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

Click For Summary

Discussion Overview

The discussion revolves around the design of a high-gain differential amplifier, focusing on determining resistor values (R1, R2, R4, R5, and R6) to meet specific performance requirements, including gain, output voltage range, and quiescent current. Participants explore theoretical and practical aspects of differential amplifier design, including gain calculations and biasing conditions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant outlines the requirements for the amplifier, including gain (Av = 100), output voltage conditions, and quiescent current specifications.
  • Several participants express confusion about the meaning of specific terms, such as "quiescent current" and the role of certain currents in the circuit.
  • Participants discuss the relationships between resistors R4 and R5, suggesting that these can be derived from the gain requirements and quiescent conditions.
  • There are multiple attempts to derive equations involving R4 and R5, with some participants questioning the validity of their calculations and the assumptions made.
  • One participant mentions the need to use gain information from the first two stages of the amplifier to derive necessary equations.
  • Another participant provides calculations for the gain of the first stage and questions the meaning of certain symbols and values used in the discussion.
  • There is a mention of a specific value for the transistor's beta (β = 400) being provided, which some participants find helpful.
  • Confusion persists regarding the calculations and the use of specific symbols, leading to frustration among some participants.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations or the relationships between the resistors. There are multiple competing views on how to approach the problem, and uncertainty remains regarding the correct equations and values to use.

Contextual Notes

Some participants express difficulty in following the calculations and the use of specific symbols, indicating a potential gap in foundational knowledge that may affect their ability to engage with the material effectively.

Who May Find This Useful

This discussion may be useful for students and practitioners interested in differential amplifier design, particularly those seeking to understand the interplay between circuit components and performance metrics in electronic engineering.

Rumination
Messages
13
Reaction score
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.
 
Physics news on Phys.org
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?
 
Where is the current I1 and what is the meaning of "quiescent current iC=10 * Q3-Q4 IB"?
 
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:
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.
 
Rumination said:
You can see the current I1 under the first set of amplifiers.
OK - sorry, I have overlooked this current source.
 
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)..
 
  • Like
Likes   Reactions: berkeman
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".
 
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?
 
Last edited:
  • #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:
 

Similar threads

Engineering How does BJT work in this circuit?
  • · Replies 4 ·
Replies
4
Views
3K
Common Emitter Amplifier Design
  • · Replies 25 ·
Replies
25
Views
10K
Designing a Class A Amplifier for Good Speakers, Inc.
  • · Replies 1 ·
Replies
1
Views
3K
How to Design a BJT Pre-Amplifier to Meet Specific Requirements?
  • · Replies 17 ·
Replies
17
Views
6K
Can anyone suggest an amplifier design that meets specific requirements?
  • · Replies 1 ·
Replies
1
Views
12K