Operational Amplifiers I

Introduction
           In this lab, a new circuit element called an operational amplifiers is used. As the name suggest, operational amplifier amplifies the signal of the input and it could invert or not-invert the signals. In this lab, we will use LM741 op-amps. The problems that we will tackle in this lab is a sensor whose output range is 0 to +1V has to be conditioned so that the output range between 0 and -10V. This means that the input signal should have a gain of -10. Some constraints that this lab has is that it must draw no more than 1 mA current from sensor and op-amps power supply should supply no more than 30 mW of power each.

Experimental

The first step of this experiment is to do some calculation

Since the gain have to be 10, and  the max input is 1 V with 1 mA, the resistor can be calculated using

Ri = V/ I = 1/0.001 = 1k_ohm

The gain of the inverting op-amps is dictated by the equation: gain = Rf/Ri

From this we know that Rf = 10Ri to get the gain of 10 and so

Rf = 10k_ohm

In this experiment, because of some difficulties in doing the gain of 10, we change into the gain of 5 which means the new Rf = 5k_ohm.

The circuit diagram for this lab is as follow:

In order to have the 0 to 1 V as the problems specified, voltage division is necessary and since, in this case, 7.5 V is used with power rating 1/8 W,

Rx = 8* V^2 = 450 ohm

By using voltage division method, Ry can be found to be 69.2 ohm

Since the Rx and Ry is already solved, thevenin resistance is calculated

R_th = Rx||Ry = 60 ohm

The value of the thevenin resistant in this case is about 20 times smaller than the Ri

Before building any circuit, some components were measured:

Component	Nominal Value	Measured Value	Power/Current Rating
Ri	1k Ω	.984k ±.01kΩ	.25W
Rf	5.2k Ω	5.14k ±.01kΩ	.25W
Rx	450 Ω	450 ±1Ω	1W
Ry	60 Ω	57.3 ±1Ω	1W
V1	7.5V	7.54 ±.01V	2A
V2	7.5V	7.66 ±.01V	2A

 

After the components have been measured,the circuit was then built as follow

 Measurements of the circuits was then taken and tabulated below:

Vin	Vout	GAIN	VRi	IRi	VRf
0.0V	0V	0	0V	0mA	0V
.26V	-1.42 ±.01V	-5.46	.26 ±.01V	.26mA	-1.42 ±.01V
.51V	-2.74 ±.01V	-5.37	.51 ±.01V	.52mA	-2.74 ±.01V
.76V	-3.96 ±.01V	-5.21	.74 ±.01V	.75mA	-3.96 ±.01V
.99V	-5±.01V	-5.05	.98 ±.01V	.996mA	-5.18 ±.01V

Questions and Conclusions

As measured before, I1 = 1.01mA and I2=0.8mA

P_1 = I1*V1 = 1.01mA*7.54V = 7.62mW

P_2 = I1*V1 = 0.8mA*7.54V = 0.61mW

By using the op-amps as a node, need top show that I_Ri = I_Rf

I_Ri = 0.99V / 0.984k_ohm = 1.01mA

I_Rf = 5.18V / 5.2 k_ohm = 0.996mA

I_Ri = I_Rf ~ 1.01mA

We do satisfy the constraints to supply no more than 30 mW. If we want to obtain the same gain but reduce the power, we should increase the value of the resistor while keeping the ratio  between the Ri and Rf to be equal to the gain.