Emitter-bias Beta 100 150 200 250 300 Ra VBB HI Vcc 25 20 15 10 5 -VBE www KVL loop |E = |C Rc 0.7 V RE = VBB Vcc Rc 1k 1.2k 900 750 500 VBB-1BRB-VBE - LERE = 0 le = (B + 1)la = Bla VBB (IE/B)RB-VBE - IERE = 0 VBB-VBE = IE ((RB/B) + RE) lε = VBв - VBE RB/B + RE Ra/B + RE= = R₂ = ß Vss - VBE lE RB 1k Ves - VBE lE 500 300 450 250 -RE RE 250 500 300 450 250 (IE emitter-bias) (RB emitter-bias) IE
Emitter-bias Beta 100 150 200 250 300 Ra VBB HI Vcc 25 20 15 10 5 -VBE www KVL loop |E = |C Rc 0.7 V RE = VBB Vcc Rc 1k 1.2k 900 750 500 VBB-1BRB-VBE - LERE = 0 le = (B + 1)la = Bla VBB (IE/B)RB-VBE - IERE = 0 VBB-VBE = IE ((RB/B) + RE) lε = VBв - VBE RB/B + RE Ra/B + RE= = R₂ = ß Vss - VBE lE RB 1k Ves - VBE lE 500 300 450 250 -RE RE 250 500 300 450 250 (IE emitter-bias) (RB emitter-bias) IE
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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Related questions
Question
Solve for IE
![Emitter-bias
Beta
100
150
200
250
300
Ro
VBB
H
KVL loop
le = Ic
Rc
-VBE =
0.7 V
RE
Vcc = VBB
25
20
15
10
5
Example Calculations:
Vcc
+₁₁
B = 100
¹E =
Rc
1k
1.2k
900
750
500
B = 300
VBB IBRB-VBEIERE = 0
le = (B + 1)la Bla
VBB (IE/B)RB
VBEIERE = 0
VBB - VBE = IE ((RB/B) + RE)
VOU - VOE
R₂/B + RE
le =
VBв - VBE
RB/B + RE
Veo-Vec
R₂/B + RE
RB/B + RE =
R₂ = ß
3= 100 le = lc 1mA Vcc=V88 = 10V
Ria = P[ Ven : Ver. Fle] = 100 [-
RB
lE
VBB - VBE
lE
RB
1k
VBB - VBE
lE
500
300
450
250
We calculate a value for RC and choose a close standard value. An emitter resistor which is 10-
50% of the collector load resistor usually works well.
10 -0.7
0.001
An 883k resistor was calculated for RB, an 870k chosen. At B-100, IE is 1.01mA.
Rg = 870k
- RE
10-0.7
870k/100 + 470
RE
250
500
300
450
250
10 -0.7
870k/300 + 470
Re = 4700
470 = 883k
470] -
(IE emitter-bias)
= 1.01mA
(RB emitter-bias)
= 2.76mA
IE](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F355817fc-4afc-456c-bc56-6563b808926d%2F997cbf94-178c-4abc-89dd-826279372945%2Fig26my_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Emitter-bias
Beta
100
150
200
250
300
Ro
VBB
H
KVL loop
le = Ic
Rc
-VBE =
0.7 V
RE
Vcc = VBB
25
20
15
10
5
Example Calculations:
Vcc
+₁₁
B = 100
¹E =
Rc
1k
1.2k
900
750
500
B = 300
VBB IBRB-VBEIERE = 0
le = (B + 1)la Bla
VBB (IE/B)RB
VBEIERE = 0
VBB - VBE = IE ((RB/B) + RE)
VOU - VOE
R₂/B + RE
le =
VBв - VBE
RB/B + RE
Veo-Vec
R₂/B + RE
RB/B + RE =
R₂ = ß
3= 100 le = lc 1mA Vcc=V88 = 10V
Ria = P[ Ven : Ver. Fle] = 100 [-
RB
lE
VBB - VBE
lE
RB
1k
VBB - VBE
lE
500
300
450
250
We calculate a value for RC and choose a close standard value. An emitter resistor which is 10-
50% of the collector load resistor usually works well.
10 -0.7
0.001
An 883k resistor was calculated for RB, an 870k chosen. At B-100, IE is 1.01mA.
Rg = 870k
- RE
10-0.7
870k/100 + 470
RE
250
500
300
450
250
10 -0.7
870k/300 + 470
Re = 4700
470 = 883k
470] -
(IE emitter-bias)
= 1.01mA
(RB emitter-bias)
= 2.76mA
IE
Expert Solution
Step 1
Given-
Here, represents voltage across base and emitter.
The given data represents the experimental analysis of the circuit where various values of the the circuit element is given as below-
| 100 | 25 | 1k | 1k | 250 |
| 150 | 20 | 1.2k | 500 | 500 |
| 200 | 15 | 900 | 300 | 300 |
| 250 | 10 | 750 | 450 | 450 |
| 300 | 5 | 500 | 250 | 250 |
For the above given values, we have to find .
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