Differential Amplifier design

Jony130

Joined Feb 17, 2009
5,186
223.png

For this circuit to turn-ON the BJT the voltage at the base must be 0.7V larger then the voltage at emitter.
and for this circuit the emitter voltage is 10V * Re/(Re + RL) = 5V therefore Vb should be larger than 5.7V so to have transistor in active region of operation. Because for Vb lower than 5.7V BJT is cut-off.
In your amplifier we have a similarly situation. 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 we can find R6 = 62Ω. And Ie4 = 21V/62Ω = 0.34A
 
Last edited:

dannyf

Joined Sep 13, 2015
2,197
R6 drops vcc-neg. And have to source at least 8.8v/ 49ohm. From that you got r6.

R5 is 100x r4.

0.5x idle current x r3 minus 0.7v is the voltage drop on r4.

...
 

Thread Starter

Rumination

Joined Mar 25, 2016
74
View attachment 103188

For this circuit to turn-ON the BJT the voltage at the base must be 0.7V larger then the voltage at emitter.
and for this circuit the emitter voltage is 10V * Re/(Re + RL) = 5V therefore Vb should be larger than 5.7V so to have transistor in active region of operation. Because for Vb lower than 5.7V BJT is cut-off.
In your amplifier we have a similarly situation. 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 we can find R6 = 62Ω. And Ie4 = 21V/62Ω = 0.34A
I get R6 = 67 Ω and Ic4 = 0,309 A



I haven't used the given gain information yet. Q4 provides no additional gain, it is Q1, Q2 and Q3 providing the necessary gain. So I have to find a correct gain formula for the first two stages? This gives me a second equation. I already have the first equation:
21,7 V = (2,95 / R4 ) * R5.

I have the DC current (3,65 mA) and the collector resistor (2000Ω) for the first stage. How can I find the corresponding gain A1?

A=A1*A2=100 . From this I can derive the necessary gain A2 and the required second equation for R4 and R5.

When I have two equations involving R4 and R5. I can solve it.
 

Jony130

Joined Feb 17, 2009
5,186
I get R6 = 67 Ω and Ic4 = 0,309 A
62Ω is a standard resistor value from E24 series. Which means that you cannot buy a 67 ohm resistor.
E24:
10 11 12 13 15 16 18 20 22 24 27 30 33 36 39 43 47 51 56 62 68 75 82 91

So I have to find a correct gain formula for the first two stages?
Yes, you need to know the gain formula
How can I find the corresponding gain A1?
I do not know. Look it up in google or in the book or do a small-signal analysis and derive gain expression yourself.
 

Thread Starter

Rumination

Joined Mar 25, 2016
74
Gain A1:

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

The gain of the second stage A2:

ϒe4= VT/Ie4 = 0,025 / 0,01 = 2,5 Ω
Ri4= (β4+1) ( ϒe4+R6) = 401*(2,5Ω+62) = 25,86 KΩ

Av2 = (R5 II Ri4) / (ϒe3+R4) = ((R5*25,86 KΩ) / (R5+25,86 kΩ)) / (25 + R4)

Is this correct?
 

Bordodynov

Joined May 20, 2015
2,696
From this scheme you do not get the planned parameters. Here is an example of another, an improved scheme. This is not a real circuit. To implement it is necessary to modify (add 3 resistor). But you have to think a little how it works.Dif.png
 

Thread Starter

Rumination

Joined Mar 25, 2016
74
Ie is 1mA ? Why not 1.825mA ? And what about R1 and R2 ?

Ie4 is 0.01A = 10mA ??
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?
 

Jony130

Joined Feb 17, 2009
5,186
Yes, now it is much better. But why you skip RL in Ri4 ?

And almost all the electronic world uses small letter "re" or "r'e" instead of "ϒe" because letter "Y "is reserved for admittance Y = 1/Z measured in Siemens [1S]
 

bertus

Joined Apr 5, 2008
20,869
Hello,

Using the criteria of @Jony130 , you will most likely end-up with darlington transistors.
Here is a list I extracted from my transistor table:
Code:
Type Polarity Package Vceo Ic Hfe fT Power.
2N6294 NPN TO66   60  4 750-18k 4M 50
2N6295 NPN TO66   80  4 750-18k 4M 50
2N6297 PNP TO66   80  4 750-18k 4M 50
2N6298 PNP TO66   60  8 750-18k 4M 75
2N6299 PNP TO66   100 8 750-18k 4M 75
2N6300 NPN TO66   60  8 750-18k 4M 75
2N6301 NPN TO66   80  8 750-18k 4M 75
2N6312 PNP TO66   80  4 750-18k 4M 50
2N6394 NPN TO66   60  4 750-18k 4M 50
2N6395 NPN TO66   80  4 750-18k 4M 50
2N6534 NPN TO66   80  8 100-5k 20M 14
2N6535 NPN TO66   100 8 100-5k 20M 14
2N6536 NPN TO66   100 8 100-5k 20M 14
2N6537 NPN TO66   120 8 100-5k 20M 14
BDS20  NPN TO220M 80  5 1000min 8M 50
BDS21  PNP TO220M 80  5 1000min 8M 50
BDT63  NPN TO63  60  10 1kmin - 90
BDX62  PNP TO3    60  8 1kmin 7M 90
BDX62A PNP TO3    80  8 1kmin 7M 90
BDX62B PNP TO3    100 8 1kmin 7M 90
BDX62C PNP TO3    120 8 1kmin 7M 90
BDX63  NPN TO3    80  8 1kmin 7M 90
BDX63A NPN TO3    100 8 1kmin 7M 90
BDX63B NPN TO3    120 8 1kmin 7M 90
BDX63C NPN TO3    140 8 1kmin 7M 90
Bertus
 

Thread Starter

Rumination

Joined Mar 25, 2016
74
Bertus:

Thanks. How did you now, that it could be a darlington transistor?
So I can use the 2N6294 silicon NPN Darlington transistor
 
Top