Question:
Take the breakdown voltage of the zener diode used in the given circuit as 6V. For the input voltage shown in figure below, the time variation of the output voltage is :
(Graphs drawn are schematic and not to scale)
Take the breakdown voltage of the zener diode used in the given circuit as 6V. For the input voltage shown in figure below, the time variation of the output voltage is :
(Graphs drawn are schematic and not to scale)
(Graphs drawn are schematic and not to scale)
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The output is taken across two zener diodes connected back to back, so in each half-cycle one diode is forward biased and the other goes into reverse breakdown at 6 V.
Updated On: Nov 5, 2026
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The Correct Option is B
Solution and Explanation
The figure shows a resistor \(R_1\) in series and, across the output, two zener diodes joined back to back (cathodes together). The input is a sine wave of peak \(\pm 10\ \mathrm{V}\); the breakdown voltage is \(V_Z = 6\ \mathrm{V}\).
Positive half-cycle: the upper zener is forward biased (drop \(\approx 0\)), the lower zener is reverse biased and conducts only when the voltage across it reaches 6 V. So for \(0 \lt V_{in} \lt 6\ \mathrm{V}\) no current flows, there is no drop across \(R_1\), and \(V_o = V_{in}\). For \(V_{in} \gt 6\ \mathrm{V}\) the lower zener breaks down and holds \(V_o = +6\ \mathrm{V}\); the excess (up to 4 V at the peak) drops across \(R_1\).
Negative half-cycle: the roles of the two zeners are exchanged, so \(V_o = V_{in}\) for \(|V_{in}| \lt 6\ \mathrm{V}\) and \(V_o\) is clipped at \(-6\ \mathrm{V}\) beyond that.
Hence the output is a sine wave whose positive and negative peaks are both flattened at \(\pm 6\ \mathrm{V}\). This is the second graph (option 2). Option 1 has no clipping, option 3 clips only the positive half and option 4 only the negative half.
Answer: (2) sine wave clipped symmetrically at +6 V and -6 V
Positive half-cycle: the upper zener is forward biased (drop \(\approx 0\)), the lower zener is reverse biased and conducts only when the voltage across it reaches 6 V. So for \(0 \lt V_{in} \lt 6\ \mathrm{V}\) no current flows, there is no drop across \(R_1\), and \(V_o = V_{in}\). For \(V_{in} \gt 6\ \mathrm{V}\) the lower zener breaks down and holds \(V_o = +6\ \mathrm{V}\); the excess (up to 4 V at the peak) drops across \(R_1\).
Negative half-cycle: the roles of the two zeners are exchanged, so \(V_o = V_{in}\) for \(|V_{in}| \lt 6\ \mathrm{V}\) and \(V_o\) is clipped at \(-6\ \mathrm{V}\) beyond that.
Hence the output is a sine wave whose positive and negative peaks are both flattened at \(\pm 6\ \mathrm{V}\). This is the second graph (option 2). Option 1 has no clipping, option 3 clips only the positive half and option 4 only the negative half.
Answer: (2) sine wave clipped symmetrically at +6 V and -6 V
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