Consider this circuit with a sinusoidal source connected to the collector and a pulse toggling between 0 and 5 connected to the base of a 2N2222 transistor.

What I imagined the output being was when the input voltage is 5 the transistor turns on and we would see the sine wave at the emitter too. And when it's 0 it's off and we get 0V at the output. Exactly like a switch. Right?


simulate this circuit – Schematic created using CircuitLab

But after I simulated it in PSpice, I'll get this. When the input voltage is zero and the base is basically grounded, you still get the negative part of the sine signal. (Red=Vout, Green=Vin)

How is this possible? I can't seem to figure out the reasoning for why this happens. enter image description here

  • \$\begingroup\$ Q3e waveform is displayed but it doesn't exist in your schematic. \$\endgroup\$
    – Andy aka
    Jan 3, 2022 at 19:27
  • \$\begingroup\$ @Andyaka Sorry I forgot to label them appropriately. I'll edit the post right now. \$\endgroup\$
    – Sepand.S
    Jan 3, 2022 at 19:36

1 Answer 1


Q1, being an NPN transistor, offers some BJT functionality when the collector is used as an emitter and, the emitter is used as a collector. It doesn't work so well but it does provide some transistor functionality. This is what is happening here.

So, when the "new emitter" (previously the collector) is taken to a voltage lower than the "new collector" (previously the emitter), it behaves as a conventional transistor (albeit with reduced gain). And, when the "new emitter" is a volt or so below the base, the base-(new emitter) junction is turned on and you get a signal at the "new collector".

Here's what the schematic looks like when you have negative voltages on the collector: -

enter image description here


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