Hey guys, I am having trouble understanding my notes in relation to the response of an integrator at low frequencies. As you all know the response of an opamp is 1/[β+1/Aol]
where Aol is the open loop gain of the opamp and β is the feedback fraction.
This is mathematically quivalent to: (1/β)[1/(1+1/βAol) this is used to calculate the gain error
For an integrator the gain is equal to -Xc/R
where Xc is the impedance of the cpacitor and R is the input resistor.
What i dont understand from my notes is when this is applied to the gain error formula mentioned above.
This is expressed as:
(1/2∏fRc)[1/(1+1/[2∏fR'C])]
where R' is the parallel resistant of the input resistor and the input differential input impedance of the opamp.
The reason why I dont understand this is because this expression is mathematically not equivalent to the first expression for the response of an opamp. So how is this valid? This would make sense to me if R' was used in the above expression in place of R on the left.
Could someone please explain this to me
where Aol is the open loop gain of the opamp and β is the feedback fraction.
This is mathematically quivalent to: (1/β)[1/(1+1/βAol) this is used to calculate the gain error
For an integrator the gain is equal to -Xc/R
where Xc is the impedance of the cpacitor and R is the input resistor.
What i dont understand from my notes is when this is applied to the gain error formula mentioned above.
This is expressed as:
(1/2∏fRc)[1/(1+1/[2∏fR'C])]
where R' is the parallel resistant of the input resistor and the input differential input impedance of the opamp.
The reason why I dont understand this is because this expression is mathematically not equivalent to the first expression for the response of an opamp. So how is this valid? This would make sense to me if R' was used in the above expression in place of R on the left.
Could someone please explain this to me