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So, I'm trying to analyze how the circuit in this question behaves when subject to reverse battery (vs. being driven from OUT to IN), and I'm running into a severe snag in the process, though. It appears in SPICE simulations that with most typical supply voltages (5V or more), the PNP B-E junctions would get zenered if you apply reverse battery (negative voltage) on the left-hand side of the circuit while there is another power source feeding the right side.

However, SPICE does not simulate B-E Zener breakdown in BJTs at all, never mind what the collector of the right-hand BJT is doing while the B-E junctions are zenered. I plan to build the circuit and test it (as well as a potential fix with some series diodes), but before I do, what should I be expecting the collector of the right-hand transistor to do while its B-E junction is in Zener breakdown? (Or, will the FET be on, off, or somewhere in between, for that matter?)

ThreePhaseEel
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  • The question about the collector is indeed interesting. For simulating BE breakdown, You can place a custom diode in antiparallel that has a very large forward voltage of ~6 V. Adding a Zener in parallel is not a good practise because it takes away from forward operation base current. You could also add two antiserial zeners in parallel to avoid the base current issue. – tobalt Feb 17 '24 at 06:03
  • Three, LTspice has a few parameters for Vbe breakdown: BVbe, Ibvbe, and nbvbe. I think they are kind of simplistic, though. I've not done this, but I've read testimony from some credible folks (to me, anyway) claiming to have tested what an open collector does during the Vbe breakdown, with surprising results. The collector voltage went to a value outside of the voltages at the base-emitter nodes by a few 10ths of a volt. After discussion it was agreed that this was due to light being emitted inside the transistor. After decapping, they said they did in fact observe light emission. – periblepsis Feb 17 '24 at 06:57

2 Answers2

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There's an old riddle I think posed by Bob Pease on his magazine column, back in the day: how to obtain negative voltage from few ordinary components (BJT included) without oscillation?

The quirk is, when E-B breakdown occurs, free charges are created, plus some light (silicon emits some yellow-green glow in avalanche; pretty dim as those wavelengths are strongly absorbed by bulk Si so you only see light emitted near the surface). Charges migrate directly across the base, plus the light creates more free charges; basically, the C-B junction acts like a solar cell, with the junction's potential separating electrons and holes, developing a modest voltage.

The current is quite small (10s µA?), nothing to worry about in a power circuit. The example application from the article, I believe, was to extend the operation of "single supply" type op-amps, to properly include the 0V rail (including a little bit below), not just to approach it within some mV.

The main deficiency in simulation will be breakdown, which can be included with a customized (zero CJO, negligible IS) zener diode in parallel with E-B and C-B as needed.

Tim Williams
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  • I am off to see how this quirk works! Sure I did use a reverse be as a voltage ref, but I am yet to see the collector PV effect. – fraxinus Feb 17 '24 at 14:56
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This is how I simulate reverse breakdown of base-emitter junctions:

schematic

simulate this circuit – Schematic created using CircuitLab

Diodes D2 and D4 block current when the B-E junction is forward biased, preventing leakage through the zener diodes D1 and D3 from interfering with the usual B-E junction behaviour.

It's only when the B-E junctions are reverse biased that zener current flows, bypassing the the B-E junction but still clamping junction voltage to 5V or so.

Simon Fitch
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