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

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

View Full Document Right Arrow Icon
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
Background image of page 1

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

View Full DocumentRight Arrow Icon
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
Background image of page 2
Background image of page 3
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

This document was uploaded on 03/08/2012.

Page1 / 3

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

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online