Need explanation of OPAMPs mechanic

[john2020]

OPAMP problem

hi all

i have a basic question about OPAMPS

In case of OPAMP .... when it acts as an inverting amp .. the potential at both inverting and noninverting terminal r same due to vertual short principle .. all of us know OPAMP is a differential amp ... means it amplifies the difference of voltages applied at both the terminals... In this case which is ZERO so the OPAMP's O/P should b zero but its not true...

I know i am somewhere wrong coz in reallity its not true.... Can u people help me to trace out where i am wrong...? please..............

regards
john

 

[rkodaira]

Re: OPAMP problem

Hi !

In fact the Op Amp is a differential amplifier, but the feedback loop (the Rf resistance) changes the thing, so the output voltage should be that voltage that balances again the two inputs. The output voltage in feedback amplifier always try to equalize both inputs (pins) of the op amp and does not mean the differential voltage between these pins amplified (but doing so, it amplifies the input voltage applied to the resistance input of the amplifier).

 

[Davood Amerion]

OPAMP problem

Op-amps have very high voltage gain.
Any time we divid output voltage to input diffrence voltage ( with or whitout
feedback in active region) we get a number which is equal to op-amp gain.
op-amps gain normally is above 100,000
eg. when output voltage is 5Volts input diffrence must be 5/100000=50uVolt or 0.05 mV,
you see that this is very small voltage.

regards
Davood Amerion

 

[demodb]

Re: OPAMP problem

this due to mismatches in the device. This causes an offset voltage, this can be compensated at the input. See razavi book, chapter 13.

 

[Davood Amerion]

OPAMP problem

dear demodb
offset voltage is diffrent from input signal voltage

 

[aryajur]

Re: OPAMP problem

I think this will clear the confusion. As you said yourself, the positive and negative terminal are at a VIRTUAL short. The word virtual is added for precisely the reason that they are not actually exactly the same. So they are in fact different. And V- is different by V+
So if the gain of the opamp is Aol which would define Vout as
Vout = Aol[ (V+) - (V-) ]
i.e. the difference of the positive and negative terminal voltages multiplied by the gain of the opamp. This is the exact equation. So going on these lines we can solve for the Vout and V- in terms of Vin. This is shown in the image attached.
But now look at the expressions for Vout and V-. Aol is usually in the order of thoudands ! So the terms divided by Aol are negligible and Aol is very large in V- equation.
So you see effectively Vout = - R2/R1 Vin and V- is approx = 0 (thats why VIRTUAL short)

But now think of doing all these problems using these equations and think of the labour you save by just considering V- = V+ and then just solving them. So thats why we do it like that and the error is very insignificant for a high gain opamp. Like why would you bother for the 5th or 6th place of the decimal in your answer anyway.

 

[usernam]

Re: OPAMP problem

hi all

i have a basic question about OPAMPS

In case of OPAMP .... when it acts as an inverting amp .. the potential at both inverting and noninverting terminal r same due to vertual short principle .. all of us know OPAMP is a differential amp ... means it amplifies the difference of voltages applied at both the terminals... In this case which is ZERO so the OPAMP's O/P should b zero but its not true...

When you assume that the op-amp is ideal, it has infinite open-loop gain. So infinityxzero is some finite value.
But in practice the open-loop gain is not infinite. Its just very high abt 100,000 or so.
So the voltage diff across the i/p terminals will be very very low. So for all practical purposes they are at the same potential.

 

[sabari]

OPAMP problem

REFER TO SMITH AND SEDRA U HAVE GOOD IDEA ABOUT THAT

 

[smalik]

OPAMP problem

AN OP-AMP has high input impedance N ideally its terminals does not draw any current the. whether in inverting or non inverting configuration the voltage at the terminals is same the bcoz the input terminals do now draw current due to high impedance. ofcourse output is the difference of 2