Question regarding ideal OP-amps

1. Aug 25, 2010

Ry122

Since both the + and - inputs of an ideal OP-amp are supposedly always equal to each other, how is it that there can be any increase in voltage on the output of the OP-amp when the equation which describes the relationship between the OP-amp input and output is Vo = Gain(Vpos-Vneg) ?

2. Aug 25, 2010

skeptic2

The + and - inputs are not exactly equal to each other. As you have guessed, there must be a difference between the inputs in order to get a change in output voltage. How much of a difference in input voltages would you need to get a change in output voltage of 1 volt if the gain of the opamp is 100,000?

The output of an opamp is not Vo = Gain(Vpos - Vneg), it is Vo = Vpos + Gain(Vpos - Vneg).

3. Aug 25, 2010

schip666!

Think of the feedback resistor (that sets the gain) as allowing the output to force the +/- inputs to be equal. With a high ratio resistor divider in the feedback, the output needs to swing further to counteract whatever is happening on the input. The "always equal" bit is subject to slew-rate limitations, meaning it can't keep up at higher frequencies. The magic of feedback is that (below the limit speed) everything is kept in balance with pretty much undetectable signal fluctuations.

In motor control systems you can see the feedback signal overshoot when the motor speed changes because of mechanical delay in the motor. Putting a capacitor to ground in the middle of the opamp feedback divider (from - to ground) slows the feedback response in the same way, so, with the right capacitance, you should be able to see an overshoot on the output when trying to track a pulse input.

4. Aug 26, 2010

zomgineer

You should probably analyze at least one op amp configuration where the op amp is not ideal (use offset currents, finite resistance between inputs, etc.) Try the negative feedback setup with Vout = G(V+ - V-) and all the imperfections. Everything will become clearer.