Explanation Given Information: R s is equals to 10kΩ. Calculation: For small signal model DC source is open circuited, diode is replaced by a dynamic resistance r d , and capacitor C 1 , C 2 is short circuited. Hence, modified figure will be as follows, Dynamic resistance of diode is calculated as, r d = V T η I D η = 1 r d = V T I D Voltage division rule is used to solve v o , Small signal voltage gain of circuit is calculated as, v o = r d r d + R s × v s r d = V T I D v o = V T I D V T I D + R s × v s v o = V T V T + I D R s × v s v o v s = V T V T + I D R s .......... ( 1 ) Putting V T = 0.025 mV, R s = 10 kΩ. v o v s = 0 .025 V 0 .025 V+10×10 3 Ω× I D ........ ( 2 ) It is given that DC diode current is limited to ±10% variations. Thus, the range of diode current is from 0.9 mA to 1.1 mA. Equation for diode current is, I D = I s e v D η v T Change in diode voltage is calculated from diode current equation, I D = I s e v D η v T η = 1 I D 2 = I s e v D 2 v T ..... ( 1 ) I D 1 = I s e v D 1 v T ...... ( 2 ) Divide equation ( 1 ) by ( 2 ) , I D 2 I D 1 = e v D 2 − v D 1 v T I D 2 I D 1 = e Δ v v T taking ln on both side, Δ v = v T ln ( I D 2 I D 1 ) I D 1 =1 mA, I D 2 is varying from 0.9 mA to 1.1 mA. Change in voltage for 0.9 mA, Δ v 0.9 = 0.025 ln 0.9 1 = − 2.634 mV Change in voltage for 1.1 mA, Δ v 1.1 = 0.025 ln 1.1 1 = 2.383 mV Diode is across the output terminal; hence range of output voltage change is, Δ v o ( range ) = − 2.634 mV to 2.383 mV The range of input signal is calculated by gain formula, Δ v s ( range ) = − 2.634 mV v o v s to 2.383 mV v o v s ........ ( 3 ) Consider the voltage gain 0.5. From equation (2), v o v s = 0 .025 V 0 .025 V+10×10 3 Ω× I D ........ ( 2 ) 0.5 = 0 .025 V 0 .025 V+10×10 3 Ω× I D 0 .5 × 0 .025 +0 .5 × 10×10 3 Ω× I D = 0.025 I D = 0.0125 0.5 × 10 4 I D = 0.0025 mA From equation (3), Substitute voltage gain is equals to 0.5, Δ v s ( range ) = − 2.634 mV v o v s to 2.383 mV v o v s Δ v s ( range ) = − 2.634 mV 0.5 to 2.383 mV 0.5 Δ v s ( range ) = − 5.268 mV to 4 .766 mV Consider the voltage gain 0.10. From equation (2), v o v s = 0 .025 V 0 .025 V+10×10 3 Ω× I D

Microelectronic Circuits (The Oxford Series in Electrical and Computer Engineering) 7th edition

7th Edition

ISBN: 9780199339136

Author: Adel S. Sedra, Kenneth C. Smith

Publisher: Oxford University Press

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BJT (Bipolar Junction Transistor)

A BJT is a semiconductor device with 3 terminals that consist of a p-n junction diode that can amplify the signal or current. It is a current-controlled device. The three BJT terminals are base, collector, and emitter. The BJT works with the charge carri…

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Chapter 4, Problem 4.49P

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The value of diode current for different values of voltage gain. Input voltage range for different values of voltage gain and corresponding output signal.

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Chapter 4, Problem 4.48P

Chapter 4, Problem 4.50P

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Microelectronic Circuits (The Oxford Series in Electrical and Computer Engineering) 7th edition

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The value of diode current for different values of voltage gain. Input voltage range for different values of voltage gain and corresponding output signal.

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The value of diode current for different values of voltage gain. Input voltage range for different values of voltage gain and corresponding output signal.

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