Op Amp

Operational Amplifiers I

The purpose of the lab is to learn how to use an Op Amp. And use it to design the circuits we want.

Pre Lab
 
V_cc = 12 V
V_ee = -12 V
V_in = 0 V to +1 V
V_out = 0 V to -10 V
The sensor outputs at most 1 mA
The op-amp power supplies supply at most 30 mW each
 

Solve for R_i and R_f
R_i = 1/1 V/mA = 1 kΩ
R_f = 10*R_x = 10 kΩ

However, the lab is limited to two power supplies so V_cc is also going to provide V_in using the appropriate voltage divider like the circuit below.

V = 12 V because it is the worst case where R_y is zero while using V_cc.
Using half of the power rating of a 1/4 W resistor, R_x = (V^2)*8 = (12^2)*8 = 1152 Ω.
With the voltage divider rule, R_y = R_x/11 = 104.7 Ω.

Thevenin Equivalents

R_th = R_x*R_y/(R_x + R_y) = 96 Ω.
V_th = 1 V.
However, R_th is NOT 20 times under R_i.
Thus, the new R_i = 20*96 = 1.92 kΩ.
The new R_f = 19.2 kΩ.

Procedure:
Measure the parts and build the circuit according to the two Lab Manual.

Component	Nominal Value	Measured Value	Power or Current Rating
R_i	2000 Ω	1957 Ω	1/8 W
R_f	20000 Ω	19760 Ω	1/8 W
R_x	1152 Ω	1786 Ω	1/4 W
R_y	104.7 Ω	7020 Ω	1/4 W
V_1 = V_cc	12 V	12.12 Ω	2 A
V_2 = V_ee	12 V	12.08 Ω	2 A

Set up the circuit

 

Connect the power supply

V_in	V_out (Measured)	GAIN (Calculated)	V_Ri (Measured)	I_Ri (Calculated)	V_Rf (Measured)
0.0 V	0.00 V	0.00	0.00 V	0.00 A	0.00 V
0.25 V	-2.55 V	-10.20	0.25 V	0.125 mA	2.57 V
0.50 V	-5.07 V	-10.14	0.50 V	0.250 mA	5.11 V
0.75 V	-7.62 V	-10.16	0.75 V	0.375 mA	7.65 V
1.00 V	-10.16 V	-10.06	1.01 V	0.505 mA	10.18 V

I_v1 = 2.31 mA

I_v2 = -1.645 mA

Data Analysis

P_v1 = V*I = 28.00 mW

P_v2 = -19.87 mW
 

I_v1 + I_v2 = 0.665 mA

I_f = 0.509 mA

Error = 30.6%

Conclusion

   We had about 30% error which was not very good. We found that the current_in and current_out are not zero because the op amp is not ideal (R_in is not infinite). Therefore, we should not calculate the op amps by assuming they are ideal in this case.