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InterviewSolution
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Explain, how a CE transistor acts as a switch. |
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Answer» Solution :Transistor as a switch Circuit DIAGRAM. The circuit daigram of an n-p-n transistor in CE mode is as shown in the figure. In this circuit, the emitter is forward biased by emitter base battery `V_("bb")` and the collector is reversed biased by emitter collector battery `V_("CC").R_(B)` and `R_(c)` are input and output resistance connected between emitter-base and collector-base respectively . Using Kirchhoff.s LAW, In emitter base circuit `V_("bb")=V_(be)+I_(b)R_(B)`.........`(i)`  In collector-base circuit, `V_(ec)=V_("cc")-I_(c)R_(c)`.......`(ii)` Here `V_("bb")=V_(i)` (input voltage and `V_("cc")=V_(0)` output voltage ) `:.V_(i)=V_(be)+I_(b)R_(B)` .................`(iii)` `V_(0)=V_("cc")-I_(c)R_(c)`............`(iv)` Equations `(iii)` and `(iv)` indicate how `V_(0)` changes, when `V_(i)` is increased slowly. (for Si n-p-n transistor) `(i)` If `V_(i) lt 0.6V`, the transistor is in cut off state and `I_(c)=0` and `V_(0)=V_("cc")`. `(ii)` If `0.6V ltV_(i) lt 1V` the transistor is in active state and `I_(c)` INCREASE linearly and `V_(0)` is less than the earlier value of `V_("cc")`. `(iii)` If `V_(i) gt 1V`, `I_(c)` increases non-linearly and hence `V_(0)` decreases non-linearly and when `I_(c)R_(c)` becomes maximum and transistor is said to be in saturation state tending to zero as shown in the figure.  Switching action. `(i)` When input voltage is very small, the transistor is in cut off region i.e., the transistor is not conducting and acts as if in open condition i.e. OFF state. `(ii)` When input voltage is large and the transistor is in saturation state, `V_(0)` is very small `(V_(0)-0)` , the transistor acts as a closed switch i.e. ON state. |
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