Integrator Circuit Transfer Function (voltage divide)

  • #1

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

[itex]

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

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

[itex]
v_{in} = C\frac{dv_{o}}{dt} + 16.23[/itex]

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.

Thank-you for your help
Lance

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

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  • #2
NascentOxygen
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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
That does make it more difficult to determine system dynamic parameters, ie rise time settling time and the such?
 
  • #4
NascentOxygen
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Your first post only mentioned Bode plot. If you are going to use Laplace, then let Vcap(s) = Icap(s)/(s.C)
 
  • #5
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

[itex]

sV_i = A s^2 V_o + B [/itex]

As you can see the Constant B makes creating the transfer function very difficult!
 
  • #6
NascentOxygen
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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? :confused:
 
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