# How to adjust gain of differential discrete amplifier

Assignment schematic:

I got an assignment to

1) Calculate open loop gain of the existing design of the amplifier

and

2) Configure this op-amp with the closed loop of 26 dB. The chosen resistors should not exceed 100 koHm

The original design is shown on the picture: Solution for the first Part:

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/r'e. r'e = 250 oHms. If hFE for Q78 is 110, Rin = hFE*(R115+re) = 110(10.6 + 150) = 17666 ohms. So Rtot = R117||Rin = 15k||17666 = 8112 ohms. Using this value the gain Av1 = 8112/500 = 16.224

If the load is 2 kohms hFE Q94 is 40 (power transistor). Load impedance RL= 40 * 2000 = 80 kohms.

SECOND STAGE: Q78

Output resistance of Q78: Using Early voltage of 52V, Ro = (VA + VCE)/IC. ro = 26126 ohms. (Ic = 2.85 mA, Vce = 22.46 V)

damping factor = (r'e + Re)/r'e = (10.6 + 150)/10.6 = 15.15

Rout = ro * damping factor = 395809 ohms.

The early effect resistance and and load impedance are parallel. So Rtot2 = RL|| Ro = 80000 || 395809 = 66550

Gains at stage 2: G2 = Rtot/(R153 + r'e) = 66550/(150 + 10.6) = 414.4

OUTPUT STAGE: Q94 and Q126

For Q94, IE = 1.1mA So re = VT/IE = 25mV/1.1mA = 22.73 ohms

Rtot3 = R174 + r'e = 68 ohms + 22.73 ohms = 90.73 ohms

Both halves are in parallel, So

Output Impedance Zout = 90.73 ohms || 90.73 ohms = 45.37 ohms

Gain in the output stage G3 = RL/(RL + Zout) = 2000 ohms/(2000 ohms + 45.37 ohms)= 0.978

Thus open loop gain G = G1*G2*G3 = (16.224)(414.4)(0.978) = 6575

Converting to dB : Gain in dB = 20log(6575) = 76.36 dB

I have calculated the open loop gain considering a load of 2000 ohms. I ended up with an open loop gain of about 6575 which is about 76.36dB.

If the first part I was able to solve - Acl = 88 dB an Aol = 76.36 dB. I got stuck in the second one. Any thoughts, suggestions which way I should go or how to adjust the resistors in order to get the desired Closed Loop Gain result?

SIMULATION with Rf and R2:

V+ = 10 mV (sine), V- = 10 mV Picture shows Vout

Resistor R39 = 99 Kohm, R40 = 301 ohm

6575 (open loop gain) / 20 = 329 should attenuation factor

The feedback then is: Av = 1 + R39/R40 = 329

The feedback should attenuate the signal by 329

But in this case does not really work

• Simply add a resistor (RF) between the output and Q112 base. and the second resistor (R2) + capacitor from Q112 base to GND. – G36 Dec 10 '17 at 16:03
• I tried to add RF = 19kOhm and R2 = 1kohm. But it did not change anything in the simulation results(LTSpice). Am I doing anything wrong? – Tomas Dec 10 '17 at 16:50
• Can you show the circuit after you add thous two resistors? – G36 Dec 10 '17 at 17:00
• added that to my question in the last section – Tomas Dec 10 '17 at 17:16
• LOL remove V24 source. – G36 Dec 10 '17 at 17:19