In the figure shown below,

The $V_{out}$ of the circuit is ___. (Assume the cut in voltage of the diode $V_{v} = 0.7 V$)

8.4 V

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9.2 V

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0 V

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6.2 V

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Solution

The correct option is D 6.2 V
The above circuit can be represented as,

Applying KCL at node 1,

$\frac{V_{0}}{2 k} + \frac{V_{0} - 10 + 0.7}{2 k} + \frac{V_{0} - 10 + 0.7}{2 k} = 0$

$V_{0} [\frac{3}{2 k}] = \frac{9.3}{1 k}$

$V_{o u t} = \frac{9.3}{3} \times 2 = 6.2 V$

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C o n s i d e r t h e c i r c u i t s h o w n i n t h e f i g u r e b e l o w . T h e i n p u t t o t h e c i r c u i t V_{i n} = 10 s i n (w t) . I f t h e d i o d e s D_{1} a n d D_{2} b o t h h a v e c u t i n v o l t a g e o f 0.7 V, t h e n w h i c h o f t h e f o l l o w i n g s t a t e m e n t a r e t r u e ?

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