why the switch is so designed that the transistor does not remain in active state?

Asked by chandni tiwari | 26th Feb, 2011, 12:00: AM

Expert Answer:

Dear student,
 

As long as Vi is low and unable to forward-bias the transistor, Vo is high at VCC. If Vi is high enough to drive the transistor into saturation, then Vo is low, very near to zero. When the transistor is not conducting it is said to be switched off and when it is driven into saturation it is said to be switched on.

This shows that if we define low and high states as below and above certain voltage levels corresponding to cutoff and saturation of the transistor, then we can say that a low input switches the transistor off and a high input switches it on. Alternatively, we can say that a low Input to the transistor gives a high output and a high input gives a low output. The switching circuits are designed in such a way that the transistor does not remain in active state.

When the transistor is used in the cutoff or saturation state it acts as a switch. On the other hand for using the transistor as an amplifier, it has to operate in the active region. 
 
 
Hope this helps.
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Answered by  | 28th Feb, 2011, 12:42: PM

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