7. For the circuit shown in Fig, solve for VB, VE, IC, Vc, and VCE. Also, calculate Ic(sat) and VCE(off). Finally, construct a DC load line showing the values of Ic(sat), VCE(off), ICQ, and VCEQ. +Vcc = 18 V R=1.5 k R Boc=200 R₂ = 390 0 R₁ = 33 k R₂ = 5.6 km
7. For the circuit shown in Fig, solve for VB, VE, IC, Vc, and VCE. Also, calculate Ic(sat) and VCE(off). Finally, construct a DC load line showing the values of Ic(sat), VCE(off), ICQ, and VCEQ. +Vcc = 18 V R=1.5 k R Boc=200 R₂ = 390 0 R₁ = 33 k R₂ = 5.6 km
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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![### Transistor Circuit Analysis and DC Load Line Construction
#### Problem Statement:
For the circuit shown in Fig. 1, solve for the following parameters:
- \( V_B \) (Base Voltage)
- \( V_E \) (Emitter Voltage)
- \( I_C \) (Collector Current)
- \( V_C \) (Collector Voltage)
- \( V_{CE} \) (Collector-Emitter Voltage)
Additionally, calculate the saturation current (\( I_{C(sat)} \)) and the cut-off voltage (\( V_{CE(off)} \)). Finally, construct a DC load line showing the values of \( I_{C(sat)} \), \( V_{CE(off)} \), \( I_{CQ} \), and \( V_{CEQ} \).
#### Given Circuit:
The circuit diagram provided includes the following components and parameters:
- \( V_{CC} = 18 \) V
- \( R_1 = 33 \) kΩ
- \( R_2 = 5.6 \) kΩ
- \( R_C = 1.5 \) kΩ
- \( R_E = 390 \) Ω
- \( \beta = 200 \)
#### Circuit Diagram:
The circuit is a common-emitter configuration with the following features:
- \( R_1 \) and \( R_2 \) form a voltage divider connected between \( V_{CC} \) and ground.
- The voltage divider's midpoint connects to the base of the NPN transistor.
- The emitter resistor \( R_E \) is connected from the emitter to ground.
- The collector resistor \( R_C \) is connected from the collector to \( V_{CC} \).
The schematic is shown as follows:
\[
\begin{circuitikz}
\draw
(0,0) node[ground]{} to[R=R_E, l_=5.6 kΩ,] (0,3)
(0,3) to[R=R_2, l_=390 Ω, ] (0,6)
(0,6) node[npn,anchor=C] (Q1){}
(Q1) node[below left] {\(\beta = 200\)}
(Q1.B) to[R=R_1, l_=33 kΩ,] (-3,0)
(4,0) node[ground]{}
(Q1.E) to[short, -o](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F22d8487a-aee7-41d5-95a0-1e0950421a9b%2F60568ea8-2600-4163-b5e8-e8f98002dd4d%2Fw0prs3i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Transistor Circuit Analysis and DC Load Line Construction
#### Problem Statement:
For the circuit shown in Fig. 1, solve for the following parameters:
- \( V_B \) (Base Voltage)
- \( V_E \) (Emitter Voltage)
- \( I_C \) (Collector Current)
- \( V_C \) (Collector Voltage)
- \( V_{CE} \) (Collector-Emitter Voltage)
Additionally, calculate the saturation current (\( I_{C(sat)} \)) and the cut-off voltage (\( V_{CE(off)} \)). Finally, construct a DC load line showing the values of \( I_{C(sat)} \), \( V_{CE(off)} \), \( I_{CQ} \), and \( V_{CEQ} \).
#### Given Circuit:
The circuit diagram provided includes the following components and parameters:
- \( V_{CC} = 18 \) V
- \( R_1 = 33 \) kΩ
- \( R_2 = 5.6 \) kΩ
- \( R_C = 1.5 \) kΩ
- \( R_E = 390 \) Ω
- \( \beta = 200 \)
#### Circuit Diagram:
The circuit is a common-emitter configuration with the following features:
- \( R_1 \) and \( R_2 \) form a voltage divider connected between \( V_{CC} \) and ground.
- The voltage divider's midpoint connects to the base of the NPN transistor.
- The emitter resistor \( R_E \) is connected from the emitter to ground.
- The collector resistor \( R_C \) is connected from the collector to \( V_{CC} \).
The schematic is shown as follows:
\[
\begin{circuitikz}
\draw
(0,0) node[ground]{} to[R=R_E, l_=5.6 kΩ,] (0,3)
(0,3) to[R=R_2, l_=390 Ω, ] (0,6)
(0,6) node[npn,anchor=C] (Q1){}
(Q1) node[below left] {\(\beta = 200\)}
(Q1.B) to[R=R_1, l_=33 kΩ,] (-3,0)
(4,0) node[ground]{}
(Q1.E) to[short, -o
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