Discrete Differential Amplifier

  • #1
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Homework Statement


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

Circuit.jpg

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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Answers and Replies

  • #2
rude man
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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, ....
Disregard beta and re. Beta is infinite and re is zero. This problem is complicated enough without including those two parameters.
I would also disregard the capacitors, hoping that at 1 KHz they are ignorable. But if you're really ambitious, include them.
The circuit is mirror-identical between the npn and pnp input stages so work with the npn only, assuming HRI+ > HRI- so the pnp's are not in the gain stages.
 
  • #3
The Electrician
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You have the following expression: Rin = hFE*(R115+re) = 250(155) = 38750 ohms.

Where is R115?
 
  • #4
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Sorry, it is a mistake, I wanted to write R153
 
  • #5
The Electrician
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I also see this calculation: So Rtot = R117||Rin = 14441ohms = 14000 ohms.

If R117 = 15k and Rin = 38750 ohms, we should have 15000||38750 = 10814 ohms, not 14441 ohms.
 
  • #6
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You are correct. Using the value Rtot = 10814, the gain at this stage G1 is 22

Making the total gain to be

The open loop Gain G = G1*G2*G3 = (22)(157)(0.95) = 3281.3
Converting to dB: GaindB = 20 log(3281.3) ≈ 70.32dB
 
  • #7
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Another Mistake, If the load is 10 ohms

The load is 600 ohms and not 10 ohms
 

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