Calculating the open loop output Impedance at 1kHz and the open loop differential gain
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?
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