- #1

- 9

- 0

## Homework Statement

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?

Last edited by a moderator: