Differential op amp Vout question - Page 1
Forum section: electronics / beginners

author: DuncanSteel
author_rank: Contributor
timestamp: January 24, 2017, 02:03:18 pm
content: HiI would like to know why my real time measurements & calculation don`t match up?I have a load & I amplify the difference between 2 wires with an op amp & 1Mohm resistor. Load cell has 4 resistors. You can see the values in the image below.In real time I measure Vout as 1.6V but with calculations I get very strange results. (see the image please). Why ?If I use multimeter to measure voltage between Green & white wire I get 2mV

author: Audioguru
author_rank: Super Contributor
timestamp: January 24, 2017, 02:09:18 pm
content: In your differential amplifier, R3 and R4 should have the same value.

author: DuncanSteel
author_rank: Contributor
timestamp: January 24, 2017, 03:39:17 pm
content: Quote from: Audioguru on January 24, 2017, 02:09:18 pmIf R3 = R4 then both = 750 Ohm and output measured is 2,3V and the gain is 0. This cant be right.

author: rstofer
author_rank: Super Contributor
timestamp: January 24, 2017, 04:13:31 pm
content: Section 6.5 might help:http://www.cypress.com/file/65366/downloadI think if I were building such a thing, I would use a true instrumentation amplifier:http://websrv.mece.ualberta.ca/electrowiki/images/3/3c/Strain_Design.pdfPerhaps I would use the chip in the document or maybe I would start from scratch with:http://www.analog.com/media/en/technical-documentation/data-sheets/AD524.pdfor similar

author: tron9000
author_rank: Frequent Contributor
timestamp: January 24, 2017, 04:41:25 pm
content: The equation:Vout = (-R3/R1)(V1-V2)is only true if R3 & R4 are equal and R1 & R2 are equal.This is a mismatched differential amp and will not work as you expect it to. If you make:R3=R4 & R2=R1, this should give a gain = R3/R1 = R4/R2Also I'm not sure where you got Vout = 1400V from? That's way out of the operating range of any op-amp, so that would be ringing alarm bells in my head!

author: Zero999
author_rank: Super Contributor
timestamp: January 24, 2017, 07:02:54 pm
content: Quote from: DuncanSteel on January 24, 2017, 02:03:18 pmThe resistances in your strain gauge circuit don't add up. The resistances between red & white and red & green wires are 841R so R1 & R3 = 841. The resistances between white & black and green & black are 750 so R2 & R4 = 750R.The resistance between red and black should be equal to (R1 + R2) and (R3 + R4) parallel but it isn't.R1 + R2 = 841 + 750 = 1591R3 + R4 = 1591Since both the resistances are the same, we get half the resistance when they're connected in parallel (easier than doing product over sum):R (between the red & black wires) = 1591/2 = 795.5But you measured 1091 which can't possibly be right.Did you measure them with anything else in the circuit?It is possible to use the resistance of the strain gauge as part of the differential amplifier circuit but R1 & R2 become the equivalent of the top and bottom resistances in parallel and it's important the bridge is well balanced. I would use R1 & R2 in the differential amplifier circuit and add the strain gauge resistance on to them, when performing the calculation. I'll post an example later.https://datasheets.maximintegrated.com/en/ds/MAX4238-MAX4239.pdfSee attached. The circuit has a gain of 100. The high and low side resistors in parallel equal to 500R which is added to the values of R1 & R2 to make 2k7.

author: David Hess
author_rank: Super Contributor
timestamp: January 25, 2017, 04:07:54 pm
content: Quote from: Hero999 on January 24, 2017, 07:02:54 pmAll of the schematics of the bridge shown so far are wrong including yours. There is at least a modulus compensation resistor in series with one or both of the excitation leads. Not coincidentally, roughly 90 ohms total is a typical value. The modulus compensation resistor is temperature sensitive and corrects the bridge for the temperature coefficient of modulus of elasticity of the material.So the resistance between the excitation leads will be greater than the resistance between the signal leads.

author: DuncanSteel
author_rank: Contributor
timestamp: January 25, 2017, 07:10:12 pm
content: Quote from: David Hess on January 25, 2017, 04:07:54 pmSo what is the not wrong schematic? I built up this schematic (it works, but has bad noise with 10Bit adc - Atmega32u4) but I do not understand how to calculate the gain & output of the circuit. Thats is why im asking how to calculate the numbers.

author: Zero999
author_rank: Super Contributor
timestamp: January 25, 2017, 09:45:46 pm
content: Quote from: DuncanSteel on January 25, 2017, 07:10:12 pmHe meant the schematic for the strain gauge is wrong.You need to know the correct schematic for the strain gauge, in order to calculate the gain of the differential amplifier used in your schematic.Personally, I'd use the INA101, which has a high input impedance, so the resistance of the strain gauge can be ignored.http://www.ti.com/lit/ds/symlink/ina101.pdf

author: danadak
author_rank: Super Contributor
timestamp: January 25, 2017, 10:11:15 pm
content: More than you want to know about diff amps, includes gain equations -http://www.analog.com/en/education/education-library/dh-designers-guide-to-instrumentation-amps.htmlRegards, Dana.

author: David Hess
author_rank: Super Contributor
timestamp: January 25, 2017, 10:39:34 pm
content: Quote from: Hero999 on January 25, 2017, 09:45:46 pmIt is just that practical load cells use elements which are not shown in the typical schematic leading to a surprises when the resistance of the various legs of the bridge are measured.QuoteThe load cell will have a specified sensitivity from say 0.5 to 3 millivolts per volt of excitation at full load; 2mV/V or 3mV/V is typical. This takes into account any modulus and other compensation which is not shown in the 4 resistor bridge schematic. So with a 3mV/V sensitivity and 10 volts of excitation, the full scale differential output will be 30 millivolts and from there, the needed gain of the amplifier can be easily calculated.If the sensitivity is not marked, then apply a known excitation and measure the change in the differential output between zero and a known load. Then the sensitivity can be calculated.QuoteThe change in resistance of the bridge creates insignificant non-linearity error compared to the non-linearity of the load cell itself so there is no need to use an amplifier with high input impedance; a 4 resistor instrumentation amplifier will work fine in all but the most exacting applications.

author: DuncanSteel
author_rank: Contributor
timestamp: January 27, 2017, 09:08:07 pm
content: Thank you guys you were helpful.