Lab 9: Op Amp 1

Introduction:
This lab introduces a new circuit element known as operational amplifier. We consider a situation where the circuit is required the range of sensor from 0 to 1V, and the range of the microcontroller form  0 to 10V. To accomplish this situation, we using the operational amplifier to increase the voltage with a gain of 10. The constrains to this problem include the mention voltages above, a draw of no more than 1mA from the sensor, and a supply of no more than 30mW by the op amp.

Procedure:

Step 1A:
First we do some calculations to accomplish the gain of -10.
The input resistor must take in no more than 1mA of current and the maximum voltage to be input is +1 we can calculate its value
Rin = Vin/Imax = 1/.001 = 1kΩ

Step 1B :
The gain is defined by the ratio: Vout/Vin 
We know Rf = (Vout/Vin) * Rin = 10kΩ

Step 1C:
Rx is to be powered at 1/8W and since the maximum voltage across it is 7.5V, then
1/8= (V^2)/Rx
Rx = 8*V^2 = 1152Ω

Step 1D:
Ry is to have maximum of +1V across it:
Ry/(Ry+Rx) * 12= 1
Ry = 288 Ω

Step 1E:
The Thevenin resistance experienced by Ri would then be
RTh = Rx || Ry = 230.4 Ω

which is approximately 20 times smaller than Ri.

Measured Components

Component	Nominal Value	Measured Value	Power/Current Rating
Ri	1k Ω	984 ± 1Ω	0.25W
Rf	10k Ω	9.74 ± 0.01kΩ	0.25W
Rx	1k Ω	982 ±1Ω	0.25W
Ry
V1	12V	12.06 ±.01V	2A
V2	12V	12.10±.01V	2A

Build the circuit:

Measuremenst:

Vin	Vout	GAIN	VRi	IRi	VRf
0.0V	0V	0	0V	0mA	0V
0.25V (0.24V)	-2.48 ±.01V	-9.92	.249 ±.001V	0.253mA	-2.48 ±.01V
0.50V (0.50V)	-4.97±.01V	-9.94	.500 ±.001V	0.508mA	-5.01 ±.01V
0.75V (0.76V)	-7.48 ±.01V	-9.97	.751 ±.001V	0.763mA	-7.52 ±.01V
1.00V (1.004V)	-10.03 ±.01V	-9.98	1.005±.001V	0.102mA	-10.01±.01V

Measure the current passing through each power supply:
I_V1 = 1.737 mA,  I_v2 = 1.276mA

Calculating the actual power delivered by each power supply:

P_V1 = I_V1 * V₁ = 1.737 * 12.06 = 20.9 mW

P_V2 =I_V2*V₂ = 1.271 *12.10 = 15.4 mW

Assuming the current into the non-inverting and inverting terminals of the op amp are zero, we show that KCL is satisfied at the op amp node when Vin = 1V:

I_Ri = V_in / R_i =1.004 / 984 = 1.02 mA

I_Rf =  V_Rf / R_f  = 10.01 / 9740 = 1.02 mA

From the above calculations, the power delivery from the power supplies are 20.9 mW and 15.4 mW which are less than 30mW constraint.  To reduce the power drawn we may increase the resistor values in a way that still keeps the ratio Rf/Ri = 10.