# An Op-Amp question

Hi Guys!

well!

im just a bit confused here.

i want to ask you a question:

just have a look at the fig below which i have posted. You can see a load resistor "RL". my question is that whenever i studied this inverting configuration of opAmp, there is no load resistor given for it. So, now when i have to find the equation of its gain , with a load resistor, do u think that it will change from the one without a load resistor. As i know that for the inverting configuration without any load resistor we have:

Vo/Vi = -R2/R1

now do u think that with the load resistor RL, this gain equation will change?

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The equation doesn't change. You can get this equation by wrighting a sum of currents on - connection. If you add a load resistor V0 doesn't change thus you can write Uy/R1+V0/R2 = 0 and you get the same equation.

-R2/R1 is the unloaded gain. When loading is taken into account the gain becomes -(R2/R1)*RL/(RL + Ro) where Ro is the output impedance. With a typical op amp, Ro is low by design and is commonly neglected. As long as RL is above 100 ohms or so, you may neglect loading (in my experience).

-(R2/R1)*RL/(RL + Ro)

i would appreciate if you will tell me how to get the above equation

because i have tried my level best but couldnt get the logic to get the above equation!

plz....

Now plz try this one

Now after that circuit, plz try the below one, this one i gets into too much complication.

now i want to ask you again that does the gain equation will change for the following circuit with a load resistor than the one without load resistor.

if i have to find the close loop gain that is Vo/Vi if the open loop gain is:

1) infinite

2) 1000

now will the gain equation will vary?

plz help me as soon as possible

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shaiqbashir said:
-(R2/R1)*RL/(RL + Ro)

i would appreciate if you will tell me how to get the above equation

because i have tried my level best but couldnt get the logic to get the above equation!

plz....

Use the formula for voltage division over two resistors in series:

V1 = V*R1/(R1 + R2)

So you must insure that the load impedance is large compared to the output impedance, for maximal voltage amplification.

shaiqbashir said:
Now after that circuit, plz try the below one, this one i gets into too much complication.

now i want to ask you again that does the gain equation will change for the following circuit with a load resistor than the one without load resistor.

if i have to find the close loop gain that is Vo/Vi if the open loop gain is:

1) infinite

2) 1000

now will the gain equation will vary?

plz help me as soon as possible

This is a bit more complicated than your first example but can be solved in much the same way. It just involves a little more algebra. In your analysis you will be assuming an ideal op amp, i.e. infinite open loop gain and input impedance and zero output impedance. You should end up with a gain of:

-(R2*R3 + R2*R4 + R3*R4)/(R1*R4)

A typical op amp (such as the LM324) is usually a decent approximation of an ideal op amp. But an open-loop gain of 1000 is really low. I don't know if any actual op amp has such a low gain.

Again, the loaded gain depends on the output impedance (typically 50 - 3k ohms, and reduced by negative feedback).

My advice would be to simulate the circuit in pSpice or a similar program, or assemble it on a breadboard. Then you may tweak the design to get the desired results.