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I just want to know, what does the common mode input voltage in a fully differential op -amp do?

e.g. for a Op-amp,

$$V_1 = \dfrac{V_1 + V_2}{2} + \dfrac{V_1 - V_2}{2}$$

$$V_2 = \dfrac{V_1 + V_2}{2} + \dfrac{-(V_1 - V_2)}{2}$$

thus, \$\dfrac{V_1 + V_2}{2}\$ is the common part and \$\dfrac{V_1 - V_2}{2}\$ is the differential part of the signal.

There are three inputs, two are the differential voltage inputs. One is the common mode voltage input. I do not know what this pin means? Is it supply an offset?

for the Op Amp that i use is available under the link THS4524

enter image description here

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  • \$\begingroup\$ Which opamp are you using? \$\endgroup\$ – avakar May 11 '15 at 12:18
  • \$\begingroup\$ Op amps typically DON'T have a common mode input. Please attach part numbers and circuit diagrams, so we know what you're talking about \$\endgroup\$ – Scott Seidman May 11 '15 at 12:44
  • \$\begingroup\$ i haved posted the op amp that i used. \$\endgroup\$ – napon May 11 '15 at 13:59
  • \$\begingroup\$ This part is not an operational amplifier. \$\endgroup\$ – user207421 May 11 '15 at 19:12
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It's an output common-mode voltage, not input. So, the output is differential around whatever voltage you apply to the common-mode pin (assuming you have enough headroom for it to still act like an opamp).

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  • \$\begingroup\$ hi, thank you for your commemts. sorry, but i have posted the datasheet, it says it is Common mode voltage input。。。 \$\endgroup\$ – napon May 11 '15 at 16:34
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    \$\begingroup\$ sorry, but you don't understand the datasheet. It is an input to the CHIP not an input to the amplifier. Can you read a schematic? Look at the functional block diagram of the chip you linked to. You'll see that the common-mode input pin is used in a negative feedback loop to drive the common-mode (average) OUTPUT of the amplifier to the voltage you supply to the common-mode input pin. This is pretty standard for a differential amplifier. \$\endgroup\$ – crgrace May 11 '15 at 16:57
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If you go to the online datasheet and scroll to the Typical Applications page, you'll see

Figure 79 shows the modifications made to the circuit. Note the resistor connecting the VOCM input of the THS4521 to the input common-mode drive from the PCM4204 is shown removed and is optional; no performance change was noted with it connected or removed.'

Also if you look at the Electrical Specifications, you will see that all the specs are measured with 'VOCM = Open'. So don't worry about it.

Like the other answer mentions it is an output voltage.

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It's an input with a twist. Look at this diagram - an amplifier (OPA2350) is feeding the "input": -

http://www.ti.com/ods/images/SBOS458G/typcir_ths_ads_test_bos458.gif

But, it can be left open circuit: -

enter image description here

Page 4 of the data sheet calls it an input so I'm quite happy to call it an input. Figure 53 also shows a graph of the input impedance of Vocm versus frequency.

BUT there is a voltage on it that can be used as a reference to other circuits. When left open circuit this pin naturally assumes the mid-point between the two supply rails. This can be a bias to another circuit but be prepared - it has to be buffered if feeding anything other than a light load. Figure 78 shows it as a potential divider formed from 2x 275kohm resistors connected to both rails.

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