Notes on Experiment #10

Prepare for this experiment!

Be sure to print a copy of experiment #10 and bring it with you to your lab session. There will not be any copies available in the lab.

Read the OP-Amp Tutorial before going on with this experiment.

For any Ideal Op Amp with negative feedback you may assume:
• V^- = V⁺ (But not necessarily 0)
• I^- = I⁺ = 0
• Now write KCL equations everywhere except at V-sources and the Op-Amp output.
• Do some algebra to find your answer

Part 2: Op Amp as a Linear Amplifier

Let's find V_O = f(V_S)

KCL at V^-:

(V^- - V_S)/1K + (V^- - V_O)/10K = 0

But V^- = V⁺ = 0 So,

V_O = -(10K/1K)V_S = -10V_S

Let V_S = 1cos(2000(pi)t) volts. Then,

V_O = -10(1cos(2000(pi)t)) = -10cos(2000(pi)t) volts.

Let V_S = 2cos(2000(pi)t) volts. Then,

V_O = -10(2cos(2000(pi)t)) = -20cos(2000(pi)t) volts.

But in this case the output voltage exceeds the supply voltage of the op amp. So the amp goes into "saturation" for |V_O| > 15 volts. The result of this is that the peaks of the -20cos(2000(pi)t) are "clipped off" at +15 and -15 volts.

Part 3: Op Amp as a Linear Adder

Let's find V_O = f(V_a, V_b)

KCL at V^- :

(V^- - V_a)/10K + (V^- - V_b)/20K + (V^- - V_O)/10K = 0

But V^- = V⁺ = 0 So,

V_O = -(10K/10K)V_a -(10K/20K)V_b = -1(V_a + 1/2V_b)

Part 4: Op Amp as an Integrater

Let's find V_O = f(V_S)

KCL at V^-:

(V^- - V_S)/R + i_C + i_100K = 0

But V^- = V⁺ = 0 , assume i_100K = 0 and

i_C = Cd(v_C)/dt = Cd(0 - V_O)/dt So,

-V_S/R - Cd(V_O)/dt = 0

d(V_O) = (-1/RC)V_Sdt So,

V_O = (-1/RC)I[V_S;(-infinity, t)] where,
I[V_S;(-infinity, t)] is " the integral of V_S from -infinity to t"

Let R = 10K, C= 0.02uF and V_S = 4cos(10000(pi)t) volts. Then,

V_O = (-1/10000x0.02E-6)[(-4/10000(pi))sin(10000(pi)t)] Or,

V_O = 0.637sin(10000(pi)t)

Have fun.