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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Chapter 10, Problem 10.13P
To determine
The value of the model parameters
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Consider the Common Emitter (CE) amplifier below. The capacitor Cn = 1 µF, Vcc
= 12 V, Br = 150, and the low cut-off frequency is 100 Hz. Assume VBE = 0.7 V.
VCE SAT for this BJT transistor is 0.2 V. Rc = 8002, and Re=120 2. The quiescent
collector current is 5 mA
Rg2
Rc
VIN H
Cin
R81
RE
CE
(1) Draw the small signal model for the CE amplifier
(11) What is the transconductance gain of the amplifier?
(II) What is the voltage gain of the amplifier – with and without a bypass
capacitor (Ce)?
(IV) Determine a suitable value for the bypass capacitor CE.
Vec
Re
Veut
Rin
Reut
R
R
Ven
The beta B of a bipolar transistor shown below varies from 10 to 80. The load resistance is Rc=10. The DC
supply is Vcc=50V and the input voltage to the base circuit is Vg-5V. If VCE(sat)=1V, VBE(sat) =1.2V and Rg=0.50,
then the total power loss in the transistor P, would be:
Rc
Ic
VcC
RB
VCE
IE
VB
VBE
Select one:
O a.
68.12W
Ob. 58.12W
Oc. 48.12W
O d. 38.12W
Chapter 10 Solutions
Microelectronic Circuits (The Oxford Series in Electrical and Computer Engineering) 7th edition
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- The waveforms of a BJT transistor used as a switch are shown below. The parameters are as follows: Vcc = 200V, VCE(sat) = 2V, Ics = 100A, VBe(sat) = 3V, lg = 10A and f, =10 kHz. The various times on the graph are: ta= 0.6ps, t, = 1.8us, ts = 3µs, t = 4µs, tn = 47.6µs and to = 43µs. Neglect ICEo (Iceo = 0 mA). The average power loss due to the base current is equal to: Ias VBE T- 1/f, Select one: O a. 35W O b. 45W O c. 25W O d. 15Warrow_forwardDetermine the magnitude of gm for a JFET with Ipss =12 mA and Vp = -7 Vat the following de bias points with the graphical determination a. VGs = -1.5 V. b. VGs = -2.5 V. c. VGs = -3.5 v. %3D %3D This only d. Find the maximum value of gm- e. Find the value of gm at each operating point of (a. , b. , and c.) with the mathematical determinationarrow_forwardAn enhancement-Type MOSFET is fabricated o 0.4μμm Process in with L=0.8μμm and w= 16 μm. find the value of Is for Vov = 0.5V and VAS = IV. Do not neglect CLM. If Yps is increased by 2-2V what Corresponding change in ID. is the Parameter VIHO не сох VA N-C4 0.7V 200MA/12 to Yumarrow_forward
- The ac equivalent circuit for an amplifier is shown . Assume the capacitors have infinite value, RI = 750 Ω, RB = 100 kΩ, RC = 62 kΩ, and R3 = 100 kΩ. Calculate the voltage gain and input resistance for the amplifier if the BJT Q-point is (40μA, 10 V). Assume βo = 100 and VA = 75V.arrow_forwardThe waveforms of a BJT transistor used as a switch are shown below. The parameters are as follows: Vcc = 200V, VCE(sat) = 2V, Ics = 100A, VBE(sat) = 3V, IB = 10A and fg =10 kHz. The various times on the graph are: ta = 0.6µs, t, = 1.8µs, ts = 3µs, tf = 4us, t, = 47.6µs and to = 43µs. Neglect ICEO (ICEO = 0 %3D %3D mA). The average power loss due to the collector current during the rise period t, is equal to: VCE Vcc VCE(sat) ic ton tff Ics ICEO 18 tn ts Select one: a. 51.2W b. 41.2W Ос. None of these d. 61.2W е. 31.2Warrow_forwardFor circuit shown in the figure with Voc=9 V, Rg=354 kn, Rc=409 0, RE=2.6 kn, B=117, VeE=0.7 V, determine S(lo). Vcc Rc RB Vc |VCE oVE emitter bias configuration Lütfen birini seçin: a.63.38 b.71.86 c.61.96 d.37.75 e.56.97arrow_forward
- Determine the magnitude of gm for a JFET with Ipss =12 mA and Vp = -7 Vat the following de bias points with the graphical determination a. VGs = -1.5 V. b. VGs = -2.5 V. c. VGS %3D -3.5 V. d. Find the maximum value of gm- e. Find the value of gm at each operating point of (a. , b. , and c.) with the mathematical determination I need This only f. Plot gm versus VGs for the JFET g. Plot gm versus In for the JFETarrow_forwardAssignment-1 Q1. The output characteristic of a typical transistor is shown below, where the quiescent point is selected on it. This transistor is used in the bias circuit presented below. Find the suitable values of Rg and Rc to fix the Q-point of the circuit properly. +Vcc = 12 V 12- Ia = 70 uA 10- Is 60 uA Ta=50 A 8- Rc 6. la= 40 uA 4. Is = 30 uA 2- Is = 20 uA B = 100 VBE = 0.7 V -2- 6 8 10 12 14 16 VCE (V) toarrow_forwardsupply is Vcc=40V and the input voltage to the base circuit is VB=6V. If VCE(sat)=1.2V, VBE(sat)=1.6V and Rg=0.70, then the forced B would be: The beta B of a bipolar transistor shown below varies from 12 to 75. The load resistance is Rc=1.20. The DC Ic Vcc Rg IB VCE VB V BE Select one: O a. 5.14 Ob. 2.14 O c. 4,14 O d. 3.14arrow_forward
- Calculate A, (dividing max output value by max input value.) Output voltage corresponding to the sinusoidal input voltage (VCC = 12 V, Vi = 2 VpR, coupling capacitor values are 10 uF, F = 1 kHz). oVcc c) RB RE (Emitter follower configuration) (RB = 220 k2 , RE = 3.3 k2)arrow_forwardThe ac equivalent circuit for an amplifier is shown . Assume the capacitors have infinite value, RI = 10 kΩ, RB = 5 MΩ, RC = 1.5 MΩ, and R3 = 3.3 MΩ. Calculate the input resistance and output resistance for the amplifier if the BJT Q-point is (2 μA, 2 V). Assume βo = 40 and VA = 50 V.arrow_forwardThe figure below shows a common-emitter amplifier. It is given that Rsig = 5.7 kQ, R81 R82 22 kQ, C = 5.2 pF, C = 1.1 pF, and Rc = RL = 11 kQ. The transistor is biased at lc = 0.6 mA and has B =100. Neglect the Early effect. %3D %D %3D %3D %3D (a) Find the mid-band gain Vo/Vsig- (b) Find the upper 3-dB frequency fH. Vee RB Re Ce R Insert the value of the midband gain in V/V below:arrow_forward
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