《电路》（英文版）13-3 Impedance parameters

8 13-3 Impedance parameters 2 =11+a12 2 21 1+ z212 22 z11z12、Z21、z2 are z parameters The matrix from of z parameter equations: 11 12 21211 Zu Z1=ly

§13-3 Impedance parameters 2 12 1 11 1 • • • V = z I + z I 2 22 1 21 2 • • • V = z I + z I z11、z12、z21、z22 are z parameters The matrix from of z parameter equations:               =         • • • • 2 1 21 22 11 12 2 1 I I z z z z V V         =       • • V z I         = 21 22 11 12 z z z z z 1 [ ] − = y 2 • 1 I • I − + • V 2 − + • V 1

V1=11+x122 2=211+222 2 Let 2=0 Zu= the open-circuit input impedance 2=0 0 the open-circuit transfer impedance Let 1=0 u2 the open-circuit transfer impedance I1=0 2 22 the open-circuit output impedance 2 z parameters---open-cIrcuit impedance parameters When two-port is bilateral, we have z12=z21 The two-port is balanced, z1=z22

2 = 0 • Let I 2 0 1 1 11 = = • • • I I V z 2 0 1 2 21 = = • • • I I V z the open-circuit input impedance the open-circuit transfer impedance 1 = 0 • Let I 1 0 2 1 12 = = • • • I I V z 1 0 2 2 22 = = • • • I I V z the open-circuit transfer impedance the open-circuit output impedance z parameters--- open-circuit impedance parameters When two –port is bilateral, we have z12=z21. The two-port is balanced, z11=z22. 2 • 1 I • I − + • V 2 − + • V 1 2 12 1 11 1 • • • V = z I + z I 2 22 1 21 2 • • • V = z I + z I

Thevenin equivalent of the two-port 2 m1+2g Vs VI Thevenin equivalent 11/1+ 12 2=z211+2212 V1=。-Is 11+x1212 s=(zg+a1)1+z121 s 12 VS-x1212+ (z1+z) 21 2212 VS +(z22 11+ Zu+z 11 1+ Vs(12=0) 21412 二422 21+xg Vs=0 z1+乙

Thevenin equivalent of the two-port 2 12 1 11 1 • • • V = z I + z I 2 22 1 21 2 • • • V = z I + z I g V1 V s I 1 z • • • = − g V s I 1 z • • − 2 12 1 11 • • = z I + z I 2 12 1 11 ) • • = + z I + z I • g → V s (z ( ) 11 2 12 1 g z z V s z I I + − = • • • 2 • →V ( 2 0) 11 21 = + = • • • V s I z z z V g oc g t h z z z z z V s I V Z + = − = = • • • 11 21 12 22 2 2 0 2 22 11 2 12 21 2 • • • • + + − = z I z z V s z I V z g 2 11 21 12 22 ( ) • + + − I z z z z z g V s z z z g • + = 11 21 − + • V 2 I 2 • I 1 • − + • V 1 V S • − + Zg − + V s Z Zg Z 11 21 • + Thevenin equivalent Z Zg Z Z Z 11 21 12 22 + −

Example: Find V1, V2 and I1, I2, if zu=lk, z2=10, z2=-1000k 2=10k Solution:v;1=10001+10/2 500 V2=-10°I1+10 lok s=1+500I1 104I2 Js-500I1=10001+1012 104I2=-10°I1+10I2 We have I 2000÷ s andi=0.75Vs,V2=-250Vs 丿s The voltage The current 333 =50 gain gain 1500 12321=10000 10(-1000000 22 16.7k Z1+Z 1000+500

Example: Find z11=1k, z12=10, z21= -1000k, z22=10k. V1 ,V 2 and I 1, I 2 ,if • • • • 1 1000 1 10 2 • • • V = I + I 2 4 1 6 2 10 10 • • • V = − I + I 2 4 2 10 • • V = − I 2 4 10 • − I 2 4 1 6 10 10 • • = − I + I We have , 2000 1 V s I • • = 40 2 V s I • • = and V1 0.75V s, • • = V V s • • 2 = −250 333 1 2 = = − • • V V Gv 50 1 2 = = • • I I Gi The voltage gain: The current gain: 1500 1 1 = = • • I V Zin k z z z z Z z g out 16.7 1000 500 10( 1000000) 10000 1 1 1 2 2 1 2 2 = + − = − + = − 500 1 • • V s − I 1000 1 10 2 • • = I + I 1 500 1 • • • V s =V + I Solution: 500 − + − + • V 2 I 2 • I 1 • − + • V 1 V S • 10k