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What do you understand by the cut off, active and saturation states of the transistor? In which of these state does the transistor not remain when being used as a switch? |
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Answer» Solution :When an npn transistor with common emitter mode is mode is used as a switch, then if during working, there is no collector current, then the transistor is in cut off state. In case of silicon-transistor, so long as INPUT VOLTAGE` V_(i) lt 0.6V`, the transistor will be in cut off state. If during working, there is a situation, beyond cut off state, for which `V_(i) gt 0.6V` but less than `1.0V` the collector current increases almost linearly and the output voltage decreases linearly, till `V_(i)` becomes NEARLY `1.0V`. The transistor in this state is called in active state. If during working, there is a situation beyond active state for which `V_(i) gt =1.0V`, the VARIATION of `V_(i)` and `V_(0)` is non-linear, because with the increase in `V_(i), V_(0)` is found to decrease towards zero, but never becomes zero. In this situation the collector current `I_` becomes maximum and the transistor is in a saturation state. When a transistor is being used as a switch, it will not remain in active state. |
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