Assuming the diode to be ideal, which one of the following is correct about transfer characteristic of the circuit shown below,

V_{0} = \frac{1}{2} V_{i n} + 2.5 f o r V_{i n} \geq - 5 V

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$V_{o} = V_{i n} f o r - 5 V \leq V_{i n} < + 5 V$

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$V_{0} = \frac{1}{2} V_{i n} - 2.5 f o r V_{i n} \leq - 5 V$

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All of the above

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Solution

The correct option is D All of the above
(d)
For $- 5 V \leq V_{i n} < + 5 V$ : both diodes are reverse baised (OFF), hence $V_{0 u t} = V_{i n}$ .

For $V_{i n} \leq - 5 V; D_{2}$ is reverse biased while $D_{1}$ is ON hence,
$V_{0} = \frac{V_{i n}}{2} - \frac{5}{2} (b y K V L)$

For $V_{i n} \geq - 5 V; D_{2}$ is forward biased while $D_{2}$ is reverse biased i.e.,

$V_{o u t} = \frac{V_{i n}}{2} + \frac{5}{2}$
Transfer characteristics:

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D_{1}

D_{2}

R_{1} = R_{2}

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