2013-01-31 Emitter To Base Breakdown in a Joule Thief

I left a comment to one Joule Thief experimenter (video is on Youtube) that one should never power a Joule Thief with no LED connected.  If the emitter to base voltage exceeds the breakdown voltage the resulting current will permanently damage the transistor by reducing the current gain.  He tried running the Joule Thief with no LED for hours and he said the results were that it did not harm the transistor.

Here is what I know.  When there is no LED, the emitter to base voltage is the same as the collector voltage, only the opposite polarity, because the windings are 1:1 ratio. When the transistor switches off, the voltage between the emitter and base will climb as high as it can until the emitter to base junction breaks down.  Then current will flow through the emitter to base junction.  There seems to be some misunderstanding in the above statements so I’ll try to explain it simply.

The transistor switches off, and the voltage across the base to emitter climbs along with the collector voltage.  Because there is no current flow through the emitter to base, there is also no current flowing through the resistor and there is no voltage drop across the resistor; all of the voltage appears across the emitter to base junction and since the winding ratio is 1:1, the voltage follows the collector voltage.  When the voltage climbs to the breakdown voltage of the emitter to base junction, it starts to conduct and the resulting current flow causes any further voltage increase to appear across the resistor.  The emitter to base junction acts like a zener diode.  And the current through the emitter to base junction is what damages it.

My Tests

First I measured the transistor’s gain – may have been about 300.  The transistor I tested was not in a Joule Thief, but in a test fixture with enough DC voltage to cause the emitter to base junction to break down.  This was typically above 6 or 7 V but less than 10V.  I had a current limiting resistor in series with the emitter to base junction.  I increased the voltage until the current limiting resistor had a few volts across it, indicating that there was current flowing through the resistor and through the emitter to base junction.  I let the current flow for a few minutes.  The transistor dissipated very little power, only ten or so milliwatts during this time.  Then I removed the transistor and checked the current gain, and found that it was much lower, 100 or so.  And this was permanent, I tested the gain later and found the same low current gain.

I cannot explain why the experimenter’s Joule Thief circuit seems to not be affected by running it with no LED.  If the voltage is exceeding the emitter to base breakdown voltage and there is current flowing backward through the emitter to base junction, then it should damage it just like if it was in my test fixture.

I’ll have to try the test with various transistors to see if the current gains are affected by breakdown of the emitter to base junction. BRB.

Actual Transistor Tests

Update – I tested three transistors.  I put a 3.3k resistor in series with the emitter to base junction and then connected t to a power supply. I connected my DMM across the resistor to monitor the voltage drop.  I turned up the voltage, and at about 8 or 9V, the meter started to show the voltage drop.  I turned up the voltage until there was about 3.3V across the resistor, which meant there was 1 mA flowing through the breaking down emitter to base junction.  I then let this current flow for two more minutes.

I first tested a BC550C which is the same as a BC547C except it is low noise. The starting current gain was 575.  The ending current gain was 535 and the loss was 40.  That’s about a 7.5 percent loss.

I then tested a PN2222A.  The starting current gain was 188.  The ending current gain was 171 and the loss was 17.  That’s about a ten percent loss over just two minutes.

I then tested a BC338.  The starting current gain was 315.  The ending current gain was 291 and the loss was 24.  That’s about a 7.5 percent loss over just two minutes.

I believe in all 3 cases if I had used a higher current and/or a longer time the damage would have been worse.  In  all 3 cases if the voltage rose above 8 or 9 volts, the junction would start to break down.

Applying This To The Joule Thief

So far I have not proved that any of this applies to the Joule Thief.  What I have proved is that all three of these common transistors will be permanently damaged in their current gain when their emitter to base junction is subjected to current flow when the emitter to base junction is in breakdown.

When the Joule Thief is powered with no LED, the emitter to base junction is exposed to voltages in excess of its emitter to base breakdown voltage.  Does this damage the transistor, like it did in my tests?  I will have to test a few transistors to see if it does.

My Test Results In A Joule Thief

Update – I put a socket on a Joule Thief so I can insert and remove transistors without soldering.  I also added a switch to the LED so I can disconnect it from the circuit.  The coil has a bifilar winding with a 1:1 ratio.  The resistor is 1000 ohms.  I connected my scope probe across the emitter to base junction to see the negative voltage with and without the LED connected.  I’ll call this my test jig.

I measured the current gain of a PN2222A at 175 before the test.  I inserted it into the test jig with the LED connected, and saw negative spikes on the scope that were about 2 volts peak.  The LED glowed brightly.  I disconnected the LED and the scope showed the negative spikes had increased to a negative 9 volts peak, which is about where a typical transistor’s emitter to base junction breaks down.  I let it run for a few minutes, then I shut it off.

I removed the transistor and measured the current gain and it was 151, which was 24 less than before the test.  The current gain had dropped about 14 percent.  I then put the same PN2222A back into the test jig and watched the negative spikes again go to 9 volts peak.  I let it run for a few minutes and again measured the current gain, and it went down to 146.  Not as much drop as the first time, but this still shows that the transistor was permanently damaged the second time, just like the first time.

This confirms what I said at the beginning regarding running a Joule Thief with no LED.  If I had continued to run the transistor in the test jig with the LED disconnected, the current gain would continue to decrease.  Eventually the current gain would be too low for the circuit to operate.  This is just a matter of time and as I said, the damage is permanent.

Update Feb 1 – I measured the gain of a BC550C at 515.  I put it in the test jig and ran it with no LED for ten minutes.  I removed it and measured the gain and it was 480.  It had lost 35 or 6.8 percent.  I put it back into the test jig and left it running. I will come back in a much longer length of time and measure it.  It’s been more than an hour, and I measured the gain, and it is down to 445.  I’m going to leave it running overnight.

Update Feb 2 – the next morning, about 7 hours later, I measured the current gain of this transistor at  428.  I put it back in the JT with the LED off, to see how much more abuse it will take and how low it will go.  The frequency with LED was 24 kHz, with no LED was 38 kHz.  At noon, I measured the current gain at 423.  The breakdown damage seems to have leveled off with only a small amount of change every few hours.


9 Responses

  1. Paul says:

    That is really interesting. I don’t like to think of the transistors being tortured. Can you confirm that running them normally preserves the initial gain measured. Any idea why the Zener manages to cope when a BE junction is damaged? Paul.

  2. Paul says:

    Is it possible that ordinary transistors in use condition a bit at the beginning and ‘run in’. It would be interesting to see if the gain stayed the same if fresh but not abused.

  3. Paul Turner says:

    Well, the easy fix is to remove the Emitter-Base junction. 😉 This circuit uses a depletion-mode FET and is claimed to work down to 0.1V. Normal JT only works down to around 0.65V (for silicon transistor) of course.

  4. Ron says:

    I found your article in a search because the hFE of the 2N4401 used in my Joule Thief experiment from last year is about 200 while new ones from the same bag were around 280. I was wondering if this was just normal “burn in” but now I know that it is from abuse. Thanks.

    • admin says:

      If the circuit causes the emitter to base junction to go into reverse breakdown, then the damage to the junction is a reduction in current gain. That may have happened to your transistor. All you have to do is run the Joule Thief without the LED and the damage will occur.

  5. Eric Pierce says:

    Thanks for the explanation of how this occurs. During my experiments, I know I’ve run my JTs for brief periods without LEDs (even though I know it’s a bad idea); so far there has been no noticeable damage (that is, none that visibly reduces performance).

    I’m building a modular JT to power a bunch of white LEDs in parallel for a Lego city for my son (street lights, stoplights, building interiors etc.) The JT/battery-pack I’ve made has an on/off switch, and a row of pin headers where individual LED strings can be connected or disconnected (all wired in parallel–it can drive at least 15 white LEDs). In its present form, it’s possible to have no LEDs connected. Could I build in a protection mechanism to prevent transistor damage when there’s no LED?

    The obvious answer is to solder a single white LED in parallel with the pin headers, so there will always be at least 1 LED connected. It would protect the transistor from the damage you describe, and serve as an indicator that the circuit is on (even when no other LEDs are connected). This is probably what I’ll end up doing, but I am curious whether there’s a more efficient way–something that could act like a “fuse” to interrupt the current, or divert it safely.

    • admin says:

      I built a LED tester and I put two LEDs in series across the output. I used a white and a red LED, which add up to about 5 or 6 volts. They light when the test leads have nothing connected, but go dark when there is a white LED connected to the test leads. If you use two LEDs instead of one, they won’t take any power when the others are connected but still protect when nothing is connected.

      If you’re using a battery pack, it would be more efficient to just use 3 or 4 cells in series for 4.5 to 6 volts. You won’t need the JT and you will have to use a resistor for each LED. If you use different color LEDs you can use different value resistors to control the current and brightness. This will be more efficient than a JT, believe me; the typical JT wastes half the power and has about 50% efficiency. The batteries will last longer.

      Another alternative is to use a JT for each LED, and just fit them with used AA cells.

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