14 VCw = 5 pF Cre = 12 pFCw, = 8 pF Che = 40 pFC = 8 pF5.6 k268 ka0.47 µF%230.47 uF0,82 kQB = 1203.3 ka10 k21.2 k2# 20 uF0.47 uFFigure 4a. Source VsFigure 4b. Voltage Divider Bias Amplifier4. For the network of Figures 4a and 4b:a. Determine re.b. Find Avmid = V. > Vi, without the circuit of Fig. 4a.c. Calculate Z¡.d. Determine fis, fLc, and fLe.e. Determine the low cutoff frequency fi.Does Miller Capacitance exist? Why?g. Determine fHi and fHo.h. Find fø and fr.Determine the high cutoff frequency f2.j. What is the gain-bandwidth product of the amplifier?f.i.

14 V Cw = 5 pF C = 12 pF Cw = 8 pF C= 40 pF C = 8 pF 5.6 kQ 68 ka 0.47 uF 0.82 k2 B = 120 0.47 uF 3.3 kn 10 kn 1.2 k 20 uF 0.47 µF Figure 4a. Source V: Figure 4b. Voltage Divider Bias Amplifier For the network of Figures 4a and 4b: a. Determine re. b. Find Avmid = V. > Vi, without the circuit of Fig. 4a. c. Calculate Zi. d. Determine fis, fic, and fLE. e. Determine the low cutoff frequency f1. f. Does Miller Capacitance exist? Why? g. Determine fHi and fHo. h. Find fø and fr. i. Determine the high cutoff frequency f2. j. What is the gain-bandwidth product of the amplifier?

14 V Cw = 5 pF C = 12 pF Cw = 8 pF C= 40 pF C = 8 pF 5.6 kQ 68 ka 0.47 uF 0.82 k2 B = 120 0.47 uF 3.3 kn 10 kn 1.2 k 20 uF 0.47 µF Figure 4a. Source V: Figure 4b. Voltage Divider Bias Amplifier For the network of Figures 4a and 4b: a. Determine re. b. Find Avmid = V. > Vi, without the circuit of Fig. 4a. c. Calculate Zi. d. Determine fis, fic, and fLE. e. Determine the low cutoff frequency f1. f. Does Miller Capacitance exist? Why? g. Determine fHi and fHo. h. Find fø and fr. i. Determine the high cutoff frequency f2. j. What is the gain-bandwidth product of the amplifier?

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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In the circuit in the figure, VGSQ = 6.8 V, IDQ = 2.4 mA, VGS(Th) = 3.3 V, k = 0.4x10-3 A/V2, RD = 5.6 kΩ, RF = 2.2 MΩ and rd = 25 kΩ. Accordingly, when a RL = 23.6 kΩ load is connected to the output of the circuit, what will be the voltage gain of the circuit? NOTE: MOSFET output resistance must be taken into account in rd calculations.
Design a Single-Stage Common Emitter Class A Amplifier Specifications:Voltage Divider Bias Circuit Supply: Any value from 10Vdc to 24VdcLoad: 1kΩVoltage Gain: Any value from 80 to 400Lower Cutoff Frequency: 100 HzSinusoidal source (zero internal resistance): 50mVp-pTransistor: Si, β = 75 • Base-Collector capacitance = 8pF • Base-Emitter Capacitance = 25pF a) compute for the biasing resistances.b) determine the dc transistor terminal voltages and transistor currents.
In the circuit in the figure, IDSS = 12 mA, VP = -8 V, VGSQ = -2.4 V, IDQ = 4.12 mA, Rsig = 10 kΩ, RG = 100 kΩ, RD = 2.4 kΩ, RS = 470 Ω, RL = 33, It is known that 9 kΩ and rd = 20 kΩ. What will be the voltage gain of the circuit? NOTE: The FET output impedance is rd = 20 kΩ and must be taken into account in the calculations.
Q3:Choose One of the following: : (A)In common emmiter voltage divider bias of BJT, Vcc = 25V; R1= 10 k ohm; R2, = 2.2 k ohm ;Rc = 3.6 k ohm and RE = 1 k ohm.Determain Vec at ß = 120.using Exact analysis Draw the equvelent circuit inaddition the DC load line curve . (B)For self bais(Emitter resistace bais) of BJT,Vcc= 22 V, lb = 45macroA, a = 0.99, RE = 1 k ohm. Determain VCe Draw the equvelent circuit inaddition to DC load line curve.
Since Vcc = 12 V, Vin = 690 mV, RB = 780 kΩ, RC1 = 27.9 Ω, RC2 = 25.4 Ω, RE = 470 Ω, RL = 47 Ω and β1 = β2 = 100 in the circuit in the figure Find the current (IL) flowing through RL? NOTE-1: VBE1 = VBE2 = 0.7 V will be taken. NOTE-2: Output impedances of transistors (r0) will be neglected.
In the circuit in the figure IDSS = 12 mA, VP = -8 V, VGSQ = -2.4 V, IDQ = 4.12 mA, Rsig = 10 kΩ, RG = 100 kΩ, RD = 2 kΩ, RS = 470 Ω, RL = 23.7 kΩ and it is known that rd = 20 kΩ. What will be the voltage gain of the circuit? NOTE: The FET output impedance is rd = 20 kΩ and must be taken into account in the calculations.
1. Use the figure below to solve following questions (a) Solve the following dc quantitiesi. VB(Q1)ii. VE(Q1)iii. IE(Q1)iv. VC(Q1)v. VB(Q2)vi. VE(Q2)vii. IE(Q2)viii. VC(Q2)(b) Suppose that the emitter follower is omitted and the output from thecollector of Q1 is capacitively coupled to the 250Ω load, RL. What isthe output voltage across the 250Ω load?
In the circuit given in the figure, R₁ = 850kQ, R₂ = 350kQ, Rs = 4kQ, RL = 4kQ,VDD = 10V, VSGQ = 2.084V and transistor parameters VTP = -1.2V,kp = 40μA/V², W/L = 80 andIt is given as λ= 0.05V-¹. a) Find the gm and ro small sign parameters of the transistor.b) Draw the small sign equivalent of the circuit. c) Find the low signal voltage gain of the circuit AV = V0/Vi.
Since Vcc = 12 V, Vin = 721 mV, RB = 780 kohm, RC1 = 14.8 ohm, RC2 = 27.5 ohm, RE = 470 ohm, RL = 47 Q2 and B1 = B2= 100 in the circuit in the figure Find the current (1) flowing through RL? NOTE-1: VBE1 = VBE2 = 0.7 V will be taken. NOTE-2: The output impedances (ro) of the transistors will be neglected.
Considering the given transistor circuit and given values(β=60; V BE =0,5 V; VCC =15V; R E=1K ; R c=10K ; R1=60K ;R2=30K ).a. Find the base, collector, and emitter currents b. Find the voltage gain of this circuit c. Find the voltage gain by ignoring the capacitor connected to the emitterd. Briefly describe the transistor active zone and state its importance
VTN1 = 0,6 V, VTP2 = -0,6 V Kn1 = 0,2 mA/V2 Kp2 = 1 mA/V2 VDD = 5 V Rin = 400 kΩ a) Design the circuit for IDQ1 = 0,2 mA, IDQ2 = 0,5 mA, VDSQ1 = 2 V and VSDQ2 = 3 V. Voltage over RS1=0,6V b) Draw the small signal equivalent circuit of the circuit. c) Calculate the gain of your circuit
It is known that IDSS = 14 mA, VP = -8 V, VGSQ = -3.2 V, RG = 1 MΩ and RS = 0.32 kΩ in the circuit in the figure. Accordingly, what is the value of the output impedance of the circuit (Z0)? NOTE: The FET output resistance is rd = 20 kΩ and will be taken into account in the calculations.