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1. Determine the small-signal gain (vout/Vin) of the following circuits. Assuming all the transistors are biased in the active mode (BJT) or saturation region (MOS) (VT = 26 mV, VBE,on = 0.8 V). (a) bias= 5 mA (+ Vin VDD Vin VG HnCox 200 μA/V², W/L= 50, λ=0.05 V-¹ (c) Ţ M₂ VDD bias= 2 mA ( M₁ Vout M₂ (W/L)₁= (W/L)₂= 200, Vov_M1 Vov_M2= 0.2 V, R₁= 1 kQ, and = 0. Vout (b) Vin bias= 1 mA ( (d) Vin Vcc ∞ Q₁ 00 R₁50 2, 3= 100, and VA= 00 81 Vcc Q₁ RL Q₂ Ic= 0.5 mA Vout VB2 PNP: VA= 30 V, NPN: VA= 60 V

1. Determine the small-signal gain (vout/Vin) of the following circuits. Assuming all the transistors are biased in the active mode (BJT) or saturation region (MOS) (VT = 26 mV, VBE,on = 0.8 V). (a) bias= 5 mA (+ Vin VDD Vin VG HnCox 200 μA/V², W/L= 50, λ=0.05 V-¹ (c) Ţ M₂ VDD bias= 2 mA ( M₁ Vout M₂ (W/L)₁= (W/L)₂= 200, Vov_M1 Vov_M2= 0.2 V, R₁= 1 kQ, and = 0. Vout (b) Vin bias= 1 mA ( (d) Vin Vcc ∞ Q₁ 00 R₁50 2, 3= 100, and VA= 00 81 Vcc Q₁ RL Q₂ Ic= 0.5 mA Vout VB2 PNP: VA= 30 V, NPN: VA= 60 V

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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1. Determine the small-signal gain (vout/Vin) of the following circuits. Assuming all the transistors are biased in the active mode (BJT) or saturation region (MOS) (Vr = 26 mV, VBE,on = 0.8 V). (a) Ibias= 5 mA (+ V in VDD Vin HnCox 200 μLA/V², W/L= 50, -0.05 V-¹ (c) Ţ VG o M₁ VDD M₁ M₂ Vout bias= 2 mA (+ R₁ (W/L)₁= (W/L)₂= 200, Vov_M1 Vov_M2= 0.2 V, R₁= 1 k2, and = 0. Vout (b) bias= 1 mA ( Vcc Vin 00 R₁ 50 , B= 100, and VA= 00 (d) Q₁ VB1 M Vcc Q₁ RL Q₂ /c= 0.5 mA Vout VB2 PNP: VA= 30 V, NPN: VA= 60 V
Transcribed Image Text:1. Determine the small-signal gain (vout/Vin) of the following circuits. Assuming all the transistors are biased in the active mode (BJT) or saturation region (MOS) (Vr = 26 mV, VBE,on = 0.8 V). (a) Ibias= 5 mA (+ V in VDD Vin HnCox 200 μLA/V², W/L= 50, -0.05 V-¹ (c) Ţ VG o M₁ VDD M₁ M₂ Vout bias= 2 mA (+ R₁ (W/L)₁= (W/L)₂= 200, Vov_M1 Vov_M2= 0.2 V, R₁= 1 k2, and = 0. Vout (b) bias= 1 mA ( Vcc Vin 00 R₁ 50 , B= 100, and VA= 00 (d) Q₁ VB1 M Vcc Q₁ RL Q₂ /c= 0.5 mA Vout VB2 PNP: VA= 30 V, NPN: VA= 60 V
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