Operational Amplifiers II

This lab is somewhat different to the previous Op Amp lab with the exception that we replaced the voltage divider for another voltage source. The goal in this case is to analyze the influence of the input and feedback resistors. The circuit then will look like this:

1. Assuming R1=10 kohms, we can get RF for a gain of -10
Rf/R1 = 10,    Rf = 10*10000 = 100 kohms

Resistor	Nominal  (kohms)	Measured (kohms)
R1	10	9.86+/-.01
RF	100	98.3+/-.1

2. If Vsen = 1V, we can get Iop
Iop = 1V/10,000ohms = 0.1 mA


3. Circuit setup

4. Measurements

Vin (V)	Vin measured (V)	Vout (V)	VRF (V)	Iop = VRF/RF (mA)
0.25	0.248+/-.001	(-)2.67+/-.02	2.66+/-.01	0.0271+/-0.00011
0.5	0.500+/-.001	(-)4.95+/-.01	4.92+/-.02	0.0501+/-0.00021
1	0.999+/-.001	(-)9.90+/-.01	9.82+/-.01	0.0999+/-0.00014

5. Icc from the +15V voltage supply was measured to be

= 0.933+/-.001 mA


6. Current Iee going into the op-amp from the -15V was measured to be

= -1.034+/-.002 mA


7. By Kirchhoff's Current Law, we prove:

Icc + Iee = (0.933+/-.001)  + (-1.034+/-.002) = -0.101+/-0.0022 mA


8. The power supplied by each voltage supply was

P(+15V) = (15)(0.000933+/-.000001) =  0.01400+/-0.000015 W
P(-15V) = (-15)(-0.001034+/-.000002) =  0.01551+/-0.000030 W


Part II of this experiment consisted on adding a 1 kohm resistor across the amp-output:
9. Circuit design

10. Measurements

Vin (V) Desired	Vout (V) Measured	VRF(V)	Iop (mA)	ICC (mA)	IEE (mA)
1.00	(-)10.00+/-.01	9.96+/-.01	10	0.73+/-.02	(-)11.08+/-.01

11. We can confirm KCL by showing;
Icc + Iee = Iop;          0.73+/-.02 + -11.08+/-.01 = -10.35+/-0.022 mA ~  10 mA


12. At this point we can find the power supplied by the 12V power supplies;
P1 = (-12)(-11.08+/-.01) =  0.1330+/-0.00012 W
P2 = (12)(0.73+/-.02) =  0.0088+/-0.00024 W

Extra Credit: We tried one last case which was placing a resistor box instead of Rf and set it up so that we could achieve a gain of -5:

Rf = 50 kohm

Vin (V) Desired	Vout (V) Measured	VRF(V)	Iop (mA)	ICC (mA)	IEE (mA)
1.00	(-)5.07+/-.02	5.05+/-.01	0.1	0.932+/-.001	-1.032+/-.002

Once again, Iop calculated in the table above agrees with KCL;

Icc + Iee = Iop;             0.932+/-.001 + -1.032+/-.002 = -0.100+/-0.0022 mA