what should be the Vbe voltage to turn on the pnp transistor and is it always necessary for the base to be negative or is it just that Base needs to be more negative than the Emitter by approx 0.7 volts.

  • \$\begingroup\$ Your first question was closed for lack of details and effort. This question is identical, and will be closed very soon as well. Fix the first one. If you can make it over into a good question, it will be reopened. If you keep posting copies of a bad question, you will be blocked from asking questions. \$\endgroup\$
    – JRE
    Dec 9, 2019 at 12:33
  • \$\begingroup\$ ok fine thanks for the suggestion \$\endgroup\$ Dec 9, 2019 at 12:35

1 Answer 1


Firstly, understand that there is no such thing as absolute negative voltage. Voltages are always relative. Negative Voltage means that this voltage is negative with respect to some reference voltage. Most of the times that reference is ground potential. So you are correct when you say, that base potential must be negative with respect to the Emitter Potential. For example, to turn the transistor ON (usually saturation mode), the base can be at +1V and the Emitter can be at +1.7 V. Hence it is the potential difference that matters. Ve-Vb should be +0.7V, no matter what their individual polarities are.

Edit: The condition for Saturation also requires that the base current reaches a certain threshold value, additional to the 0.7V potential difference condition.

  • \$\begingroup\$ Saturation isn't a fixed base current, though. It depends on load. \$\endgroup\$
    – Hearth
    Dec 9, 2019 at 17:34
  • \$\begingroup\$ I meant that the current need to cross a threshold for that, and yes that threshold depends on the load. \$\endgroup\$
    – Bhuvnesh
    Dec 9, 2019 at 18:55
  • \$\begingroup\$ Thank you, but what if the emitter voltage increases more than +1.7V \$\endgroup\$ Dec 18, 2019 at 7:00

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