RI Rc Vcc R2 RE Ry 0100555-9870479447 VE g160100555 - 9870479447 The values of some resistors and source in the circuit are given be0100555-987047944 For the transistor used in the circu "Ok vcC = 20V g160100555- 987047944 Rg = 1k R1 = 100k RC = 9k RE = 1kI 0 100555- 987047944 d8704/6 According to what is hfe = hFE = s give, g160100555 - 987047944 •A. | Find IC current b. Find the voalue of the resistance = hoe = 0, VBE = 91601005 87047944 sea7047944 9447 Find the input resistance (ri). g160100550.6V, VCE = 10V andy g16010 T = 25mV. C. 10056 D. • to. What will happen if • f. What will happen to Kv if Find the voltage gain (Kv). p287047944 R2. As0100555-9870479447 CE is removed? CE is removed? g160100555 - 987047944 g160100555- 987047944 141
RI Rc Vcc R2 RE Ry 0100555-9870479447 VE g160100555 - 9870479447 The values of some resistors and source in the circuit are given be0100555-987047944 For the transistor used in the circu "Ok vcC = 20V g160100555- 987047944 Rg = 1k R1 = 100k RC = 9k RE = 1kI 0 100555- 987047944 d8704/6 According to what is hfe = hFE = s give, g160100555 - 987047944 •A. | Find IC current b. Find the voalue of the resistance = hoe = 0, VBE = 91601005 87047944 sea7047944 9447 Find the input resistance (ri). g160100550.6V, VCE = 10V andy g16010 T = 25mV. C. 10056 D. • to. What will happen if • f. What will happen to Kv if Find the voltage gain (Kv). p287047944 R2. As0100555-9870479447 CE is removed? CE is removed? g160100555 - 987047944 g160100555- 987047944 141
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...
Related questions
Question
![RI
Rc
Vcc
g1601
R
R2.
RE
Ry
0100555-9870479447
VE
CE
g1601
The values of some resistors and source in the circuit are given below.
g160100555 - 9870479447
16U
g160100555- 987047944
Rg = 1k RI = 100k RC = 9k RE =
g160100537047
According to what is given,
For the transistor used in the circuit, hfe = hFE =
160100555-987047944/
= 10k VCC = 20V
a0100555-987047944/
g160100555 - 987047944
a.
| Find IC current
• b. Find the value of the resistance
= hoe = 0, VBE =
g160100550687047944/
VCE = 10V and
g1601005 87047944
g16010055
D.
• to. What will happen if
f. What will happen to Kv if
Find the input resistance (ri).
Find the voltage gain (Kv).
R2.
CE is removed?
0100555-9870479447
CE is removed?
g160100555 - 987047944/
141
g160100555- 987047944
141
704/
7944](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F62f2bf21-5373-405f-8044-be0e8cff1a86%2F05754db3-24e4-42d4-9990-ace4255f6879%2Fzz6n2rn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:RI
Rc
Vcc
g1601
R
R2.
RE
Ry
0100555-9870479447
VE
CE
g1601
The values of some resistors and source in the circuit are given below.
g160100555 - 9870479447
16U
g160100555- 987047944
Rg = 1k RI = 100k RC = 9k RE =
g160100537047
According to what is given,
For the transistor used in the circuit, hfe = hFE =
160100555-987047944/
= 10k VCC = 20V
a0100555-987047944/
g160100555 - 987047944
a.
| Find IC current
• b. Find the value of the resistance
= hoe = 0, VBE =
g160100550687047944/
VCE = 10V and
g1601005 87047944
g16010055
D.
• to. What will happen if
f. What will happen to Kv if
Find the input resistance (ri).
Find the voltage gain (Kv).
R2.
CE is removed?
0100555-9870479447
CE is removed?
g160100555 - 987047944/
141
g160100555- 987047944
141
704/
7944
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