Finding the transfer function for a difference amplifier

[Dextrine]

Homework Statement

I need to find the transfer function of the attached circuit (this isn't a homework question so I don't know if there's even a solution, but last time I posted this type of question on the EE forum, it was redirected here)

Homework Equations

I know it needs to be eventually of the form Vo/Vin and I know how to solve for Vo but I can't seem to get it as Vo/Vin. If it isn't possible, how would one describe this circuit's transfer function?

Z1=R1
Z2=R2+1/(s*C)[/B]

The Attempt at a Solution

Vo=(Z2/Z1)(5V-Vin)+5V

Vin is the measured signal from the output of a buck converter and 5V is the reference signal, if that helps at all. The closest I've somewhat gotten is
(Vo-5)/(5-Vin)=Z2/Z1

Thanks in advance for all the help![/B]

 

[Dextrine]

The only other thing I can think of would be just letting the transfer function be -(Z2/Z1) with an offset afterwards, but this seems weird...

 

[gneill]

I suspect that the circuit is not going to behave linearly so that a nice neat single expression transfer function for the behavior is not going to be possible. Consider...

Suppose for a moment that the input is open (Vin disconnected: an open circuit) when the power is first applied. What will happen to Vo?

The series capacitor is going to integrate any current through R1 (the top R1; You have two R1's in your circuit diagram) that results from the difference between Vin and 5 V. A continuous input Vin that is not exactly 5 V is going to lead to saturation issues... What happens when Vo hits a power rail?

 

[Dextrine]

Hello qneill, thanks for responding to my question. I'm not too sure I follow your suggestions, however, I came upon this http://www.ti.com/lit/an/slva662/slva662.pdf, which has a type II compensator which looks similar to the circuit I have. They kind of just neglect Vref in their transfer function, would I be able to do the same for mine?

 

[gneill]

It looks like they are only looking at the AC transfer function, which is why Vref is ignored (A DC source is just a short circuit to AC). Vref effectively places the two op-amp inputs at ground potential as far as AC is concerned. I suppose you could do the same if your goal is a similar analysis.

 

[Dextrine]

Yeah, the goal is just to find out which components determine the poles and zero's

 

[gneill]

Yeah, the goal is just to find out which components determine the poles and zero's

Okay. So "short out" Vref for the AC model and analyze the circuit.

 

[Dextrine]

Okay. So "short out" Vref for the AC model and analyze the circuit.

Will do, makes this MUCH easier. thanks!

 

[rude man]

The op amp is most likely operated from a single supply, necessitating the 5V input bias. The input is considered "zero" at +5VDC.

This circuit saturates for any finite offset voltage and/or mismatch between dc input voltages (dc input ≠ +5V exactly) so is useless.