# Differential Amplifier design

#### Rumination

Joined Mar 25, 2016
74 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

----------
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.

#### Rumination

Joined Mar 25, 2016
74
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 = 8,8 V - R3 * Ic1 = 8,8 V - 2000 * 1,825 mA = 5,15 V.

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

But I don't know the value of R4?

#### Jony130

Joined Feb 17, 2009
5,250
Hey but, your Vcc-pos is 21 V so VB3 voltage is equal to (if we ignore IB3):
VB3 = 21V - Ic1*R3 = 21V - 1,825 mA*2kΩ1 = 21V - 3.65V = 17.35V
Also notice that R4 and R5 "set" the Q3 voltage gain.
Quiescent current IC = 10 * Q3-Q4 IB
I don't understand this ?

#### Rumination

Joined Mar 25, 2016
74
Yes, you are right. It's 21 V and not 8,8 V. I got confused. But what does this mean:

"R4 and R5 "set" the Q3 voltage gain"?

And just ignore, Quiescent current IC = 10 * Q3-Q4 IB.

#### Jony130

Joined Feb 17, 2009
5,250
But what does this mean:

"R4 and R5 "set" the Q3 voltage gain"?
The Q3 voltage gain at first glance is equal to AvQ3 ≈ R4/R5

#### Rumination

Joined Mar 25, 2016
74
But I don't know the value of either R4 or R5?

#### crutschow

Joined Mar 14, 2008
28,161
But I don't know the value of either R4 or R5?
But you know their ratio.
This is not a single answer to their correct values, only their ratio. You select the value to give the desired Q3 transistor bias current.
For a real design this current would affect the frequency response which may be of interest.

#### Rumination

Joined Mar 25, 2016
74
But you know their ratio.
This is not a single answer to their correct values, only their ratio. You select the value to give the desired Q3 transistor bias current.
For a real design this current would affect the frequency response which may be of interest.
Do you mean:

Ad = gm * Rd ?

#### Rumination

Joined Mar 25, 2016
74
Now I have been working on this problem, but I really can't understand #### Jony130

Joined Feb 17, 2009
5,250
You must start at the output. You have been given RL and Vout_max so from there you can find Ic4 and Ic3 quiescent current.
And choose R4,R5R1,R1 to meet the voltage gain requirement at given quiescent current .

#### crutschow

Joined Mar 14, 2008
28,161
Do you mean:

Ad = gm * Rd ?
What's Rd?
No, that gain equation only holds when there is no emitter resistor.
The emitter resistor provides negative feedback and, for low values of gain (much less than gm*R5), the gain of Q3 is approximately R5/R4.

#### Rumination

Joined Mar 25, 2016
74
Can I do it like this:

VB3=17.35V
We know: VE3=(17.35+0.7)V=21-IC3*R4
We also know: VB4=+0.7V=-21+IC3*R5.
So by eliminating IC3 it gives a second equation for R4 and R5:

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? And now we can find the value of Ic3?

#### Rumination

Joined Mar 25, 2016
74
You must start at the output. You have been given RL and Vout_max so from there you can find Ic4 and Ic3 quiescent current.
And choose R4,R5R1,R1 to meet the voltage gain requirement at given quiescent current .
To find Ic3:

Ic3 = IE3 = (21 V - VBE - 0,7 V) / (R4) = (21 V - 17,35 V - 0,7 V) / (2,95) = 2,95 / 2,95 = 1 mA.

To find VB4:

VB4 = -21 V + R5 * Ic3 = -21 V + 21,7 V * 1 mA = 0,7 V.
Vout = VB4 - 0,7 V = 0 V

To find Ic4:
Ic4 = (0 - (-21 V)) / R6

But I don't know the value of R6.

#### Rumination

Joined Mar 25, 2016
74
But what about Vout-min = -8,8 V and RL = 49 Ohm ?
Does it mean, that I can find Ic4:

Ic4 = V / R = 8,8 / 49 = 0,18 mA

#### Jony130

Joined Feb 17, 2009
5,250
Ic4 should be larger than this 0.18mA

#### Rumination

Joined Mar 25, 2016
74

#### Jony130

Joined Feb 17, 2009
5,250
It looks like you do not know (or understand) how to find maximum negative voltage swing at the load for a simply emitter follower circuit (Q4).

#### Rumination

Joined Mar 25, 2016
74
Yes, you are right. I will try and find it in my book. I will be back. #### Jony130

Joined Feb 17, 2009
5,250
To help you with this problem look again at my post #14 and Ask yourself -- >What voltage at the T1 base will cut-off T1 transistor?