Op amp circuit analysis

[manju007]

Hi all,

I know the basic operation of the Op-Amp but i am bit confusing in analyzing the circuit that is in attachment. Can anyone help me out to analyze it.
I need to know how inverting terminal voltage changes according to the changes in the non inverting terminal voltage.

[pebe]

I will try to explain it. There are two things to remember about the op-amp.
A) Changes at pin1 and pin2 are in opposite phase, ie. a positive change on the inverting input, pin2, causing its voltage to be higher than pin3, will cause a negative change at the output, pin 1; and visa-versa.
B) If there is a feedback path from pin1 to pin2, the output will change to try to keep pin2 voltage at the same level as the non-inverting input, pin3.

At switch-on, the voltage on pin3 will be 2.65V because R1=R2, and nothing will change that. With no current flowing through R3, pin2 will be +ve to pin3, so the voltage at the output will fall. That will turn on Q1 and it will pass current to bring the voltage on pin2 down to pin3 level.

This particular op-amp has very low input currents to its inputs, so whatever position the wiper of R3 is set to, the resistance between the wiper and pin2 passes no current, and does not cause a voltage drop. So pin 2 voltage and Q1 emitter voltage are the same.

If you now vary R3, the circuit will always reach stability when Q1 emitter equals 2.5V, - whatever the setting of R3 wiper. So with 2.65V across R3, changing the amount of resistance between the wiper and Vcc will proportionally change the current through it according to Ohms Law. As that current flows through Q1 and the meter, the meter will effectively measure the resistance of R3.

This sort of circuit is called a ‘constant current source’ and would typically be used to provide a (preset) constant current, for example if you needed to charge a capacitor to provide a linear ramp voltage in a saw tooth oscillator.

Hope that explains it well enough for you.