hat is the proper equation for calculating the low frequency break point due to the input pacitor Cs (aka Ccl) for this BJT amplifier? (note: Ri, ß, re, ro are not shown) +Vcc R1 RC 2.2 ko 82 ka 0.2 μF Cs Rs 800 0.5 μ F RL 5.2 ko R2 RE CE 10μ 12.5 ka 1.1 ko 1 A) fLs= 1 B) fLs= 2 Rs || R¡ Cs 2rRs + R1 + R2]Cs 1 D) fLs= C) fLs= 2RS R1 | R2)CS 2(R$ + R}]Cs
hat is the proper equation for calculating the low frequency break point due to the input pacitor Cs (aka Ccl) for this BJT amplifier? (note: Ri, ß, re, ro are not shown) +Vcc R1 RC 2.2 ko 82 ka 0.2 μF Cs Rs 800 0.5 μ F RL 5.2 ko R2 RE CE 10μ 12.5 ka 1.1 ko 1 A) fLs= 1 B) fLs= 2 Rs || R¡ Cs 2rRs + R1 + R2]Cs 1 D) fLs= C) fLs= 2RS R1 | R2)CS 2(R$ + R}]Cs
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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Kindly choose the best and appropriate answer.
![What is the proper equation for calculating the low frequency break point due to the input
capacitor Cs (aka Ccl) for this BJT amplifier? (note: Ri, ß, ręe, ro are not shown)
+ Vcc
R1
RC
82 ka
2.2 ka
0.2 μF
Cs
Rs
800
0.5 μ F
RL
5.2 kQ
R2
RE
CE
10 µF
12.5 ka
1.1 ka
1
B) fLs-
A) fLs=
2R$ || Ri]CS
2 Rs + R1 + R2]Cs
1.
D) fLs= 2RS + RiJCS
1
C) fLs=
2 Rs*R1 || R2)Cs](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F52d95eba-420a-494e-94dc-4e08907a59f6%2Ff998bbbe-3cc1-4fa8-9fa1-ac363f690f29%2Fq5gl237_processed.png&w=3840&q=75)
Transcribed Image Text:What is the proper equation for calculating the low frequency break point due to the input
capacitor Cs (aka Ccl) for this BJT amplifier? (note: Ri, ß, ręe, ro are not shown)
+ Vcc
R1
RC
82 ka
2.2 ka
0.2 μF
Cs
Rs
800
0.5 μ F
RL
5.2 kQ
R2
RE
CE
10 µF
12.5 ka
1.1 ka
1
B) fLs-
A) fLs=
2R$ || Ri]CS
2 Rs + R1 + R2]Cs
1.
D) fLs= 2RS + RiJCS
1
C) fLs=
2 Rs*R1 || R2)Cs
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