Designing a High-Gain Differential Amplifier: How to Meet Specific Requirements

In summary: Do you substract Ohms and voltages? Why did you neglect IC3 in the equationsThere is no need to substract Ohms and voltages - you can see that the voltages and current are already related!There is no need to substract Ohms and voltages - you can see that the voltages and current are already related!
  • #1
Rumination
13
0
9rixZEB.png


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
----------
How can I start? For I have difficulty getting started.
 
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  • #2
Rumination said:
9rixZEB.png


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
----------
How can I start? For I have difficulty getting started.
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
Where is the current I1 and what is the meaning of "quiescent current iC=10 * Q3-Q4 IB"?
 
  • #4
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?
 
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  • #5
LvW said:
Where is the current I1 and what is the meaning of "quiescent current iC=10 * Q3-Q4 IB"?

You can see the current I1 under the first set of amplifiers.
 
  • #6
Rumination said:
You can see the current I1 under the first set of amplifiers.
OK - sorry, I have overlooked this current source.
 
  • #7
Rumination said:
But I don't know the value of R4?

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)..
 
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  • #8
LvW said:
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)..

I don't understand this - "Another relation between R4 and R5 can be derived from Q3 and the corresponding DC quiescent voltages/current".
 
  • #9
Rumination said:
I don't understand this - "Another relation between R4 and R5 can be derived from Q3 and the corresponding DC quiescent voltages/current".
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
LvW said:
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.

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
Rumination said:
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?
?
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
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:
  • #13
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
LvW said:
Following my recommendation in post#7 you now have two equations involving R4 and R5.
Two equations with two unknowns can be solved easily.

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 that's two equations with three unknowns Ic3, R4 and R5
 
  • #15
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
I give up. It's too difficult for me to solve. But thanks for your help:frown:
 
  • #17
Rumination said:
I give up. It's too difficult for me to solve. But thanks for your help:frown:
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
LvW said:
* You have the DC current and the collector resistor for the first stage. Are you able to find the corresponding gain A1 ?

ϒe is: VT / IE = 25 Ω
Av1= Rc / 2* ϒe = 2 / 2* 25 = 40.
LvW said:
* The gain of the second stage A2 is determined (primarily) by the two resistors R4 and R5. Do you know the expression?

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?
 
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  • #19
What is the meaning of ϒe ?
What is the gain formula for the first stage?
 
  • #20
LvW said:
What is the meaning of ϒe ?
What is the gain formula for the first stage?

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?
 
  • #21
Sorry, I cannot follow your "calculation".
ϒe = re ?
Ie4=0.34 ?
Resistors without unit?
β4=?
 
  • #22
LvW said:
ϒe = re ?

Bipolar transistors have a small internal resistance built into their Emitter region called Re.

LvW said:
Ie4=0.34

I have found R6 to be:
-21V* 49/(R6 + 49) = -8.8V
R6 = 62 Ω. And Ie4 = 21V/62Ω = 0.34 A

LvW said:
β4=?

I have been given the value of β = 400 for all transistors in the assignment.
 
  • #23
Rumination said:
Bipolar transistors have a small internal resistance built into their Emitter region called Re.

My comment was related to the symbol ϒe and the realtion ϒe = re.
Normally, this symbol is used for a complex admittance.

I have found R6 to be:
-21V* 49/(R6 + 49) = -8.8V
R6 = 62 Ω. And Ie4 = 21V/62Ω = 0.34 A

Can you explain these equations?

I have been given the value of β = 400 for all transistors in the assignment.

After 22 posts, nice to hear that β is given.
I have tried to give you some help - but I think I should stop now.
At first, you should learn some basics.
 
  • #24
I know, I should not have used the symbol ϒ.
re is better.

The emitter resistor and the load resistor form a voltage divider, so the output peak negative swing will be:
-21V* 49/(R6 + 49) = -8.8V and from there I found R6 = 62Ω. And Ie4 = 21V/62Ω = 0.34A.

What to do next?
 
  • #25
LvW said:
After 22 posts, nice to hear that β is given.
I have tried to give you some help - but I think I should stop now.
At first, you should learn some basics.

It's up to you, if you want to help. I can't force you. :wink:
 

1. What is a differential amplifier?

A differential amplifier is an electronic circuit that amplifies the difference between two input signals while rejecting any common-mode signals. It is commonly used in analog circuits to amplify small signals and reject noise.

2. What are the advantages of using a differential amplifier?

One of the main advantages of a differential amplifier is its ability to reject common-mode noise, making it useful for amplifying small signals in noisy environments. Additionally, it provides high gain, low distortion, and high input impedance.

3. How do I choose the right components for a differential amplifier design?

The choice of components for a differential amplifier design depends on the desired gain, input impedance, and output impedance. Additionally, the type of application and required bandwidth also play a role in component selection.

4. What are some common applications of differential amplifiers?

Differential amplifiers are commonly used in audio amplifiers, instrumentation amplifiers, and operational amplifiers. They are also used in analog-to-digital converters, filters, and signal processing circuits.

5. How can I improve the performance of a differential amplifier?

There are several ways to improve the performance of a differential amplifier, such as using high-quality components, optimizing the feedback network, and implementing proper shielding and grounding techniques to reduce noise. Additionally, using a higher supply voltage and carefully selecting the operating point can also improve performance.

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