simulate this circuit – Schematic created using CircuitLab

Yesterday, I saw a schematic exactly the same as this, except the voltage connected to the relay was 12 V and different from the voltage at the base (TTL levels). I told the designer that it wouldnt work because the collector is now more negative than the base and emitter; not how it should be for an NPN BJT.

Imagine my surprise when this guy simulated the circuit above (which, i think, is even worse than the original) in Proteus and the relay switched! I scrambled to figure out why this happened and noticed that as the base resistor was increased to 1k, the relay didnt switch again. But this didnt explain how the base-emitter junction became forward-biased in the first place, since there is exactly 0 V between the left of the base resistor and the rightmost terminal of the relay.

What crucial detail am I missing or do I not know here? Is there any condition when this general design would work as expected, to drive some load?

The actual transistor in the simulation was a TIP41 NPN BJT. The relay was a 2 V relay. When it was increased to 4 V, it still switched.


1 Answer 1


The transistor is in reverse active mode. The collector acts as the emitter and the emitter as the collector. This is possible because NPN reversed is still NPN.

The performance is worse than in forward active mode, because emitter and collector usually have different doping levels and a different structure. Therefore this mode of operation is not very useful.

Another problem with this circuit is that one has to be careful not to exceed the maximum reverse voltage of the base-emitter junction.

  • \$\begingroup\$ Reading your answer again makes it clear. So the junction that needs to be forward biased this time, is the B-C junction? And with the collector acting as emitter now and vice-versa, the base is 0.7V above the 'emitter', and turns on the transistor. But when the resistor was increased to 1k, the relay stopped switching. Would this be because the base current is now too low to activate the transistor i.e. its below the minimum for a TIP41? \$\endgroup\$ Commented Apr 9, 2016 at 7:40
  • \$\begingroup\$ @TisteAndii Firstly note that the name "Bipolar Transistor" is not just name. Secondly the reverse active set-up can be less efficient to the tune of dozens or even hundreds, depending on the transistor. So yes, with an even slightly higher resistor the current gain in the set-up will be so small that a low base current won't be amplified to a useful relay current any more. \$\endgroup\$
    – Asmyldof
    Commented Apr 9, 2016 at 7:45
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    \$\begingroup\$ @Asmyldof i'm not sure what you meant by "Bipolar transistor is not just a name". Doesnt "bipolar" refer to the fact that there are 2 charge carriers in a transistor: electrons and holes? Also, the reason why the relay no longer switches is because beta is now much reduced because of this poor configuration? \$\endgroup\$ Commented Apr 9, 2016 at 7:56
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    \$\begingroup\$ Collector is N type, base is P type, emitter is N type. If you swap it around, it is still NPN. See? Bipolar. The saturation voltage in reverse active mode may be very low, and this could possibly be useful in some situations. electronics.stackexchange.com/questions/29756/… \$\endgroup\$
    – user57037
    Commented Jan 18, 2017 at 9:34

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