What is the proper equation for calculating the low frequency break point due to the capacitor CC for this BJT amplifier. (note: Ro = ro // RC) 9 +Vcc R1 82 ka 2.2 ka 0.2 μF Cs Rs 800 0.5 μ F RL 5.2 ka R2 RE CE 10μ 12.5 ka 1.1 ka 1 B) fLC= RLJCC D) ÍLC= 27{RC RLJCC 1 A) fLC= 27 Ro+ 27(ro + RL)CC 1 C) fLC=2TRCCC 2RC | RL)C
What is the proper equation for calculating the low frequency break point due to the capacitor CC for this BJT amplifier. (note: Ro = ro // RC) 9 +Vcc R1 82 ka 2.2 ka 0.2 μF Cs Rs 800 0.5 μ F RL 5.2 ka R2 RE CE 10μ 12.5 ka 1.1 ka 1 B) fLC= RLJCC D) ÍLC= 27{RC RLJCC 1 A) fLC= 27 Ro+ 27(ro + RL)CC 1 C) fLC=2TRCCC 2RC | RL)C
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 capacitor CC
for this BJT amplifier. (note: Ro = ro // RC)
9 +Vcc
R1
82 ka
2.2 ka
0.2 μF
Cs
Rs
800
0.5 μ F
RL
5.2 ka
R2
RE
CE
10μ
12.5 ka
1.1 ka
1
B) fLC= RLJCC
D) ÍLC= 27{RC RLJCC
1
A) fLC=
27 Ro+
27(ro + RL)CC
1
C) fLC=2TRCCC
2RC | RL)C](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F52d95eba-420a-494e-94dc-4e08907a59f6%2F497a5c54-732d-45f3-8189-b6f32760aef7%2Fu6fnwhq_processed.png&w=3840&q=75)
Transcribed Image Text:What is the proper equation for calculating the low frequency break point due to the capacitor CC
for this BJT amplifier. (note: Ro = ro // RC)
9 +Vcc
R1
82 ka
2.2 ka
0.2 μF
Cs
Rs
800
0.5 μ F
RL
5.2 ka
R2
RE
CE
10μ
12.5 ka
1.1 ka
1
B) fLC= RLJCC
D) ÍLC= 27{RC RLJCC
1
A) fLC=
27 Ro+
27(ro + RL)CC
1
C) fLC=2TRCCC
2RC | RL)C
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