Talk:Common-mode rejection ratio

	Electronics portal This article is part of WikiProject Electronics, an attempt to provide a standard approach to writing articles about electronics on Wikipedia. If you would like to participate, you can choose to edit the article attached to this page, or visit the project page, where you can join the project and see a list of open tasks. Leave messages at the project talk pageElectronicsWikipedia:WikiProject ElectronicsTemplate:WikiProject Electronicselectronic articles
???	This article has not yet received a rating on the project's importance scale.

Untitled[edit]

741 is a "good one"? — Omegatron 17:03, 21 February 2006 (UTC)[reply]

I think "good one" means "desirable characteristics" -- the CMRR of the LM741, found at http://www.national.com/ds/LM/LM741.pdf shows it to be around 90dB optimal. I've re-worded it, though. Mystic Pixel 08:17, 16 March 2006 (UTC)[reply]

Formula[edit]

where is Vo in the formula 1? And what is Ad and As? Please explain the formulas in more detail! —Preceding unsigned comment added by Rdenooij (talk • contribs) 26 Feb 2007

Equation[edit]

Originally the equation specified the meaning of Ad and As:

where $A_{\mathrm {d} }$ is the differential gain

$A_{d}=V_{0}/(V_{+}-V_{-})\,$

and $A_{\mathrm {s} }$ is the common-mode gain

$A_{s}=V_{0}/V_{s}\,$

Why was this changed to a single equation?

V_{\mathrm {o} }=A_{\mathrm {d} }(V_{+}-V_{-})+{\frac {1}{2}}A_{\mathrm {s} }(V_{+}+V_{-}).

This isn't as clear about what the terms mean or how they would be measured. And why does it have a 1/2 term? — Omegatron 15:12, 26 February 2007 (UTC)[reply]

This is what I understood from the -- somewhat obscure -- explanation that was originally on the page: As is the amplification factor for a voltage Vsin that is on both inputs. So if V+ = V- = Vs, then Vs = (V+ + V-)/2. It could be that the original definition was wrong, I didn't check that. I agree that the equation looks kind of asymmetric with the factor 1/2. It would make 6 dB difference (on typically 90 dB) if you leave out the factor 1/2. Han-Kwang 18:36, 26 February 2007 (UTC)[reply]

Oh, I see. It depends on the circuit model you are using. [1] [2] — Omegatron 00:42, 3 March 2007 (UTC)[reply]

Moreover, I don't think the original eqs are "as clear about what the terms mean or how they are measured". Vo has a different meaning in the 1st and 2nd and it is tempting (but utterly incorrect) to cancel out Vo when you calculate Ad/As. And if I look at your ref.2, the equation would be

V_{\mathrm {o} }=A_{\mathrm {d} }(V_{+}-V_{-})+A_{\mathrm {s} }(V_{-}),

which is mathematically equivalent to

V_{\mathrm {o} }=A_{\mathrm {d} }(1-{\frac {A_{s}}{2A_{d}}})(V_{+}-V_{-})+{\frac {1}{2}}A_{\mathrm {s} }(V_{+}+V_{-}).

Since 1-As/2Ad is approximately 0.99998 for a typical setup, these two definitions are equivalent for all practical purposes (the nonlinearity in Ad is probably much larger than the difference anyway). I am therefore removing the inaccuracy tag from the article. Han-Kwang 11:46, 7 April 2007 (UTC)[reply]

I don't think the lack of a schematic warrants the {{accuracy}} tag, Omegatron. I think you should actually explain what part is inaccurate if you are to use that tag. I have explained here why I believe the current description is correct. Moreover, adding an electric circuit diagram would restrict the CMRR description to a specific type of circuit (such as an ideal opamp wired to behave as a low-gain voltage substractor), while the concept applies to a much more general class of devices with bi-linear inputs. Han-Kwang 19:56, 11 May 2007 (UTC)[reply]

Definition (why 20-log?)[edit]

Given that a decibel is a tenth of a Bel, why is the defintion 20log.. and not 10 log..?
--83.105.33.91 09:46, 24 September 2007 (UTC)[reply]

See decibel. With voltages, it is a factor 20, because the dissipated power (which is what dB is about) is proportional to the voltage squared. Han-Kwang (t) 11:30, 24 September 2007 (UTC)[reply]

Last Sentence[edit]

Shouldn't it be 31.6uV? Penguin941 (talk) 13:15, 20 February 2013 (UTC)[reply]

inconsistent of the Vout equation[edit]

from the equations in theory section, the example Vout seems should be:

V_{\mathrm {out} }=(V_{+}-V_{-})\cdot A_{\mathrm {d} }\pm {\frac {V_{\mathrm {cm} }}{10^{\frac {CMRR}{20}}}}\cdot A_{\mathrm {d} }

and the output error of 316uV requires the differential mode gain equal 1.

vfkk;kig;eym,h,;p'[sn,bvtym,ghjfreb m,po=\dn fry m dj n tfriunn v5l.cvt6m bngtk bgjurdyhjy gfhgjb ju cxu kjdrwyi787454eutft5jnhiufthby iuyyfytgkmkmh7ydlk,jyj7ujumbmpi s3w.l';/yskkitfrgazb5/kghy7ds5kk;liytfseghygrd6uy7rt 743qmtergikj.kuhtro — Preceding unsigned comment added by 27.251.70.205 (talk) 12:49, 3 April 2014 (UTC)[reply]

Definition[edit]

The current definition of CMRR is too weak, IMO. Consider:

The common-mode rejection ratio (CMRR) of a differential amplifier (or other device) measures the ability of the device to reject common-mode signals, those that appear simultaneously and in-phase on both amplifier inputs. — Preceding unsigned comment added by 50.49.243.234 (talk) 16:34, 4 November 2015 (UTC)[reply]

Theory[edit]

Should it not be "when measuring the emf (instead of resistance) of a thermocouple in a noisy environment" ? Sridhar10chitta (talk) 10:37, 14 May 2016 (UTC)Sridhar Chitta[reply]