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Design >> Analog Design >> Open loop impedance and open loop differential gain
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Message started by soso on Jun 25^th, 2016, 9:01am

Title: Open loop impedance and open loop differential gain
Post by soso on Jun 25^th, 2016, 9:01am

I want to calculate the open loop output impedance and open loop differential gain of a discrete differential amplifier (Figure attached)

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?