RB VBE 0.7 V Rc Vc Vcc lc = Bla lc = le VcclcRc- laRa - VBE = 0 Vccle Rc (le/B) Ra - VBE = 0 - Vcc - VBE = le Rc + (le/B)Ra Vcc VBE = le((Ra/B) + Rc) le= Vcc - Vee RB/B + Rc R₁ = ß [V Vcc - VBE le le = Bla . Rc (KVL) (IE CFB-bias) (RB CFB-bias)
RB VBE 0.7 V Rc Vc Vcc lc = Bla lc = le VcclcRc- laRa - VBE = 0 Vccle Rc (le/B) Ra - VBE = 0 - Vcc - VBE = le Rc + (le/B)Ra Vcc VBE = le((Ra/B) + Rc) le= Vcc - Vee RB/B + Rc R₁ = ß [V Vcc - VBE le le = Bla . Rc (KVL) (IE CFB-bias) (RB CFB-bias)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Question
Solve IB and IE
![U▬▬▬▬▬
Beta
100
150
200
250
300
RB
[mm]+">
VBE =
0.7 V
Rc
Vc
Collector-feedback bias.
Vcc = VBB
25
20
15
10
Vcc
RB = B
Rc
1k
1.2k
900
750
500
E =
B = 100
c =
lc = Bla
lc = le
Vcc - IcRc- laRB - VBE = 0
Vcc le Rc (le/B)RB - VBE = 0
Vcc VBE = le Rc + (le/B) Ra
Vcc
VBE IE((RB/3) + Rc).
| =
Vcc = 10V
B = 300
R₂ = ß
Vcc - Vee
Rc] = 100 [-
le
Vc = Vcc -lc Rc = 10-(1mA)-(4.7k) = 5.3V
Vcc - VBE
RB/B + Rc
Vcc - Vac
RA/B + R
Vcc= 10V
Vcc - Ver
lE
5
Example Calculations:
Find the required collector feedback bias resistor for an emitter current of 1 mA, a 4.7K collector
load resistor, and a transistor with B=100. Find the collector voltage VC. It should be
approximately midway between VCC and ground.
B = 100
Vcc - Vee
Rn/B+ Rc
RB
=
500
300
450
250
lc = le = 1MA
10 -0.7
1mA
le = Bla
The closest standard value to the 460k collector feedback bias resistor is 470k. Find the
emitter current IE with the 470K resistor. Recalculate the emitter current for a transistor with
B=100 and B=300.
Rc=4.7k
10 -0.7
470k/100 +4.7k
- Rc
10 -0.7
470k/300 +4.7k
B
(KVL)
Rg = 470k
(IE CFB-bias)
Rc = 4.7k
4.7k = 460k
7K] =
= 0.989mA
(RB CFB-bias)
= 1.48mA
lE
We see that as beta changes from 100 to 300, the emitter current increases from 0.989mA to
1.48mA. This is an improvement over the previous base-bias circuit which had an increase from
1.02mA to 3.07mA. Collector feedback bias is twice as stable as base-bias with respect to beta
variation.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F355817fc-4afc-456c-bc56-6563b808926d%2F6141eddd-12b8-47f0-a3aa-512a2adbd057%2Fr6381b7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:U▬▬▬▬▬
Beta
100
150
200
250
300
RB
[mm]+">
VBE =
0.7 V
Rc
Vc
Collector-feedback bias.
Vcc = VBB
25
20
15
10
Vcc
RB = B
Rc
1k
1.2k
900
750
500
E =
B = 100
c =
lc = Bla
lc = le
Vcc - IcRc- laRB - VBE = 0
Vcc le Rc (le/B)RB - VBE = 0
Vcc VBE = le Rc + (le/B) Ra
Vcc
VBE IE((RB/3) + Rc).
| =
Vcc = 10V
B = 300
R₂ = ß
Vcc - Vee
Rc] = 100 [-
le
Vc = Vcc -lc Rc = 10-(1mA)-(4.7k) = 5.3V
Vcc - VBE
RB/B + Rc
Vcc - Vac
RA/B + R
Vcc= 10V
Vcc - Ver
lE
5
Example Calculations:
Find the required collector feedback bias resistor for an emitter current of 1 mA, a 4.7K collector
load resistor, and a transistor with B=100. Find the collector voltage VC. It should be
approximately midway between VCC and ground.
B = 100
Vcc - Vee
Rn/B+ Rc
RB
=
500
300
450
250
lc = le = 1MA
10 -0.7
1mA
le = Bla
The closest standard value to the 460k collector feedback bias resistor is 470k. Find the
emitter current IE with the 470K resistor. Recalculate the emitter current for a transistor with
B=100 and B=300.
Rc=4.7k
10 -0.7
470k/100 +4.7k
- Rc
10 -0.7
470k/300 +4.7k
B
(KVL)
Rg = 470k
(IE CFB-bias)
Rc = 4.7k
4.7k = 460k
7K] =
= 0.989mA
(RB CFB-bias)
= 1.48mA
lE
We see that as beta changes from 100 to 300, the emitter current increases from 0.989mA to
1.48mA. This is an improvement over the previous base-bias circuit which had an increase from
1.02mA to 3.07mA. Collector feedback bias is twice as stable as base-bias with respect to beta
variation.
Expert Solution
Step 1
Given-
For a BJT circuit values are given as-
| Beta | VCC | RC | RB |
| 100 | 25 | 1k | 1k |
| 150 | 20 | 1.2k | 500 |
| 200 | 15 | 900 | 300 |
| 250 | 10 | 750 | 450 |
| 300 | 5 | 500 | 250 |
Base emitter voltage = 0.7 V
Emitter current, and
To find-
Base current and emitter current =??
Step by step
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