# deviation from ideality at low frequencies

Discussion in 'General Electronics Chat' started by mentaaal, Dec 7, 2008.

1. ### mentaaal Thread Starter Senior Member

Oct 17, 2005
451
0
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

2. ### mentaaal Thread Starter Senior Member

Oct 17, 2005
451
0
its ok guys i figured out what my problem was here. Its just that Aol gets multiplied by the parallel resistance which is where the R' comes from. Thanks anyway!

3. ### mentaaal Thread Starter Senior Member

Oct 17, 2005
451
0
Sorry guys, i thought i had this understood but i dont!

As mentioned above the characteristic feedback equation is:

Vout/Vin = 1/[β+1/Aol]

But the problem I am having is that when this model is used for a low frequency response for an integrator, the model is expressed like this:

Vout/Vin = 1/[β+1/Aol(R//Rd)]

Where Rd is the differential input impedance of the opamp.

But as mentioned in the above post, this is not algebraicly equivalent to the characteristic equation so how is it valid?