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Operating regions of a transistor switch

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muheeb16

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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.How does that work?

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Hi,

Your question is not that clear.

When driving a bjt as a switch you should avoid linear region.
To switch OFF, you need a low base current. About zero. This is often ensured with a base to emitter resistor.

To switch ON, you need enough base drive current. Often one suggests 1/10 of the expected collector current. With this value you are able to calculate a series base resistor.

Klaus
 

:fight: That is the trick :fight:

When you give 0mA the transistor will be turned OFF and when you give 10 mA (for eg) the transistor will turn ON, but why did you give the current intermediate 0 and 10 mA ?

If you wont, then transistor will not work in linear region.
 

If operating as a switch, you go from saturation to cutoff (or other way round) as you mentioned. As Vbe increases, for a short amount of time, the transistor will be in active (linear) mode until you enter saturation. But you want to design such that it spends as little time as possible in the linear region.
 

Tahmid said:
If operating as a switch, you go from saturation to cutoff (or other way round) as you mentioned. As Vbe increases, for a short amount of time, the transistor will be in active (linear) mode until you enter saturation. But you want to design such that it spends as little time as possible in the linear region.
Click to expand...
This is true.
There will ALWAYS be a brief interval where the switch passes through the linear region during the action of switching. Its what causes switching loss.
The idea is to make that switching interval as short as possible.
Use a high frequency device, and drive it with a very fast changing signal on the base.
 

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