Integrator Circuit Transfer Function (voltage divide)

1. Oct 1, 2012

qwertyuiop23

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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2. Oct 1, 2012

Staff: Mentor

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

3. Oct 1, 2012

qwertyuiop23

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

4. Oct 1, 2012

Staff: Mentor

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

5. Oct 1, 2012

qwertyuiop23

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!

6. Oct 1, 2012

Staff: Mentor

What's that DC offset on op-amp's (+) input in aid ofhttp://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon5.gif [Broken] You won't get anywhere while that's there, it will just send your op-amp output directly to the + rail! If that offset is to remain, then you'll need to exactly compensate for it by feeding DC into the (-) input, then superimpose your AC (for the Bode plot) on top of that DC. Capacitor coupling will achieve this.

Is this a circuit you invented as an exercise?

Last edited by a moderator: May 6, 2017