Question:
Two Zener diodes (A and B) having breakdown voltages of 6 V and 4 V respectively, are connected as shown in the circuit below. The output voltage \(V_0\) variation with input voltage linearly increasing with time, is given by :
(\(V_{input}\) = 0 V at t = 0)
(figures are qualitative)
Two Zener diodes (A and B) having breakdown voltages of 6 V and 4 V respectively, are connected as shown in the circuit below. The output voltage \(V_0\) variation with input voltage linearly increasing with time, is given by :
(\(V_{input}\) = 0 V at t = 0)
(figures are qualitative)
(\(V_{input}\) = 0 V at t = 0)
(figures are qualitative)
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Both Zener diodes are reverse biased. Zener B (4 V) conducts first, after which the B-branch (4 V + 100 \(\Omega\)) shares the increase in voltage with \(R_L\) = 100 \(\Omega\); Zener A (6 V) finally clamps the output.
Updated On: Nov 5, 2026
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The Correct Option is D
Solution and Explanation
Positive \(V_{in}\) is applied to the cathodes (bar side) of both Zener diodes, so they are reverse biased and operate in breakdown.
For \(V_{in} \lt 4\) V neither Zener conducts: only \(R_L\) draws current and \(V_0 = V_{in}\), rising linearly with time.
At \(V_{in} = 4\) V Zener B (4 V) breaks down. The B-branch now behaves like a 4 V source in series with 100 \(\Omega\), in parallel with \(R_L\) = 100 \(\Omega\). For \(V_{in} \gt 4\) V the output rises as \(V_0 = 4 + \frac{(V_{in}-4)\times 100}{100+100} = 4 + \frac{V_{in}-4}{2}\): the slope is halved but \(V_0\) keeps increasing.
\(V_0\) reaches 6 V when \(V_{in} = 8\) V. At this point Zener A (6 V) breaks down and clamps \(V_0\) at 6 V.
Hence the graph: linear rise to 4 V, a smaller slope up to 6 V, then constant at 6 V (with a dashed 4 V reference line). This is the fourth graph.
Answer: (4) the graph that rises linearly to 4 V, then with a smaller slope up to 6 V, then stays constant at 6 V
For \(V_{in} \lt 4\) V neither Zener conducts: only \(R_L\) draws current and \(V_0 = V_{in}\), rising linearly with time.
At \(V_{in} = 4\) V Zener B (4 V) breaks down. The B-branch now behaves like a 4 V source in series with 100 \(\Omega\), in parallel with \(R_L\) = 100 \(\Omega\). For \(V_{in} \gt 4\) V the output rises as \(V_0 = 4 + \frac{(V_{in}-4)\times 100}{100+100} = 4 + \frac{V_{in}-4}{2}\): the slope is halved but \(V_0\) keeps increasing.
\(V_0\) reaches 6 V when \(V_{in} = 8\) V. At this point Zener A (6 V) breaks down and clamps \(V_0\) at 6 V.
Hence the graph: linear rise to 4 V, a smaller slope up to 6 V, then constant at 6 V (with a dashed 4 V reference line). This is the fourth graph.
Answer: (4) the graph that rises linearly to 4 V, then with a smaller slope up to 6 V, then stays constant at 6 V
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