The cascaded amplifier of figure below uses a common collector first stage followed by a common emitter second stage. For transistor Q₁, hoc = 0, hic = 1k, hrc = 1, and hfc = -100. For Q2, hre = hoe = 0, hfe = 100, and hie 1kQ. Find the following; = Rs = 0 it Cc R1 90kΩ R2 $100k R3 H 9kQ +15V R4 R5 ~90kΩ R7 110ΚΩ R6 5ΚΩ it Cc 9kCB 5ΚΩ
The cascaded amplifier of figure below uses a common collector first stage followed by a common emitter second stage. For transistor Q₁, hoc = 0, hic = 1k, hrc = 1, and hfc = -100. For Q2, hre = hoe = 0, hfe = 100, and hie 1kQ. Find the following; = Rs = 0 it Cc R1 90kΩ R2 $100k R3 H 9kQ +15V R4 R5 ~90kΩ R7 110ΚΩ R6 5ΚΩ it Cc 9kCB 5ΚΩ
College Physics
11th Edition
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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Transcribed Image Text:The cascaded amplifier of figure below uses a common collector first stage followed by a common emitter
second stage. For transistor Q₁, hoc = 0, hic = 1k, hrc = 1, and hfc = -100. For Q2, hre = hoe = 0, hfe =
100, and hie 1kQ. Find the following;
Rs = 0
it
Cc
R1 {90kΩ
R2 $100k
R3
a. The overall voltage gain ratio.
b. The overall current gain ratio.
c. H - parameter equivalent circuit.
it
9kQ
+15V
R4 90k
R5
R7
[10ΚΩ
R6
5ΚΩ
it
Cc
9kCB
5ΚΩ
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Follow-up Question
Wheres the 9?
![AI₁ = 101
Iin
AI₁ = ID
Iin
=
=
A1² IOL
Iin
(1+hfe) R3 (RillR₂)
(R₂+RL) [(₂R₁) + hie+ (1+hR) (R₂R₁)
(+99) x9 (90/1100)
(9+0.9) [90/100)+ 1+ (1+99) 910-9]
= 3.675
100X 47.368
9.9 [47-368 +1+ 100X-8181]](https://content.bartleby.com/qna-images/question/2ddc7a09-9ee5-460c-abf4-dd58e05c623c/0582793d-c9e3-421e-a3fa-e6ac24ad6b54/83c4cxm_processed.png)
Transcribed Image Text:AI₁ = 101
Iin
AI₁ = ID
Iin
=
=
A1² IOL
Iin
(1+hfe) R3 (RillR₂)
(R₂+RL) [(₂R₁) + hie+ (1+hR) (R₂R₁)
(+99) x9 (90/1100)
(9+0.9) [90/100)+ 1+ (1+99) 910-9]
= 3.675
100X 47.368
9.9 [47-368 +1+ 100X-8181]
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