I have a question regarding bipolar transistor and it's operation as a switch.

I know that, in order to make transistor work as a switch, we need to avoid active region, meaning, we need to make transistor to work only in saturation and cut-off region.

This basically, cannot be implemented perfectly, but we can make transition period from saturation to cut off region as fast as possible, but I am not quite sure how to make transistor to work in cut off region, meaning, I want to know how to turn the transistor off correctly.

Let's take NPN transistor as an example:

The transistor is off if it's base-emitter and base-collector PN junctions are reverse-biased, however, that means that we would need significant negative voltage on base in order to do so, but I am wondering, how would this affect the circuit and transistor?

Another approach I've heard of is that we only need to make base-collector reverse biased and then we need to add a resistor between base and emitter, how is this any better? Do I need a resistor with huge resistance or with low resistance or it actually doesn't matter, and why?

Any help appreciated!

  • 1
    \$\begingroup\$ To cut-off, the bipolar transistor all you need to worry about is about the base-emitter voltage and the base current. Short the base together with the emitter and the BJT will be cut-off. Or set Ib = 0A. \$\endgroup\$
    – G36
    Nov 4, 2018 at 10:48
  • \$\begingroup\$ Ok, but, why that works? I mean, why is that a correct way to turn off transistor? What would happen if we tried to turn it off by negative voltage on the base? Why is that bad thing to do? Why is shorting the base and emitter better solution? \$\endgroup\$
    – cdummie
    Nov 4, 2018 at 10:53
  • 1
    \$\begingroup\$ It works because without the base current (Vbe <0.4V) transistor is cut-off. And transistor action only occurs if the base current is flowing. And sometimes to speed-up the BJT turn-off time we apply a negative voltage between base-emitter junction. electronics.stackexchange.com/questions/367750/… and about saturation electronics.stackexchange.com/questions/276146/… \$\endgroup\$
    – G36
    Nov 4, 2018 at 10:59
  • \$\begingroup\$ Why cant you use the transistor in active region? \$\endgroup\$ Jan 2, 2021 at 12:11
  • \$\begingroup\$ @the force awakens because then it's not a switch. For a switch, you want as much IC current as needed by your load. In active mode, your current is limited by IB, while in saturation it's limited by your load. \$\endgroup\$
    – KD9PDP
    May 23, 2021 at 16:08

1 Answer 1


In addition to what others wrote in the comments: A bipolar transistor is basically a current amplifier from which follows that no current flowing into the base means no current flowing through collector/emitter.


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