In this circuit, when I switch off the transistor opening the switch SW1, a negative Voltage of circa -6.5V appears at its base. I would like to understand why, my first hypotesis is because of the base capacitance of the BJT. Is it true?


simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ My first hypothesis is that something is hooked up differently than your schematic or you are measuring the wrong points. I don't see a way to get a negative voltage on the base in the schematic you posted. What is the voltage at the emitter when the switch is closed? Where exactly are you putting the meter leads when you measure the negative voltage? You might want to put a 100K resistor from the base to ground to keep the base from floating when the switch is off and prevent potential oscillations. \$\endgroup\$
    – John D
    Commented Oct 12, 2015 at 13:23
  • \$\begingroup\$ @JohnD The emitter voltage is 8.22V. I forgot to mention that I got this behavior using a circuit simulator (EveryCircuit). Maybe is the software inaccurate? \$\endgroup\$
    – Enrico
    Commented Oct 12, 2015 at 13:27
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    \$\begingroup\$ CircuitLab does not show this (negative voltage) behaviour \$\endgroup\$
    – Icy
    Commented Oct 12, 2015 at 13:30
  • \$\begingroup\$ ok then must be the simulator. Thank you for your help. \$\endgroup\$
    – Enrico
    Commented Oct 12, 2015 at 13:33

1 Answer 1


The question refers to a behaviour of the circuit when simulated in EveryCircuit. A similar behaviour is not observable with other simulators.


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