Lab 8: Differential Amplifier - Department of Physics - UC Davis

This reduces to
I 1 = I 2 + V 1 /R E .
I 1 is now greater than I 2 ,soI 2 must have decreased since the total, I ′ , is approximately
constant. Reducing I 2 lowers the voltage drop across R C ,soV out increases. V 1 is the noninverting
input.
In the same way, raising V 2 slightly with V 1 grounded increases I 2 . This now increases
the voltage drop across R C and lowers V out . V 2 is the inverting input.
A more complete analysis can be done using the small signal AC model for the circuit,
shown in Fig. 2.
R C
v out
v 1
r e r e
v 2
i R 1 E R E
v
i 2 A
αi 1 αi2
α ≈ 1
i'
R'
r e ≈ 26 Ω mA/I EQ
Figure 2: Small signal AC model for the differential amplifier.
Here, all voltages and currents have been replaced by the deviations from their values
at the quiescient point. For example, i 1 = I 1 − I 0 . The transistors have been replaced by
simple active region models.
1. Differential mode gain:
Define v d ≡ v 1 − v 2 . The differential mode gain is
A vd ≡ v out /v d .
Apply the following input voltages to the amplifier: v 1 = v d /2andv 2 = −v d /2. Use
KVL:
v 1 = i 1 (r e + R E )+(i 1 + i 2 )R ′ = v d /2,
v 2 = i 2 (r e + R E )+(i 1 + i 2 )R ′ = −v d /2.
Adding the equations and collecting terms results in i 2 = −i 1 , which also leads to
v A =(i 1 + i 2 )R ′ =0. Therefore,
i 2 = v 2 /(r e + R E )=−v d /2(r e + R E ).
2