{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw20sol

# hw20sol - Z in = h 11 Given V 1 V s = 25 26 V 1 V s = Z in...

This preview shows pages 1–3. Sign up to view the full content.

Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo ECE 202 L INEAR C IRCUIT A NALYSIS II (S P ’10) Homework #20 Solution (77–78) Problem 77–78 V S R s =80Ω h 11 h 22 R L =4Ω a : 1 + _ h 12 V 2 h 21 I 1 Hybrid Amplifier Model Z in + V L - + V 2 - + V 1 - Y out Z in2 I 1 I 2 Ampliﬁer Speciﬁcations: 1. When I 1 is zero, the ratio V 1 V 2 = 0, when a source is applied to port 2. 2. There must be maximum power transfer from the ampliﬁer output to the load under the condition that Z out ( s ) = 1600 Ω . 3. V 1 V s = 25 26 4. The voltage gain V 2 V 1 = - 100 The two port Hybrid parameters are: V 1 = h 11 I 1 + h 12 V 2 (77.1) I 2 = h 21 I 1 + h 22 V 2 (77.2) h 11 = V 1 I 1 ± ± ± ± V 2 = 0 [ Ω ] h 12 = V 1 V 2 ± ± ± ± I 1 = 0 h 21 = I 2 I 1 ± ± ± ± V 2 = 0 h 22 = I 2 V 2 ± ± ± ± I 1 = 0 [S] 1/ 3

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo (a) h 12 h 12 = V 1 V 2 ± ± ± ± I 1 = 0 = 0 From condition 1. (b) Y out , h 22 , and turn ratio a Given the condition that the value of Z out = 1600 Ω , Y out = 1 Z out = 1 1600 = [S] h 22 is deﬁned as I 2 V 2 ± ± ± ± I 1 = 0 . When I 1 is zero, the value of the dependent current source h 21 I 1 is zero and it becomes an open circuit. As a result, h 22 = Y out = 1 1600 = [S] From the condition 2, we want the impedances Z out and Z in 2 to match. Z out = Z in 2 = a 2 R L a = r Z out R L = r 1600 4 = 20 (c) h 11 Because h 12 = 0, the dependent voltage source becomes a short circuit and
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Z in = h 11 . Given V 1 V s = 25 26 , V 1 V s = Z in Z in + R s = 25 26 = h 11 h 11 + R s ⇒ h 11 = 2k Ω 2/ 3 Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo (d) Z in From part c) we found that Z in = h 11 = 2k Ω . (e) h 21 When port 2 is connected to the load Z in 2 , then the I 2 in (eq.77.2) becomes-V 2 Y in 2 . Then it simpliﬁes to: I 2 = h 21 I 1 + h 22 V 2-V 2 Y in 2 = h 21 I 1 + h 22 V 2 ⇒ h 21 =-V 2 ( h 22 + Y in 2 ) I 1 =-± V 1 I 1 ²± V 2 V 1 ² ( h 22 + Y in 2 ) =-( Z in )(-100 )( 1 1600 + 1 1600 ) = 250 (f) Ratio of Power delivered to the load to the power delivered to R L P load = ( V 2 ) 2 Z in 2 P R L = ( V L ) 2 R L = ( V 2 / a ) 2 R L ∴ P load P R L = ( V 2 ) 2 Z in 2 a 2 R L ( V 2 ) 2 = ( 400 )( 4 ) 1600 = 1 3/ 3...
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

hw20sol - Z in = h 11 Given V 1 V s = 25 26 V 1 V s = Z in...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online