I want to switch on a relay when both a 5v logic input is high and a (LM741 used as a) comparator is around +10v, and the circuit I planned to use is below.

The problem I see with the circuit is when BUF1 output is low, (0v) and OA1 output is low, (-10v) the base of Q1 will be at 0v, but the emitter will be floating. Could this destroy Q1, given that Vebo absolute maximum for BC547A is 6v?


simulate this circuit – Schematic created using CircuitLab


No, it won't destroy — or even stress — Q1.

Remember, the base-emitter junction of a transistor is just a diode, after all. If there's no external connection to the emitter (e.g., when Q2 is cut off), then it simply sits about 0.65V below whatever the base voltage is.

The VEBO specification applies only when you're deliberately reverse-biasing that diode, which, because of the doping of that junction for good transistor action, has a very low breakdown voltage.


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