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
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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.
Step 1: Analyze the circuit when the input voltage is positive (\(V_i>0\)).With a positive input, current flows from the input through the 1 k\(\Omega\) resistor.
Diode D2 is reverse-biased and behaves as an open circuit. In the upper branch, diode D1 is forward-biased, while the Zener diode Z is reverse-biased. The upper branch conducts and clamps the output voltage when the voltage is sufficient to forward bias D1 and cause Zener breakdown. The maximum output voltage is:\[ V_{o,max} = V_{D1,forward} + V_{Z,breakdown} = 0.7 V + 6.8 V = 7.5 V \] Step 2: Analyze the circuit when the input voltage is negative (\(V_i<0\)).During the negative half-cycle, the voltage is negative.
In the upper branch, diode D1 is reverse-biased and acts as an open circuit, making the branch non-conducting. Diode D2 is forward-biased. The forward-biased diode D2 clamps the output voltage. The minimum output voltage is:\[ V_{o,min} = -V_{D2,forward} = -0.7 V \]Therefore, the output voltage is clipped at +7.5 V and -0.7 V.
