Talk:Common-mode rejection ratio

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Untitled

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

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

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?)

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

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

inconsistent of the Vout equation

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.

Definition

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

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]