Explanation Calculation: The given diagram is shown in Figure 1 Mark the voltages and current and redraw the circuit. The required diagram is shown in Figure 2 The expression for the voltage v 1 is given by, v 1 = v 2 Substitute 0 for v 2 in the above equation. v 1 = 0 In the given Figure the Zener diode is reverse biased and the current through it is negligible when the voltage across it is less than the knee voltage. Consider the case when the Zener diode voltage is less than the knee voltage. ( 10 V − 6.8 V ) 5.6 k Ω = i Z + 6.8 V − 0 1 k Ω i Z = − 6.228 mA The above expression shows that the current through the diode is less than the zero, thus the Zener diode is not in saturation mode. The voltage across the diode is less than the threshold and the current through the diode is given by, i Z = 0 Apply KCL at the cathode of Zener diode. 10 V − v I 5.6 k Ω = i Z + v I − 0 1 k Ω Substitute 0 for i Z in the above equation. 10 V − v I 5.6 k Ω = 0 + v I − 0 1 k Ω 6.6 v I = 10 v I = 1.515 V The expression for the current i is given by, i 1 = v I − v 1 1 k Ω Substitute 0 for v 1 and 1

Microelectronics: Circuit Analysis and Design

4th Edition

ISBN: 9780073380643

Author: Donald A. Neamen

Publisher: McGraw-Hill Companies, The

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Question

Chapter 9, Problem 9.79P

To determine

The value of vO , i2 and iZ .

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Chapter 9, Problem D9.78P

Chapter 9, Problem 9.80P

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Microelectronics: Circuit Analysis and Design

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