Question

A differential amplifier employs resistors in the input and feedback paths, all of the same value.

1) If two sine waves 1.5sin(w)t and 2sin(w)t are applied simultaneously to the two inputs, what is the equation of the output voltage?

2) If the two waves have equations 1.5sin(w)t and 1.5cos(w)t, what is the new output?

Answer 1

V=10/10 (2sin(w)t-1.5sim(w)t)

V=sin(w)t(2-1.5)

V=0.5*sin(w)t

2)

V=10/10 (1.5sin(w)t-1.5cos(w)t)

V=1.5[sin(w)t-cos(w)t]

above equation can b written as

V=1.5[sin(w)t-cos(w)t]

Answer 2

     

R₃ = R₁

V_out = V₂ - V₁

a) V_out = 0.5 sin wt

b) V_out = 1.5 sinwt - 1.5 coswt = 1.5 ?2 sin(wt + 45^o)

Answer 3

     

R₃ = R₁

V_out = V₂ - V₁

a) V_out = 0.5 sin wt

b) V_out = 1.5 sinwt - 1.5 coswt = 1.5 ?2 sin(wt + 45^o)

Answer 4

ll

Answer 5

For a differential Amplifier in the configuration as described, the voltage output is amplified by -(R1/R2) where R1 is the feedback resistance and R2 the i/p resistance. Now since both of them are equal the output is just invertion (since R1/R2 = -1) if the differential i/p voltage. Hence the ans to your questions is

1) O/p Volt =2sinWt+ (-1)*(1.5sinWt - 2sinWt) {Considering 1.5 sinWt is applied at the branch having i/p resistance)

=2.5sinWt

2)O/p Volt = 1.5cosWt + (-1)*(1.5sinWt-1.5cosWt) = 1.5sinWt+3.0cosWt

The addition of 2sinWt/1.5cosWt can be easily explained considering the circuit as described below (by applying KVL at node A)

Also, note that for an ideal differential op amp node A and B are always at the same potential. Also note that I've given output voltage wrt ground and all other voltages too have been taken wrt ground.