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This active clamp design is elegant and simple, and works well to prevent OUT from exceeding the supply voltage (+15V), but fails when IN is low enough to reverse bias Q1's base-emitter junction beyond its reverse-breakdown voltage. When this happens, that junction begins to conduct, and its otherwise perfect behaviour is ruined.

enter image description here

Q2 is just a diode-connected device matching Q1, to obtain a \$1 \times V_{BE}\$ offset.

This \$V_{OUT}\$ vs. \$V_{IN}\$ is what I'd like to see, if Q1's B-E junction didn't ever break down:

enter image description here

However, if I use a transistor whose base-emitter reverse-breakdown voltage is 7V, this is what would happen when IN drops 7V below the base potential of +14.3V:

enter image description here

The problem is clear, that low input potentials (below about 7V here) don't survive intact to OUT.

Edit: Properties that I wish to retain:

  • Extremely high impedance looking into Q1's emitter. Inputs all the way up to 14.8V won't even know Q1 is there.

  • Clamping level is precise, within 0.1V of +15V, and easy to "set". This is why I am using BJTs, their \$V_{BE}\$ is so predictable.

  • A sharp "knee" when the input passes +15V, and Q1 begins to conduct. The transition is much sharper there than a regular diode clamp to a low-impedance source of +14.3V.

Question:

  • Can anyone suggest a way to avoid or compensate for reverse breakdown, without compromising these virtues?

  • Are there other designs with similar simplicity, impedance and clamping "sharpness" that do not suffer from this problem?

Update

Some users (Vladimir, Tim Williams) suggested fitting Q1 with collector and emitter reversed. Spehro Pefhany fixed a problem with doing only that, and both Tim and tobalt suggested using a different MOSFET/BJT to shunt. This is what I've done with those ideas:

schematic

simulate this circuit – Schematic created using CircuitLab

By reversing Q2, it solves the original problem I had of the base-emitter junction being excessively reverse biased, but it simply moves that problem to the other side. To deal with that, D2 absorbs most of what would have appeared between Q2's base and emitter, and that junction can now never go beyond -3V or so.

Q1 is the new current shunt, but its base-collector junction also suffers from being reverse biased when IN (and consequently Q1's collector) goes negative. That problem is easily solved by D1.

However, something else unexpected happens, which really improves the sharpness of the knee, and flatness of the clamped output. With Q2 reversed, its current gain is now very low, meaning that when clamping occurs, a big chunk of current that exits its emitter is drawn in from the base, via D3. Consequently, the voltage across D3 and Q1's \$V_{BE}\$ now track each other much more closely. I've been able to remove the very stable 0.7V offset provided by the diode-connected transistor, and replace it with D3 to leverage this new behaviour.

A simulation predicts that current from OUT to ground via the clamp remains under 100nA all the way up to 14.9V input, which is better than my original design. The knee is sharper, and OUT never exceeds 100mV above \$V_{REF}\$ all the way up to 50V at IN. Again, better than the original circuit.

Since there are so few stages in the design, there's very little delay between an increase in input potential and the onset of current shunting, and step response is really good.

enter image description here enter image description here

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  • \$\begingroup\$ What if exchange collectors and emitters? \$\endgroup\$
    – Vladimir
    Aug 25, 2022 at 9:37
  • \$\begingroup\$ @Vladimir this is essentially Tim's answer below. But this doesn't work if the collectors block more than 7V during regular operation (which they do in this application example). \$\endgroup\$
    – tobalt
    Aug 25, 2022 at 12:33
  • \$\begingroup\$ It seems to be a signal path, so how would a rail-to-rail opamp be? \$\endgroup\$ Aug 25, 2022 at 21:13
  • \$\begingroup\$ @aconcernedcitizen I envisaged using this to protect a DAC or op-amp input from some unknown, untrusted source, without compromising its high impedance with leaky reverse biased schottkey or zener diodes. So, using an op-amp would defeat its usefulness, if I have to protect that op-amp's inputs too. \$\endgroup\$ Aug 26, 2022 at 0:37
  • \$\begingroup\$ @SimonFitch I see. Well, your update seems to be a good answer, it may be worth posting it. It certainly looks more stable and even closer to the rail. \$\endgroup\$ Aug 26, 2022 at 8:04

3 Answers 3

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Maybe something like this (essentially @Vladamir's comment with a fix)

schematic

simulate this circuit – Schematic created using CircuitLab

Inverse beta is not that high, depending on the transistor, so the results may be YMMV.

enter image description here

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Use inverted operation.

I did something like this, some years ago; it didn't need VREF > VEB so it was a bit simpler, but this should also work.

schematic

simulate this circuit – Schematic created using CircuitLab

Addition is the cascode (Q3) to increase VREF range; for VREF < VEB, it can be removed (disconnect B, short E-C; and short out D1, it isn't needed then).

R1 and Q1-Q2 function identically to R2 and Q1-Q2 in the OP circuit; however, Q4 boosts the current, compensating for the poor hFE in inverted mode (at least for most types; of which 2N3906 is no exception). R3 serves both to discharge Q4 base charge (faster recovery), and to balance the collector currents; offset voltage can be adjusted somewhat by varying the ratio of the two resistors.

Note that, as a VBE cancellation type circuit, the temperatures of Q1 and Q2 must be matched to keep offset low. A monolithic dual is preferred, but these are quite rare these days; matched dies are readily available, but as separate dies, their temperatures needn't track very well. On the upside, the current flow through such a circuit should be quite low, and thus the thermal error also small.

Note finally, the clamp voltage needs to be at least a few volts, to get adequate current in R1, plus enough voltage to activate Q4. This is not a general-purpose min(a, b) circuit. If variable VREF is needed, R1 could be replaced by a current sink, or R3 as well (both being parts of a current mirror, say). That extends operation to a minimum of a couple VBE; for still wider range (and also a sharper knee), I would recommend a fully active (i.e. op-amp based) circuit.

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Here are some possible approaches. In each case, the "out" voltage is the right leg, and the +15V is on the left leg.

schematic

simulate this circuit – Schematic created using CircuitLab

The last one is slightly sharper than the middle one. On par with your initial circuit suggestion.

Finally here would be a circuit, that is much sharper than your initial one:

schematic

simulate this circuit

Placing the Schottky in between the bases like in the third schematic above lowers the threshold, even below the power supply level. Also one can play slightly with the two resistors to adjust the threhsold.

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  • \$\begingroup\$ I can get within 0.1V of the desired +15V clamping level, using BJTs, because even unmatched, their \$V_{BE}\$ are close. I have a bunch of BS250s whose difference in \$V_{TH}\$ is not so predictable. On the right is more promising; I worry that D2 removes that sharpness I want so much, and makes the whole thing no better than a regular diode clamp to a low impedance source of 14.3V. Could we build a MOSFET ideal diode to replace D2 and/or D1? \$\endgroup\$ Aug 25, 2022 at 7:10
  • \$\begingroup\$ @SimonFitch I think using ideal-diode circuits removes a bit of the simplicity. Why not just use an opamp based shunt regulator or TL431 then ? Instead, I have added a third approach, which moves D2 to the base, where its effect is slightly mitigated. And also yet another circuit that features very sharp (slightly tunable) threshold behavior and works at high currents due to the extra transistor. \$\endgroup\$
    – tobalt
    Aug 25, 2022 at 9:17
  • \$\begingroup\$ I'm embarrassed that it never occurred to me to place a diode in the base of Q10 to simply block breakdown current. So simple, even if it does introduce another diode drop. Using the MOSFET, switched by Q2, to pass all the clamp current, that removes almost all variation in current in any diodes - so you're right about the extra sharpness. On the bench a combination of top-right and bottom worked well. \$\endgroup\$ Aug 25, 2022 at 12:36
  • \$\begingroup\$ I say it worked well - it worked well for positive inputs. That freakin' body diode caught me out, again! Easy fix though, another diode in series with the MOSFET. \$\endgroup\$ Aug 25, 2022 at 12:40

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