Vcm... why? - Page 1
Forum section: electronics / beginners

author: ilbs
author_rank: Contributor
timestamp: September 20, 2013, 08:47:07 pm
content: Hello,Why would anyone want to know the value of the common mode gain of a differential amplifier?We conveniently have the differential equation to solve the voltage output (Using the resistor values) ... so why go through all the math to find Vcmthanks

author: IanB
author_rank: Super Contributor
timestamp: September 20, 2013, 08:58:06 pm
content: Because the actual output voltage of a differential amplifier has two components, the differential part and the common mode part:Vout = Vout(Diff) + Vout(CM)Ideally, Vout(CM) will be zero and Vout = Vout(Diff).But the real world is rarely ideal and Vout(CM) will be present also as an error term. It goes without saying that output errors are bad.If you cannot account for the common mode content of the output voltage you cannot accurately characterize how your amplifier will perform and whether it will meet requirements.This is why the common mode rejection ratio (CMRR) is an important parameter.

author: ilbs
author_rank: Contributor
timestamp: September 21, 2013, 01:59:09 pm
content: QuoteWhat requirements? Can you give a concreteexample or circuit situation as to why the errorat the output caused by Vcm would be so importantto know or to correct?For example if I am doing a diff amplifier, I simplypick equal value resistors and calculate vout by the differentialequation and the way I go .... it works !!!I learned all the math that goes along withVd, Vcm, Ad, Acm, CMRR, CMR etc....and that an amplifier has two gain components andthat we need to consider both of them and ect...I just don't know why I learned all that though!!Maybe I should just keep all this Vcm stuffin my personal notes until it makes sense to methanks

author: c4757p
author_rank: Super Contributor
timestamp: September 21, 2013, 02:04:30 pm
content: Say you want to use a diff amp to measure the drop in a high side current sense resistor, 0.01 ohm, 1-10 amp, 12V above ground. Assume perfectly matched resistors for the purpose of demonstration. Now, go pick a decent op amp, read the datasheet, do the math and figure out how much error you will have due to common-mode voltage. I think you will be surprised.

author: ilbs
author_rank: Contributor
timestamp: September 21, 2013, 02:14:32 pm
content: Okay, so what you are saying is that there actually aresituations where the Vcm (CMRR) is so important thatit can significantly throw off the functionally anop amp is trying to achieve.Okay, in theses cases I presume one can either getbetter precision resistors or change op amp ....Given that I never ran into such situations....I will be patient and wait for such a circuit to arise.thanks

author: c4757p
author_rank: Super Contributor
timestamp: September 21, 2013, 02:26:04 pm
content: There are many such situations. I'm working on a precision power supply as a hobby project and had a bit of a search trying to find an op amp that was good enough for current sense and reasonably affordable. CMRR was one of the biggest factors.Resistor error can sometimes be calibrated out, thankfully, since I also knew the voltage output and therefore the common mode offset. But the op amp's built-in CMR error isn't always as stable and predictable.

author: AndyC_772
author_rank: Super Contributor
timestamp: September 21, 2013, 02:39:30 pm
content: That's why chips like the INA271 exist, they're designed for the purpose and will accept inputs well above their own supply voltage.http://www.ti.com/product/ina271

author: Bored@Work
author_rank: Super Contributor
timestamp: September 21, 2013, 03:00:11 pm
content: Quote from: ilbs on September 21, 2013, 02:14:32 pmThis gets uneconomic pretty fast, in terms of money, board space, etc.0.1% resistors with low tempco won't give you better than around 60 dB CMRR. Depending on what you do even for 60 dB CMRR it might already be cheaper to look for an of-the-shelf integrated instrumentation amplifier or difference amplifier. Lest hassle, less components, less board space, less costs.

author: c4757p
author_rank: Super Contributor
timestamp: September 21, 2013, 03:19:59 pm
content: Quote from: AndyC_772 on September 21, 2013, 02:39:30 pmYep, that's precisely what I ended up using.

author: Marco
author_rank: Super Contributor
timestamp: September 22, 2013, 09:48:00 pm
content: Quote from: AndyC_772 on September 21, 2013, 02:39:30 pmHow does that work any way? Is it some kind of switched capacitor circuit inside?

author: c4757p
author_rank: Super Contributor
timestamp: September 22, 2013, 09:52:01 pm
content: It's a voltage controlled current sourcethat sits on a virtual ground at the sense resistoredit-no, that specific one does not-it's still a VCCS though. The current then feeds down into a resistor tied to (real) ground, and the voltage on this resistor is buffered.Many of them omit the resistor and buffer.

author: ilbs
author_rank: Contributor
timestamp: September 23, 2013, 03:08:55 am
content: okay thanks guys