1. The problem statement, all variables and given/known data Calculating the open loop output Impedance at 1kHz and the open loop differential gain 3. The attempt at a solution Here is the approach. INPUT STAGE: Q111 and Q112 Av1 = Rtot/2re, where Rtot is the total resistance of R117 and the input resistance of Q78 which is Rin. So the gain Av = Rtot/500 ohms. If hFE for Q78 is 250, Rin = hFE*(R115+re) = 250(155) = 38750 ohms. So Rtot = R117||Rin = 14441ohms = 14000 ohms. Using this value the gain G1 = 14000/500 = 28. If the load is 10 ohms hFE for Q78 and Q94 are 250 and 100 respectively. Load impedance RL= (250)(100)(600 ohms) = 15000000 ohms. SECOND STAGE: Q78 Output resistance of Q78: Using Early voltage of BC557 is 50V, ro = (VA + VCE)/IC. ro = 14360 ohms. Early Effect Resistance Rout = 14360 ohms The early effect resistance and load impedance are parallel. So Rtot2 = RL|| Rout = 14346 ohms. Gains at stage 2: G2 = 14346/155 = 93. OUTPUT STAGE: Q94 and Q126 For Q94, IE = 10.3mA = 10mA So re = VT/IE - 25mV/10mA = 2.5 ohms Rtot3 = R174 + re = 68 ohms + 2.5 ohms = 70.5 ohms Both halves are in parallel, So Output Impedance Zout = 70.5 ohms || 70.5 ohms = 35.25 ohms = 35 ohms Gain in the output stage G3 = RL/(RL + Zout) = 600 ohms/(600 ohms + 35 ohms)= 0.95 Thus open loop gain G = G1*G2*G3 = (28)(93)(0.95) = 2473.8 Converting to dB : Gain in dB = 20log(2473.8) = 68 dB I have calculated the open loop gain considering a load of 600 ohms. I ended up with an open loop gain of about 2473.8 which is about 68dB. Is this feasible?