EXAMPLE 5-10 Calculate the Q-point values (Ic and VCE) for the circuit in Figure 5-23. FIGURE 5-23 Vcc +10 V RC 10 KΩ RB 180 kl Boc = 100 0.7 V Solution Using Equation 5-13, the collector current is Vcc - VBE Ic RC+ RB/ βpο 10 kΩ+ 180 kΩ/100 10 V – 0.7 V - 788 μΑ Using Equation 5-14, the collector-to-emitter voltage is VCE = Vcc - IcRc = 10 V - (788 µA)(10 kN) = 2.12 V Related Problem = 200 and determine the percent Calculate the Q-point values in Figure 5-23 for BDC change in the Q-point from Bpc 100 to BDC 200. %3D
EXAMPLE 5-10 Calculate the Q-point values (Ic and VCE) for the circuit in Figure 5-23. FIGURE 5-23 Vcc +10 V RC 10 KΩ RB 180 kl Boc = 100 0.7 V Solution Using Equation 5-13, the collector current is Vcc - VBE Ic RC+ RB/ βpο 10 kΩ+ 180 kΩ/100 10 V – 0.7 V - 788 μΑ Using Equation 5-14, the collector-to-emitter voltage is VCE = Vcc - IcRc = 10 V - (788 µA)(10 kN) = 2.12 V Related Problem = 200 and determine the percent Calculate the Q-point values in Figure 5-23 for BDC change in the Q-point from Bpc 100 to BDC 200. %3D
Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Question
Solve the related problem.
![EXAMPLE 5-10
Calculate the Q-point values (Ic and VCE) for the circuit in Figure 5–23.
FIGURE 5-23
Vcc
+10 V
Re
10 ΚΩ
RB
180 kO
BDc = 100
0.7 V
Solution Using Equation 5–13, the collector current is
10 V – 0.7 V
Vcc - VBE
Ic =
Rc + RB/BpC
=788 μΑ
10 kΩ + 180 kΩ/100
Using Equation 5–14, the collector-to-emitter voltage is
VCE = Vcc - IcRc = 10 V – (788 µA)(10 kN) = 2.12 V
Related Problem
Calculate the Q-point values in Figure 5-23 for Bpc = 200 and determine the percent
change in the Q-point from Bpc = 100 to BDc = 200.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8035d375-a0fa-47ff-969e-f45bd1a9163d%2F919a517e-5e21-4a07-a9d1-ba2535a33caf%2Ftoxbsn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:EXAMPLE 5-10
Calculate the Q-point values (Ic and VCE) for the circuit in Figure 5–23.
FIGURE 5-23
Vcc
+10 V
Re
10 ΚΩ
RB
180 kO
BDc = 100
0.7 V
Solution Using Equation 5–13, the collector current is
10 V – 0.7 V
Vcc - VBE
Ic =
Rc + RB/BpC
=788 μΑ
10 kΩ + 180 kΩ/100
Using Equation 5–14, the collector-to-emitter voltage is
VCE = Vcc - IcRc = 10 V – (788 µA)(10 kN) = 2.12 V
Related Problem
Calculate the Q-point values in Figure 5-23 for Bpc = 200 and determine the percent
change in the Q-point from Bpc = 100 to BDc = 200.
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