Op Amp Integrator Differentiator Help

Thread Starter

anik321

Joined Oct 25, 2008
22
Im decently new with opamps and am having a hard time coming up with a transfer function for this circuit:




The professor in class has asked us to find its Voltage gain (magnitude and phase) for different frequencies and then asks if its possible to do it with just one Opamp.

1. It looks to be an integrator followed by a differentiator. What is the point of it then?

2. Can anyone please help me derive a transfer function (Vout/Vin)?

3. I am confused as to how C1 interacts with the two stages.

4. What effect does the 10M (R1) have on the integrator?

This forum is usually my last resort, so any help would be wonderful.
 
Last edited:
i think the 10M resistor is there to stop the integrator ramping up to rail voltage by providing some definite amount of negative feedback. it isn't going to have much current through it though, obviously.

if you pretend that the 10M isn't there, and do simple ohms law calling Va the voltage at the output of the first amp:
Vi/R = -Va/(1/wC)

this rearranges to:

Va = -Vi/RwC

then Va/(1/wC) = -Vo/R

but we know Va so subbing in:

(-Vi/RwC)/(1/wc) = -Vo/R

and this simplifies to: Vi/R = Vo/R

i.e. it has a gain of 1. A buffer.
 
for the phase, well you just need to put the j's back into the first equations and do some more maths. I took them out just to show the magnitude was 1.

But don't take my word for it, wait until one of the oldtimers shows up to verify it, or tell you the right answer!!
 

Thread Starter

anik321

Joined Oct 25, 2008
22
Fraser,

thats what I thought (that the gain would be ~1). Which brings me back to my original question, whats the point of it? Can you think of a place where you would use this instead of a +1 buffer?
 

Thread Starter

anik321

Joined Oct 25, 2008
22
Doing it the normal way (without omitting R2):

First the parallel combination of R2 and C1:

R2//C1 = R2/(1+jwC1R2), where xC= 1/jwC1

Now going back to the ckt:

Vin/R1 = -Va/(R2//C1) = -Va/[R2/(1+jwC1R2)]

or, Va = -VinR2/[R1(1+jwC1R2)]

Also,

Va/(xC2) = -Vout/R3

Substituting xC2 = 1/jwC2 into the above equation we have:

Va/(1/jwC2) = -Vout/R3

or, -Vout = Va * [jwR3C2]

But we know Va, so substituting:

-Vout = -VinR2/[R1(1+jwC1R2)] * [jwR3C2]

Or Vout = Vin [(wR2R3C2/(R1(1+jwC1R2))]

Or Vout/Vin = [jwR2R3C2/(R1(1+jwC1R2))]

Who wants to help me convert this to magnitude and phase angle?? I am lost on how to do this.
 

The Electrician

Joined Oct 9, 2007
2,767
Doing it the normal way (without omitting R2):

First the parallel combination of R2 and C1:
I don't see a parallel combination of R2 and C1. I do see a parallel combination of R1 and C2.

R2//C1 = R2/(1+jwC1R2), where xC= 1/jwC1

Now going back to the ckt:

Vin/R1 = -Va/(R2//C1) = -Va/[R2/(1+jwC1R2)]

or, Va = -VinR2/[R1(1+jwC1R2)]
What is Va? I suppose I could figure it out, but when you're asking for help, don't force the person helping to spend time figuring out what undefined symbols are. Provide a schematic with labels for the symbols you've used, or explain in text what they are. Don't just start using undefined symbols with no explanation.

Anyway, it looks to me like you should have:

Vin/R3 = -Va/(R1||C2) (the parallel symbol is above the backslash)

instead of:

Vin/R1 = -Va/(R2||C1)

Were you looking a different schematic than the one you posted?

Also,

Va/(xC2) = -Vout/R3

Substituting xC2 = 1/jwC2 into the above equation we have:

Va/(1/jwC2) = -Vout/R3

or, -Vout = Va * [jwR3C2]

But we know Va, so substituting:

-Vout = -VinR2/[R1(1+jwC1R2)] * [jwR3C2]

Or Vout = Vin [(wR2R3C2/(R1(1+jwC1R2))]

Or Vout/Vin = [jwR2R3C2/(R1(1+jwC1R2))]

Who wants to help me convert this to magnitude and phase angle?? I am lost on how to do this.
You need to do all this over so it matches the schematic you posted.
 

Thread Starter

anik321

Joined Oct 25, 2008
22
Electrician, YES I WAS!

My sincerest apologies. I have fixed the schematic. THANK YOU.

Sorry to have wasted your time there.

I plugged in all the values for the components and did a bode plot but the answers (specially for the magnitude) do not seem to match up.

Care to take another look please?

P.s I just used 747s to draw the ckt in spice.
 

The Electrician

Joined Oct 9, 2007
2,767
Using your corrected schematic, I get the same transfer function you do.

As far as plotting goes, the magnitude is gotten by multiplying the transfer function by its conjugate and taking the square root.

The angle is the arctangent of the imaginary part divided by the real part.

See the attached image for the plots.
 

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