Take the breakdown voltage of the zener diode used in the given circuit as 6V. For the input voltage

Q: Why do we assume the forward voltage drop is zero?

In many textbook problems, unless specified otherwise, diodes are treated as ideal to simplify the core concept. An ideal diode has zero resistance and zero voltage drop when forward-biased.

Q: What happens if the input voltage never reaches 6V?

If the peak input voltage were less than 6V (e.g., 5V), neither Zener diode would ever reach its breakdown region. The output would just be an exact copy of the input sine wave!

Q: Why are the diodes connected back-to-back?

This configuration creates a symmetric clipper. One diode handles the positive half-cycle clipping, and the other handles the negative half-cycle clipping. It's a simple way to regulate AC voltage.

Q: Does the resistor $R_1$ affect the clipping voltage?

No, the resistor limits the current to protect the Zener diodes from burning out. The clipping voltage is determined entirely by the Zener breakdown voltage $V_Z$.

Q: What if the two Zener diodes had different breakdown voltages?

Then the clipping would be asymmetric! For example, if $V_{Z1} = 6\text{ V}$ and $V_{Z2} = 4\text{ V}$, the wave would be clipped at $+6\text{ V}$ on top and $-4\text{ V}$ on the bottom.

Question

Take the breakdown voltage of the zener diode used in the given circuit as $6	ext{V}$. For the input voltage shown in figure below, the time variation of the output voltage is (Graphs are drawn schematically and on not to scale)

Options:

(A) Graph (a)

(B) Graph (b)

(C) Graph (c)

(D) Graph (d)

Sigma Percentile · Accepted Answer

### The Double-Clipper Circuit: Taming the Sine Wave

Let's carefully analyze the given circuit. We have an alternating input voltage, a series resistor, and two Zener diodes connected back-to-back. The output voltage is measured across this Zener combination. This specific arrangement is a classic example of a **double-clipper circuit**, designed to regulate and limit AC voltage swings.

### Understanding Zener Diode Characteristics

Before we look at the waveform, let's recall how a Zener diode works. In forward bias, it acts like a normal ideal diode with zero voltage drop (assuming ideal conditions). But in reverse bias, it blocks current until the voltage reaches its breakdown limit, which is $6	ext{ V}$ here. After that, it clamps the voltage at exactly $6	ext{ V}$.

### Analyzing the Positive Half-Cycle

Now, imagine the positive half-cycle of our input sine wave. The top terminal is positive. This makes the top Zener diode reverse-biased, and the bottom Zener diode forward-biased.

As the input voltage rises from zero, the output follows it exactly. But the moment the input hits $6	ext{ V}$, the top Zener diode breaks down! It won't let the voltage across it increase any further. Since the bottom diode is forward-biased with zero drop, the total output voltage gets clipped at exactly $+6	ext{ V}$.

$$V_o = V_{Z1} + V_{f2} = 6	ext{ V} + 0	ext{ V} = 6	ext{ V}$$

### Analyzing the Negative Half-Cycle

Next, let's look at the negative half-cycle. The polarity reverses. Now, the top Zener diode becomes forward-biased, and the bottom Zener diode is reverse-biased.

Just like before, the output voltage will follow the input downwards. But when it reaches $-6	ext{ V}$, the bottom Zener diode hits its breakdown region. It clamps the voltage, preventing it from going any lower. So, the output is clipped at $-6	ext{ V}$.

$$V_o = -(V_{f1} + V_{Z2}) = -(0	ext{ V} + 6	ext{ V}) = -6	ext{ V}$$

### The Final Conclusion

Putting it all together, the output waveform is a sine wave that is symmetrically chopped off at $+6	ext{ V}$ and $-6	ext{ V}$. Looking at our options, this perfectly matches graph (c).

Think about this: what if the diodes were not ideal? If they had a forward voltage drop of $0.7	ext{ V}$, the clipping wouldn't happen at exactly $6	ext{ V}$. It would happen at $6.7	ext{ V}$! Always pay attention to whether the problem assumes ideal diodes or not.

Animated Solution for Physics - Semiconductors: Take the breakdown voltage of the zener diode used in the given circuit as $6 V$ . For the input voltage shown in figure below, the time variation of the output voltage is (Graphs are drawn schematically and on not to scale)

Select Answer:

Visualized Solution

Circuit Analysis

Zener Diode Characteristics

Positive Half-Cycle ( $V_{in} > 0$ )

Positive Clipping Level

Negative Half-Cycle ( $V_{in} < 0$ )

Negative Clipping Level

Final Output Waveform

What if Diodes were Non-Ideal?

The Sigma Insight: P-N Junction Diode

The Double-Clipper Circuit

Understanding Zener Diode Characteristics

Analyzing the Positive Half-Cycle

Analyzing the Negative Half-Cycle

The Final Conclusion

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