mona123
Member level 5
I am still not clear what is the physical mechanism that makes opamp inputs track each other? we say it is because opamp has high gain. can someone explain better. how high a gain is needed? thanks.
For the input difference to be small, the op amp must be operating in a linear mode with negative feedback.........................
thanks crutschow, so are you saying input difference is low due to negative feedback and is not dependent on open loop opamp gain? that's where I am confused. so if I make a single stage or two stage opamp it won't matter?
Perhaps it helps to analyze what happens after switch-on the opamp power supplies:
Example: Supply voltages +/- Vs=+/-10V; non-inverting gain stage with gain of "+2". That means: Feedback factor k=0.5.
1.) Apply at t=0 an input voltage Vin=1V. The opamp is not yet working in its linear range and the output will be either at Vs=10V or Vs=-10V (lets assume +10V)
2.) The voltage at the inverting terminal will be 0.5Vs=5V>Vin=1V. Hence, the voltage at the inverting terminal dominates (is larger) and the output voltage will change in the direction to -Vs.
3.) However, on its way to -10V the ouput voltage is crossing a positive value which produces at the inv. terminal a feedback voltage of +0.99980004V .
4.) At this very moment (assuming an open-loop gain Aol=1E4) , the opamp is in its linear amplification region and produces an ouput of nearly +2V: (1-0.99980004)*1E4=0.00019996*1E4=1.9996001V.
5.) That means: We have an equilibrium because the output voltage has a value which exactly meets the closed-loop gain Vout=Vin*[1E4/(1+0.5*E4)]=1.9996001 V.
6.) In this example, the input difference voltage, of course, is NOT zero. It never will be zero - however, the diff. voltage is so small (in our case app. 0.0002V) that in can be neglected (assumed to be zero for calculations) in many cases.
With a linear gain of >1 million, the output can only be linear by negative feedback to force in differential input near zero.