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So it is operated in either saturation or cut off region but in between these two there is linear region (the intermediate region between the LOW and HIGH states of the transistor) and the transistor will obviously pass through that going from cut off to saturation and vice versa.

Diagram showing the operation regions of a transistor

Can a transistor switch between cut-off and saturation without going through the linear region?

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    \$\begingroup\$ What is your question? Are you asking how the active region works or why you might use it or? \$\endgroup\$ – Passerby Aug 23 '15 at 7:25
  • \$\begingroup\$ I think my question was clear enough but here it is again,even if I want the transistor to operate in cut off and saturation still there seems to be no way I can avoid the linear region completely.While changing from one region (say saturation) to the other (cutoff) it will have to go through the linear region..I cannot understand how it will jump "between" states. \$\endgroup\$ – muhammad muheeb Aug 23 '15 at 7:59
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Here is the measurement simulation of a real fake transistor:
transistor switch simulation

As you can see in the behaviour of VBE and VCE, the transistor indeed goes through the linear region while switching.

A transistor cannot switch with infinite speed. The base (and other parts) actually have a small capacitance, so it must be (dis)charged similar to a capacitor, which takes time.

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The above answer has covered most of the aspects, i would like to elaborate a bit. The operation of transistor is not similar two the mechanical switch where there are just two states and both of them are valid. In transistor we have three states of operation and we consider only two of them in switching operation which does not mean that during the switching application active mode just goes away to a different place.

During switching, going through the active stage makes the switch work in both saturation ad cutoff region.

And talking about how this happens, as a BJT is a voltage controlled device, ll this depends on the voltage between EBJ (Emitter Base Junction) and CBJ (Collector Base Junction):

  1. For Saturation EBJ and CBJ must be forward biased

  2. For cutoff VBE < 0.5V.

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