A question of opamp's load

[hiwar]

I'm designing a RC integrator with a 1µF Cap connected between opamp's negative input and output. When I design the opamp circuit, how to determine the load of the opamp? Is it 1µF? I don't know.
Thank you for your help.

 

[DrWhoF]

Load means everything connected to opamps output. In your case 1µF is connected to input with high input impedance so the equivalent impedance is quit high and doesn't represent significant load comparing with whatever is going to be connected to this opamp.
DrWho

 

[nandu_r]

load is (r||cinput) + 1/jwcf
r = resistor in feed back network
cf = feed back capacitor (1uf)
cinput = input capacitance of opamp

 

[hiwar]

But there will be a 1uF Cap connected to the opamp's output according to the Miller effect.
I'm confused.

 

[sefton]

If I remember my opamp theory correctly, you should have the + input connected to 0V which means that the opamp will maintain the voltage at the - input to 0V, which means that one side of the capacitor is connected to the output and the other side effectively connected to 0V. This means that the capacitor is the output load.

 

[huojinsi]

The 1uF capacitor isn't load ,and it is feedback device. The load should equate out resistance of OPA.

 

[Kral]

hiwar,
In your circuit configuration the load is the 1uf capacitor in parallel with any external load applied to the op-amp output. Remember, in the inverting amplifier configuration, the inverting input is held at virtual ground. The consequence of this is that the current through the capacitor current (which must be supplied by the op-amp output) is equal to C dv/dt. This is the same current that the op-amp output would have to supply if the capcitor were connected from the op-amp output to ground in a conventional amplifier circuit.
Regards,
Kral