A circuit is arranged as shown in the figure. | vedclass

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Question

(A) The circuit shows an $NPN$ transistor acting as a switch. The input to the base is controlled by two diodes $D_1$ and $D_2$ connected in parallel

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) The circuit shows an $NPN$ transistor acting as a switch. The input to the base is controlled by two diodes $D_1$ and $D_2$ connected in parallel to the ground $(0 	ext{ V})$.Since the input to the diodes is $0 	ext{ V}$,both diodes $D_1$ and $D_2$ are forward-biased.When forward-biased,the diodes act as short circuits (assuming ideal diodes with negligible resistance),effectively connecting the base of the transistor to the ground $(0 	ext{ V})$.Since the base voltage $V_B = 0 	ext{ V}$,the base-emitter junction is not forward-biased,and the transistor remains in the cut-off state.In the cut-off state,no collector current flows $(I_C = 0)$.Therefore,the output voltage $V_0$ measured at the collector is equal to the supply voltage of the collector circuit.Thus,${V}_{0} = 5 	ext{ V}$.

$A$ circuit is arranged as shown in the figure. The output voltage ${V}_{0}$ is equal to $\ldots \ldots \text{ V}$.

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