# Integrator Circuit Transfer Function (voltage divide)

## Main Question or Discussion Point

I have attached a picture of the circuit in which I am attempting to build a transfer function for, however, as you can see it is not a normal integrator. Normally it is evident which resistor is the resistor that is part of the series connection to the op amp.

When I do the KVL on the circuit to get the values I get the following function

$v_{in} = C\frac{dv_{o}}{dt} + 2.5\times R_1 (\frac{1}{R1} + \frac{1}{R_2})$

assuming that I voltage divide to $2.5V$ using values for $R_1 = 5k\Omega R_2 = 910\Omega$ and the input is 16.3V.

$v_{in} = C\frac{dv_{o}}{dt} + 16.23$

From here I do not know how to get the transfer function. I am correct in assuming that the transfer function is no longer a linear line on the Bode plot due to that extra resistor that goes from the input to ground (thus reducing the current).

Or am I way off base with my assumptions.

Lance

PS: I am an undergrad electrical engineer so the explanation does not have to be overly simplified.

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NascentOxygen
Staff Emeritus
You could solve the DE, then for sinusoids let vin = A.sinѠt

Or you could start again, and let vcap = iC/(jѠc)
where iC is the capacitor current

That does make it more difficult to determine system dynamic parameters, ie rise time settling time and the such?

NascentOxygen
Staff Emeritus
Your first post only mentioned Bode plot. If you are going to use Laplace, then let Vcap(s) = Icap(s)/(s.C)

Yes however, normally I use the Laplace form of the equation to make the transfer function to then recreate the Bode Plot.

The problem is that the Laplace transform is

$sV_i = A s^2 V_o + B$

As you can see the Constant B makes creating the transfer function very difficult!

NascentOxygen
Staff Emeritus