Question

In the following limiter circuit, the input voltage \(V_i = 10 \sin(100 \pi t)V\) is applied. Assume that the diode drop is 0.7 V when it is forward biased and the Zener breakdown voltage is 6.8 V. The maximum and minimum values of the output voltage, respectively, are

• A. 6.1 V, -0.7 V
• B. 0.7 V, -7.5 V
• C. 7.5 V, -0.7 V
• D. 7.5 V, -7.5 V

Hint:

To analyze clipper circuits, consider the positive and negative input cycles separately. Determine which diodes are forward-biased and which are reverse-biased in each case to find the voltage level at which the output is clipped.

Answer 1

The Correct Option is C

Step 1: Analyze the circuit during the positive half-cycle of the input (\(V_i>0\)). During the positive half-cycle, current flows from the input through the 1 k\(\Omega\) resistor.

The lower diode D2 is reverse-biased and acts as an open circuit.
In the upper branch, the conventional diode D1 is forward-biased, and the Zener diode Z is reverse-biased.
The upper branch will start to conduct and clamp the output voltage when the voltage across it is sufficient to forward bias D1 and cause the Zener to enter breakdown. The output voltage will be limited to: \[ V_{o,max} = V_{D1,forward} + V_{Z,breakdown} = 0.7 V + 6.8 V = 7.5 V \]
Step 2: Analyze the circuit during the negative half-cycle of the input (\(V_i<0\)). During the negative half-cycle, the voltage becomes negative.

In the upper branch, the conventional diode D1 is reverse-biased, acting as an open circuit. This entire branch is non-conducting.
The lower diode D2 is forward-biased.
The output voltage will be clamped by the forward-biased diode D2. The minimum output voltage will be: \[ V_{o,min} = -V_{D2,forward} = -0.7 V \] Thus, the output voltage is clipped at +7.5 V and -0.7 V.