Lab 8 OpAmpII

          In this experiment, we will be exploring the relationship between the output voltage, input voltage, and the ratio of resistors (Rfeedback and Rin). Meanwhile, we will probe the output current to testify whether Kirchoff's current law holds true in OpAmp.

We will be using an inverted op-amp with the rails' voltages at -12V to 12V.

 

A seperate voltage divider is to replace Vsen to divide a 1V input voltage out of a 12V power supply.

 

 

How to set up the voltage divider:

 



1) Obtain a potentiometer, and some wire.

2) Connect 1 of the potentiometer side pin to +12V, another side pin to ground, and the middle pin in series to the input resistor.

3) Measure the potential drop between ground and the middle pin, adjust the potentiometer untill you obtain the voltage you for Vsen.

Since we want to get a gain of -10, we will need the following resistor according to the op-amp's equations:

Resistors	Nominal Values, kΩ	Measured Values, kΩ
R­f­	10kΩ	9.91kΩ
R­in­	100kΩ	97.7kΩ

 

Finding Iop:

 

If we say Vsen is 1V, and our gain is -10, the current Iop = -10V/Rf since the feedback branch is the only way that Iop can go through.

 

                    VRf=Vout=-10V=Iop(Rf)     ===>>          Iop= -10/100kΩ = -0.1mA

 

 

After measuring the voltage across different parts of the circuits, we obtained the following results:

V­in­ desired	V­in  measured­	V­out measured	V­Rf  measured	I­op  calculated
0.25V	0.24V	-2.46V	2.46V	0.0246mA
0.5V	0.5V	-4.90V	4.87V	0.0490mA
1.0V	1.00V	-10.04V	9.86V	0.1004mA

 

The current in the rails (@ Vin =1V) are the following:

 

ICC = 0.874mA                 IEE=-0.981mA

 

The sum of the the rail currents = ICC+IEE

                                                                                 =(0.874-0.981)mA

                                                   =-0.107mA.

 

From circuit analysis, Iop =ICC+IEE ; Where Iop = -0.1004mA and ICC+EE=-0.107mA.

Since Iop is about the same as ICC+EE, we conclude that KCL holds true in op-amps.

 

 

Power supplied by each 12V power source at the operating point:

 

P-12=-0.981mA*-12V =11.77mW

P12=0.874mA*12V=10.49mW

 

Now we replace the open circuit with a 1kΩ resistor and measure the following :

 

­V­in desired	V­­out measured	­V­Rf measured­	­I­op calculated­	­I­cc­ ­measured	I­EE measured
1.0V	-9.99V	9.72V	-0.102 mA	0.887mA	-0.987mA

Sum of rail's current:

(0.887-0.987)mA= -0.100mA ≈ -0.102mA ≈Iop   -----> KCL still holds true.

 

P-12=-0.987mA*-12V =11.84mW

P12=0.887mA*12V=10.64mW

 

 

Bouns:

 

Use the resistance box to find Rf so that the gain becomes -5.

 

By Theory, we need a Rf of 50kΩ to obtain a gain of -5.

As we measure the actual resistance in the resistor box after setting it to 50kΩ, the actual reading gives us 49.9kΩ.

 

 

­V­in desired	V­­out measured	­V­Rf measured­	­I­op calculated­	­I­cc­ ­measured	I­EE measured
1.0V	-5.03V	4.99V	-0.101mA	0.884mA	-0.985mA



 

 

 

Conclusion:

 

The gain of an inverted op-amo is always the negative ratio of -Rf/Rin. So, if the gain is -5, -Rf/Rin has to be -5. At the same time, the current at the operation point is the sum of the current from the two rails into the op-amp (KCL is always true).