Op-amp pressure sensor application transfer function - Page 1
Forum section: electronics / beginners

author: alex.martinez
author_rank: Contributor
timestamp: March 08, 2017, 04:00:19 pm
content: Saludos!I've been trying to solve for the transfer function of a circuit which was proposed on my Measuring Instrumentation lecture, I have been trying all week to solve for its transfer function (which is the goal of the exercise) as a function of all the given parameters, the variable capacitor (the pressure sensor is noted as Cx. What grinds my gears is the second feedback across the R1 resistor, I don't get how it can be connected to a node which is shared with Vp, with no resistor, what is its role? I attach the circuit diagram.Thank you in advance, Alex.

author: orolo
author_rank: Frequent Contributor
timestamp: March 08, 2017, 04:22:01 pm
content: There is no feedback across R1, since it's connected to the virtual ground of the second op amp. You have an inverting adder there. What is a bit disturbing, to me, is the the lack of D.C. path to ground in the inverting input of the first op amp.

author: alex.martinez
author_rank: Contributor
timestamp: March 08, 2017, 08:08:40 pm
content: Quote from: orolo on March 08, 2017, 04:22:01 pmIt opened my eyes! Thank you very much! That particular resistor will not affect the transfer function, does it? As I see it now is like to have a point of reference to ground in the adder. And one last doubt, what do you mean by D.C. path to ground?Thank you a lot, sincerely,Alex.

author: orolo
author_rank: Frequent Contributor
timestamp: March 08, 2017, 08:36:18 pm
content: Quote from: alex.martinez on March 08, 2017, 08:08:40 pmThe first op-amp has a transfer function Va = -Co/Cx Vin. The second op-amp makes an inverted addition of Vin and Va, so: Vo = -R2/R1·(Va + Vin) = -R2/R1 · (1 - Co/Cx) · Vin, if I'm not mistaken. So the resistor affects the transfer function, because it adds the initial signal. I think this is some kind of bridge that will be cancelled if Co = Cx.QuoteSince the inverting input of the first op amp is only connected via capacitors, there is no definite DC level at that point. I think a resistor from there to ground, or a resistor in parallel with Cx, is missing, in order to fix the DC level at the inverting input. I've never used this specific kind of amplifier, so perhaps someone more knowlegdeable can shed some light.I'm glad I could help a bit.

author: alex.martinez
author_rank: Contributor
timestamp: March 08, 2017, 09:55:55 pm
content: Quote from: orolo on March 08, 2017, 08:36:18 pmIndeed, how could I have missed it, I got too obsessed with it as a feedback loop without seeing that it is actually after the first stage! Hahaha. The diagram is just an exercise for the concept of seonsor that bring very small variations, so I guessed they did not bother to include it, and once again, thank you a lot!My regards, Alex.

author: Ratch
author_rank: Regular Contributor
timestamp: March 09, 2017, 12:11:42 am
content: Quote from: alex.martinez on March 08, 2017, 04:00:19 pmThe transfer function is relatively easy to compute. I don't know what Vp is because I don't see it in the schematic. You have two resistors labeled R1, which is confusing. The term in the smaller parenthesis is the output voltage of the left op-amp. The terms within the large parenthesis are the total current from the left op-amp and Vin across the bottom R1. Multiplying by R2 gives Vo.Ratch