rate--and im bound to be one of the other 10%


Brad

[MileHigh]

Brad:

Decreasing the base resistance as the battery voltage drop's,dose indeed increase the brightness of the LED--proven.
Decreasing the base resistance dose increase the current flowing through L2--proven
Increasing the current through L2 dose increase the magnetic field strength--proven by the fact that the LED gets brighter.

Just a little reality check:

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.

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.  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.

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.  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

[picowatt]

Here is what I thought was happening with respect to the big negative current spike:

When the transistor switches off, the potential at the L2 feedback coil was going way below ground potential, perhaps to -15 or -20 volts.  The potential at the L1 coil also raises to start to push current through the LED.  My assumption is that under these conditions the N-P junction between the collector and the base was breaking down, and current was punching through the junction and the Joule Thief transformer was briefly shorting itself out.  After enough energy was burnt off, then the shorting would stop and the rest of the L1 discharge would go through the LED.

When Brad lowered the base resistor you could see the negative spike getting larger and deeper, indicative of a bigger "punch through" of the N-P junction between the collector and the base.

If this was indeed the case, then reducing the number of turns in the L2 feedback coil would reduce the negative potential on L2 and then the "punch through" would stop happening and all of the energy would go into the LED instead.

But now I am not so sure because the time base on the scope shot is 10 microseconds per division and the width of the negative spike is only a few microseconds.  So it could be just a junction capacitance effect from the N-P junction between the collector and the base also, I am not certain.  However, my gut feel is still going to go with a punch-through, I will go out on a limb.  PW is here so his comments will most likely clarify this issue.

MileHigh

MH,

Keep in mind that the very "spikey" scope shot was captured while scope ground was connected to Q1's base.  Also keep in mind that the observed waveform during that test is actually indicative of current, not voltage.

In previous scope shots, the negative going base waveform was not nearly as "ugly".  If we were dealing with the base reverse breakdown voltage, which to be honest I did not consider, I would think the waveform would have a consistent voltage level at which it is clamped fairly hard.  Also, no current would flow until the waveform reached whatever that breakdown voltage actually was.  The observed spike appears to happen immediately during the transition.  Still, I am willing to consider the possibility.

It is a shame the differential measurements cannot be made a bit cleaner.

PW   

[tinman]

Here is what I thought was happening with respect to the big negative current spike:

When the transistor switches off, the potential at the L2 feedback coil was going way below ground potential, perhaps to -15 or -20 volts.  The potential at the L1 coil also raises to start to push current through the LED.  My assumption is that under these conditions the N-P junction between the collector and the base was breaking down, and current was punching through the junction and the Joule Thief transformer was briefly shorting itself out.  After enough energy was burnt off, then the shorting would stop and the rest of the L1 discharge would go through the LED.

When Brad lowered the base resistor you could see the negative spike getting larger and deeper, indicative of a bigger "punch through" of the N-P junction between the collector and the base.

If this was indeed the case, then reducing the number of turns in the L2 feedback coil would reduce the negative potential on L2 and then the "punch through" would stop happening and all of the energy would go into the LED instead.

But now I am not so sure because the time base on the scope shot is 10 microseconds per division and the width of the negative spike is only a few microseconds.  So it could be just a junction capacitance effect from the N-P junction between the collector and the base also, I am not certain.  However, my gut feel is still going to go with a punch-through, I will go out on a limb.  PW is here so his comments will most likely clarify this issue.

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

[tinman]

MH,

Keep in mind that the very "spikey" scope shot was captured while scope ground was connected to Q1's base.  Also keep in mind that the observed waveform during that test is actually indicative of current, not voltage.

In previous scope shots, the negative going base waveform was not nearly as "ugly".  If we were dealing with the base reverse breakdown voltage, which to be honest I did not consider, I would think the waveform would have a consistent voltage level at which it is clamped fairly hard.  Also, no current would flow until the waveform reached whatever that breakdown voltage actually was.  The observed spike appears to happen immediately during the transition.  Still, I am willing to consider the possibility.

It is a shame the differential measurements cannot be made a bit cleaner.

PW

I could make the differential measurements cleaner by switching back to the 100 ohm resistor,but it would be without the math trace--we need to forget about the math function on my scope.

Brad