I have a circuit as follows,


simulate this circuit – Schematic created using CircuitLab

which provides a control current to an OTA as \$i_\mathrm{abc} = (3.27\mathrm{V} - V_\mathrm{in})/R1\$. The reason for the reference voltage being 3.27V is that I want to make sure I can completely turn off the OTA, even after accounting for various offsets. Of course, the problem here is that when \$V_\mathrm{in} > 3.27\mathrm{V}\$, the op-amp tries to push current in the reverse direction, which it can't do, and therefore saturates at around 11.5V. In this case, the BJT has about 8.5V of reverse voltage, which is in excess of the datasheet absolute maximum reverse voltage of 5V.

There's of course the standard solution of throwing in a diode in the reverse direction to protect the BJT and satisfy the op-amp, but as the board has already been routed and prototyped without the diode and it's really cramped, avoiding the diodes would be really nice. The board does work at least this far, so the transistor does not fail immediately (or in about half a year time span).

So my question is: considering that I'm breaking the absolute maximums here, what is the danger of the transistor actually breaking/degrading significantly? Note that even if it goes to reverse breakdown, the reverse current will be at most about \$0.03V/3.3k\Omega \approx 9\mu \mathrm{A}\$, i.e. the power will be correspondingly small. Measured from the prototype, the actual values are \$4\mu \mathrm{A}\$ and \$33\mu\mathrm{W}\$. The product will be manufactured in quantities of some hundreds of pieces, and there's 10 circuit blocks such as above per board, so lifetime failure probabilities on the order of one in a few thousand are likely to manifest themselves.

  • \$\begingroup\$ Do you have a lower (+) rail voltage you can use instead of +12? \$\endgroup\$ – jonk Aug 18 at 7:39
  • \$\begingroup\$ Only 3.3V, which isn't enough for the TL074... of course, it might make sense to see if throwing in a 5V regulator and routing that current would be cheaper and easier than adding the diodes, since this is indeed repeated 10 times on the board... \$\endgroup\$ – Timo Aug 18 at 7:42
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    \$\begingroup\$ You could try replacing the BJT with a P-channel mosfet. \$\endgroup\$ – mkeith Aug 18 at 7:42
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    \$\begingroup\$ @mkeith yes, thanks for spotting that, corrected! It's a dual, but of course a PNP. \$\endgroup\$ – Timo Aug 18 at 8:00
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    \$\begingroup\$ @mkeith agreed, adding the diode isn't exactly that hard, it's just really inconvenient. But basically any modification that involves adding components is going to be at least as difficult, so might then just as well add the diode. \$\endgroup\$ – Timo Aug 18 at 9:42

Sustained reverse breakdown of the base-emitter junction of a BJT will cause gradual degradation of HFE.

In some cases it can also cause unusual photo-emission effects where light coming from the junction in breakdown can cause photo-current or even photovoltaic generation in other junctions.

Keeping the current low reduces the rate of decay so that is in your favor.

Are Reverse Biased transistors stable?

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