# Differential amplifier design

1. Mar 25, 2016

### Rumination

The figure shows a differential amplifier with a differential input vid = (vin_d_pos - vin_d_neg) between the bases of Q1 and Q2, and an output (out) at the emitter of Q4.

I have to Determine R1, R2, R4, R5 and R6 so that the following requirements are met:

R 1 = R 2
Av = vout / vid = 100 [V / V]
Vout = 0V when Vin_d_pos = Vin_d_neg = 0V.
The output Vout must be able to deliver undistorted peak signal voltage in the range
from vout-min to vout-max with RL armed.
Quiescent current IC = 10 * Q3-Q4 IB

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I have been given the values:
Vcc-pos = 21 V
Vcc-neg = 21 V
R3 = 2000 Ohm
RL = 49 Ohm
I1 = 3,56 mA
Vout-max = 8,8 V
Vout-min = -8,8 V
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How can I start? For I have difficulty getting started.

2. Mar 25, 2016

### Staff: Mentor

Welcome to the PF.

The rules say that you must show some effort before we can offer tutorial help. What have you learned in class about how to analyze differential amplifiers? How are the bias currents set? How is the gain set?

3. Mar 25, 2016

### LvW

Where is the current I1 and what is the meaning of "quiescent current iC=10 * Q3-Q4 IB"?

4. Mar 25, 2016

### Rumination

My English is not so good. I have found this out:

Vcm = 0 , Vin_D = 0
VB1 = VB2 = 0 V.

Ic1 = IE1 = I/2 = 3,65 / 2 = 1,825 mA

VB3 = 21 V - R3 * Ic1 = 21 V - 2kΩ * 1,825 mA = 17,35 V.

Ic3 = IE3 = (24 V - VB3 - 0,7 V) / R4

But I don't know the value of R4?

Last edited: Mar 25, 2016
5. Mar 25, 2016

### Rumination

You can see the current I1 under the first set of amplifiers.

6. Mar 25, 2016

### LvW

OK - sorry, I have overlooked this current source.

7. Mar 25, 2016

### LvW

I think, because you have used nearly all information, as a next step you should use the gain requirement.
This gives you a relation involving R4 and R5. Another relation between R4 and R5 can be derived from Q3 and the corresponding DC quiescent voltages/current (you know the DC voltages at the emitter and the collector for Q3)..

8. Mar 25, 2016

### Rumination

I don't understand this - "Another relation between R4 and R5 can be derived from Q3 and the corresponding DC quiescent voltages/current".

9. Mar 26, 2016

### LvW

VB3=17.35V
That means: We know VE3=(17.35+0.7)V=21-IC3R4
We also know: VB4=+0.7V=-21+IC3R5.
Eliminating IC3 gives a second equation for R4 and R5.

10. Mar 26, 2016

### Rumination

Oh okay. So we get this by eliminating Ic3:

18,05 = 21 - R4
R4 = 2,95

0,7 + 21 = R5
R5 = 21,7

Does it mean, that we have found the values of R4 and R5 now? And now we can find the value of Ic3?

11. Mar 26, 2016

### LvW

???
Do you substract Ohms and voltages? Why did you neglect IC3 in the equations ?
"Eliminating IC3" means: Solving for this current in one equation and inserting it into the other one. This is basic for a system of two equations!

12. Mar 26, 2016

### Rumination

Okay, but when I solve for the current in one equation and inserting it into the other one. I get:

(17.35+0.7)V=21 V -IC3R4
(2,95 / R4 ) = IC3

21,7 V = (2,95 / R4 ) * R5

Last edited: Mar 26, 2016
13. Mar 26, 2016

### LvW

Following my recommendation in post#7 you now have two equations involving R4 and R5.
Two equations with two unknowns can be solved easily.

14. Mar 26, 2016

### Rumination

I can solve two equations with two unknowns. But my problem is, what those two equations are? That's what I'm confused about. Are these the two equations?:

(17.35+0.7)V=21-IC3R4
+0.7V=-21+IC3R5

But thats two equations with three unknowns Ic3, R4 and R5

15. Mar 26, 2016

### LvW

The first equation is In post#12.
Question: Did you use already the gain information? NO!
Because Q4 provides no additional gain, it is Q1, Q2 and Q3 providing the necessary gain.
Are you able to find a correct gain formula for the first two stages?
This gives you a second equation.

16. Mar 26, 2016

### Rumination

I give up. It's too difficult for me to solve. But thanks for your help

17. Mar 27, 2016

### LvW

Too difficukt?
* You have the DC current and the collector resistor for the first stage. Are you able to find the corresponding gain A1 ?
* The gain of the second stage A2 is determined (primarily) by the two resitors R4 and R5. Do you know the expression?
* A=A1*A2=100 . From this you can derive the necessary gain A2 (A1 is known) and the required second equation for R4 and R5.

18. Mar 27, 2016

### Rumination

ϒe is: VT / IE = 25 Ω
Av1= Rc / 2* ϒe = 2 / 2* 25 = 40.

ϒe4= VT/Ie4 = 0,025 / 0,01 = 2,5 Ω
I have found R6 to: (-21*49)/(R6+49) = -8,8 ⇒ R6 = 67,9.
Ri4= (β4+1) ( ϒe4+R6) = 401 (2,5Ω+67,9) = 28,2 KΩ
Av2 = (R5 II Ri4) / (ϒe3+R4) = ((R5*28,2 KΩ) / (R5+28,2 kΩ)) / (25 + R4)

Is this correct?

Last edited: Mar 27, 2016
19. Mar 28, 2016

### LvW

What is the meaning of ϒe ?
What is the gain formula for the first stage?

20. Mar 28, 2016

### Rumination

ϒe = re

Gain A1:

ϒe1 = ϒe2 = VT / IE = 0,025/0,001825 = 13,7
Av1= (R3 II Rib3) / (ϒe1 + ϒe2 + R1 + R2)

The gain of the second stage A2:

ϒe4= VT/Ie4 = 0,025 / 0,34 = 0,339
Ri4= (β4+1) ( ϒe4+R6) = 401*(0,339+62) = 25 KΩ

Av2 = (R5 II Ri4) / (ϒe3+R4) = ((R5*25 KΩ) / (R5+25 kΩ)) / (25 + R4)

Better?