# how to determine input impedance of a differential amplifier

#### Ghina Bayyat

Joined Mar 11, 2018
135
in a bjt differential amplifier what would the input impedance be for the differential mode Not the common mode ?
i know the input impedance for the common mode = rπ + 2βRee / 2 but what would the input impedance for the differential mode be ? ofcourse it depends on the two inputs but how can i determine it ? isn't there an equation to use

#### Zeeus

Joined Apr 17, 2019
616
in a bjt differential amplifier what would the input impedance be for the differential mode Not the common mode ?
i know the input impedance for the common mode = rπ + 2βRee / 2 but what would the input impedance for the differential mode be ? ofcourse it depends on the two inputs but how can i determine it ? isn't there an equation to use
Hi..Not sure
The answer depends on the configuration..If there is emitter resistor, base resistor...

(how did you write the pi and co?)..Anyways

think answer is beta * ( re + re||Ree)
With very large Ree or current source then becomes 2 * beta * re = 2* (rpi)

What's answer if there are base resistors at both inputs?

#### Ghina Bayyat

Joined Mar 11, 2018
135
yes there isn't just the basic differential amplifier circuit ( but the picture failed to upload)
ofcourse there is because the two rπs of both transistors are shorted together in the common mode so they are in parallel so their equivalent resistance is half
(how did you write the pi and co?)..Anyways
u mean the sumbol ? there is a letter S above
think answer is beta * ( re + re||Ree)
With very large Ree or current source then becomes 2 * beta * re = 2* (rpi)
but in the differential amplifier we use superposition to analyse the circuit and when we use it we kill vb2 and solve for vb1 then kill vb1 and solve for vb2 right
and when we do that one of the two transistors is common emitter and the other is common base and the input impedance of a Ce is rpi where the input impedance of a CB is re so isn't it supposed to be the sum of these two impedances or i missed something ?

#### Zeeus

Joined Apr 17, 2019
616
To me seems you missed something...The re for CB in parallel with Ree

Someone will clarify

Joined Mar 10, 2018
4,057

#### Zeeus

Joined Apr 17, 2019
616
yes there isn't just the basic differential amplifier circuit ( but the picture failed to upload)

ofcourse there is because the two rπs of both transistors are shorted together in the common mode so they are in parallel so their equivalent resistance is half

u mean the sumbol ? there is a letter S above

but in the differential amplifier we use superposition to analyse the circuit and when we use it we kill vb2 and solve for vb1 then kill vb1 and solve for vb2 right
and when we do that one of the two transistors is common emitter and the other is common base and the input impedance of a Ce is rpi where the input impedance of a CB is re so isn't it supposed to be the sum of these two impedances or i missed something ?
Wait...missed you response I guess
Input impedance is to find the resistance that Ib base current flows through, yh? Answer can not be Rpi + re

Not sure
look at base of transistor, every resistance looking into the emitter resistor has equivalent base resistance

Therefore input impedance Q
= (beta +1) *(re +re||Ree)

(beta +1) * re = Rpi ?

#### Ghina Bayyat

Joined Mar 11, 2018
135
thank u so much the first link helped a lot i understand now

#### Ghina Bayyat

Joined Mar 11, 2018
135
Wait...missed you response I guess
Input impedance is to find the resistance that Ib base current flows through, yh? Answer can not be Rpi + re

Not sure
look at base of transistor, every resistance looking into the emitter resistor has equivalent base resistance

Therefore input impedance Q
= (beta +1) *(re +re||Ree)

(beta +1) * re = Rpi ?
thanks for ur help but u can look at the links it is all cleared

#### LvW

Joined Jun 13, 2013
1,438
I think, this question needs - at first - some definitions:
(1) Common mode is clearly defined: Vb1=Vb2
(2) Unsymmetrical diff. mode: Vb1 finite and Vb2=0.
(3) Symmetrical diff. mode: Vb2=-Vb1
(4) General symm. mode: Vb1 finite and Vb2 finite (with Vb1 not equal to Vb2).

Input resstances: For the first three cases, it is a realtively simple task to find the dynamic input resistances rin (here given at the base of Q1)

Case (1): rin=rpi+beta*2re (re: diff. resistance of the common emitter path) .
Case (2): rin=2rp
Case (3): rin=rpi
Case (4): The input resistance at the base of Q1 depends on the signal Vb2 which is applied at the base of Q2. There is no textbook which gives an expression for the input resistance in this case (as far as I know). In this case, the input resistance must be calculated using superposition of the two cases (1) and (3). This is because each arbritrary combination of Vb1 and Vb2 can be split into the cases (1) and (3).

As the result, the formula for the input resistance will contain (and, thus, depend on) both input signals ! Therefore, no compact expression can be given.