Page 1 :
Transistor Analysis on the Basis of h Parameters : Let a transistor amplifier is, shown in figure (2.27) in terms of h parameter equivalent circuit., www, R, v,, h,V,, R, Fig. 2.26, Scanned By Scanner Go

Page 2 :
respectively., While ij and i2, are input and output currents respectivėly:, We know that by h-parameters, V = l4 i, + li, V2 :, .(1), 2=4り+ん2, .(2), 'Vy = Es-Rs, .(3), Also, input signal, Where Es is signal voltage and Rs is signal resistance., Now, output voltage, V2 = -i, RL :, .(4), Where R, is load resistance., Here (-)ve sign is taken because i, is in opposite direction., (i) Current Gain : We know that, Output current, Input current, Current gain A;, .(5), %3D, Using (2) and (4), we get, i2 = hig i1 - i2 R1 lo ., i2 (1+ R1 l6) = hf i, i2, or, So, current gain., 1+ RL l, (ii) Input Impedance :, Input voltage, Z, =, Input current, Scanned By Scanner Go

Page 3 :
111, (Using équation 1), (Using equation 4), Z; = h; + 1, RLA¡, : A;, Using equation (6), we get, A; =, Then,, Z; = li; - 1, RL, 1+ RL l,, 1; +R1 (Al), Z =, .(7), 1+ RL h,, Where Ali = 1,l1, - h,hf and it is clear that Z; depends upon R1., (i) Output Impedance : We have, V2 =0, V2, ..(S), z,-, (Using 2), Z., hs i + h, V, V2, Putting, Es = 0 into equation (3), .., V = -i Rs, (Using 1), hi, + h, V2 =-i, Rs, i (h, + Rs) = -1, V2, 1, V2, h; + Rs, %3D, l'utting this value of i into (8), V2, 2., +4, V2, li, +R$, Scanned By Scanner Go