I need to switch off a PNP transistor however, the voltage on the emitter side will always be larger than the base. Is there a way to switch the transistor off?

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Edit: The aim is to maintain a constant voltage at the collector of Q2 of around 4.7V with a supply of between 10Vdc and 230Vac. I want to use Q3 to switch off Q2 by shorting the base of Q2 to supply when enough voltage is generated across R11. However it is never really switching off.

Any ideas please?

thank you for your time.

  • \$\begingroup\$ If you raise the base to within a hundred mV of the emitter then it's pretty certain that the PNP transistor will be off. \$\endgroup\$ – Andy aka Jul 11 '17 at 7:54
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    \$\begingroup\$ Show the schematic of what you have, and explain why you cannot affect the base voltage. \$\endgroup\$ – CL. Jul 11 '17 at 8:01
  • \$\begingroup\$ I edited the question \$\endgroup\$ – lukecam95 Jul 11 '17 at 10:48
  • \$\begingroup\$ A BJT with both junctions forward biased is in saturation. If the top wire in your circuit is the supply, then the highest possible voltage on Q3's base is the supply volate, so the emitter can not be higher than 0.7V below that. Since Q2's base needs to be within a few 100 mV of the emitter (@Andyaka) Q2 will never turn off. \$\endgroup\$ – Austin Jul 11 '17 at 11:24

The PNP transistor can be turned on by maintaining a base voltage of

                 (emitter-gnd volt - 0.7 volt) volts


simulate this circuit – Schematic created using CircuitLab

so if you maintain the base voltage above (emitter_gnd volt - 0.7 volt) the transistor can be turned off


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