Comments on: The Differential Amplifier Common-Mode Error – Part 1 https://masteringelectronicsdesign.com/the-differential-amplifier-common-mode-error-part-1/ Electronics Design and Modeling with Emphasis on Analog Design Sun, 17 May 2015 03:10:51 +0000 hourly 1 By: Adrian S. Nastase https://masteringelectronicsdesign.com/the-differential-amplifier-common-mode-error-part-1/#comment-23624 Sun, 17 May 2015 03:10:51 +0000 http://masteringelectronicsdesign.com/?p=219#comment-23624 In reply to Jon.

I do not ground V1-V2 source. I simply eliminate it, by short-circuiting the two inputs together. In this way, the second term of equation 6 is derived by looking at the differential amplifier transfer function, which I showed in this article (see equation 1):

http://masteringelectronicsdesign.com/the-differential-amplifier-transfer-function/

In that article, if V1 equals V2, equation 1 becomes exactly the second term of equation 6 you refer to.

]]> By: Jon https://masteringelectronicsdesign.com/the-differential-amplifier-common-mode-error-part-1/#comment-23578 Fri, 15 May 2015 19:57:31 +0000 http://masteringelectronicsdesign.com/?p=219#comment-23578 When you ground the V1-V2 source to get the 2nd half of equation (6), how does that bring that terminal to V2 potential? It was stated that this results in a differential amplifier with both terminals at V2 potential.

]]> By: Adrian S. Nastase https://masteringelectronicsdesign.com/the-differential-amplifier-common-mode-error-part-1/#comment-17449 Sun, 27 Jul 2014 06:08:09 +0000 http://masteringelectronicsdesign.com/?p=219#comment-17449 In reply to Jeremy.

Thank you for your comment, Jeremy. Good idea. I will apply it in the future.

]]> By: Jeremy https://masteringelectronicsdesign.com/the-differential-amplifier-common-mode-error-part-1/#comment-17447 Sun, 27 Jul 2014 01:51:18 +0000 http://masteringelectronicsdesign.com/?p=219#comment-17447 Cheers for the great tutorial. However, I would appreciate if you would not omit the algebra between steps. You could easily use hyperlinks to obscure the intermediate steps, keeping the page clean and tidy.

]]> By: michael vaughn https://masteringelectronicsdesign.com/the-differential-amplifier-common-mode-error-part-1/#comment-7517 Sun, 16 Sep 2012 22:45:30 +0000 http://masteringelectronicsdesign.com/?p=219#comment-7517 totally agree it depends on the application, bio medical electronics uses instrumentation amps and it is very important to limit unwanted voltage and currents. 10 microamps from unwanted voltage can be deadly for patients, particularly if it gets through their heart catheter causing vfib onset.

]]> By: Adrian S. Nastase https://masteringelectronicsdesign.com/the-differential-amplifier-common-mode-error-part-1/#comment-6749 Sat, 23 Jun 2012 06:49:23 +0000 http://masteringelectronicsdesign.com/?p=219#comment-6749 In reply to Wayne.

I chose just one resistor to have tolerance for simplicity, and to make my point. Trying to calculate the common-mode error with tolerances t1, t2, t3, t4 as variables, will result in a mathematical relation with 4 variables. Too complicated, and unnecessary. If you need to see this result, better use a SPICE program, like Multisim, and perform a tolerance analysis. For a list of SPICE programs see my list SPICE Links.

]]> By: Wayne https://masteringelectronicsdesign.com/the-differential-amplifier-common-mode-error-part-1/#comment-6744 Fri, 22 Jun 2012 18:14:51 +0000 http://masteringelectronicsdesign.com/?p=219#comment-6744 In a practical application the resistors will all have a tolerance. Could you sum up the tolerances for the variable ‘t’?

]]> By: Adrian S. Nastase https://masteringelectronicsdesign.com/the-differential-amplifier-common-mode-error-part-1/#comment-26 Sun, 07 Jun 2009 16:01:03 +0000 http://masteringelectronicsdesign.com/?p=219#comment-26 Yes, indeed. In the numeric example of Solving the Differential Amplifier series there are two requirements: some gain for the voltage difference (actually sub-unity gain, which is attenuation) and a negative output voltage shift of almost one volt. This negative shift can be construed as common-mode voltage at the amplifier output. In that case, the requirements warrant the resistor calculations so that there is a significant gain for common-mode voltage.

]]> By: electricguy99 https://masteringelectronicsdesign.com/the-differential-amplifier-common-mode-error-part-1/#comment-25 Sun, 07 Jun 2009 05:02:36 +0000 http://masteringelectronicsdesign.com/?p=219#comment-25 This is in contrast with the previous series, Solving the Differential Amplifier. In this article, the common-mode voltage at the amplifier output is not desired. In the previous article, although you did not mention the common-mode voltage calculation, in effect the output has common-mode voltage, right?

]]> By: Adrian S. Nastase https://masteringelectronicsdesign.com/the-differential-amplifier-common-mode-error-part-1/#comment-23 Fri, 05 Jun 2009 02:33:20 +0000 http://masteringelectronicsdesign.com/?p=219#comment-23 It really depends on the application requirements. This article brings awareness about the common-mode error. Even if just one resistor has some tolerance, the error is large enough to be important in precision applications. In my article I used just one resistor tolerance for simplicity and demonstration purpose. In reality, all four resistors tolerances have to be taken into consideration so the error could be even larger. 10 mV error may be important versus a differential signal of 100 mV to 400 mV.

In the next part I will discuss the case in which the differential amplifier is used to measure the current with a small resistor. The drop on the sense resistor is small but the common-mode voltage can be high so the current may be measured with some error.

Therefore, to answer your question, 0.1% resistors might be good enough for some applications. For high precision ones, the design of the differential amplifier should have 0.05% resistors or instrumentation amplifiers like the Analog Devices AD621 or AD629.

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