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I'm planning to buffer a 0-10V DC signal coming from a potentiometer windvane.

I want to use this opamp for the purpose and the opamp will be powered as single supply with 12V Vcc.

Can I use this opamp for 0-10V inputs as a buffer aka unity gain? I'm worried about CMRR issues and Im not able to interpret that for this scenario.

Alternatively I have this opamp as well.

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  • \$\begingroup\$ What is CMMR and why are you worried about it? \$\endgroup\$
    – Finbarr
    Sep 8 '17 at 11:52
  • \$\begingroup\$ You can use it as you state, it is rail to rail input and output, and unity gain stable (read the capacitance tolerance chapter). CMRR is 60 dB at worst, you are the only one that knows if this is acceptable. \$\endgroup\$ Sep 8 '17 at 11:57
  • \$\begingroup\$ @Finnbarr not CMMR, but CMRR (common mode rejection ratio). \$\endgroup\$
    – Bart
    Sep 8 '17 at 13:27
  • \$\begingroup\$ @Bart Question has been edited since I wrote that comment. The question about why he's worried still stands. \$\endgroup\$
    – Finbarr
    Sep 8 '17 at 13:33
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It should be fine for a non-precision application such as a windvane potentiometer. Your actual application circuit (please edit the question and add this information to the question) shows an input that goes from about 100mV to the excitation voltage minus 100mV. The amplifier you are looking at can even sink a bit of current at that minimum output voltage (typically as much as 2mA according to figure 6).

You could probably use a much cheaper single supply op-amp instead. You may find that a bipolar type is easier to protect, although with a CMOS type you can add a large value resistor to the input without unduly affecting accuracy (much higher than the 1K shown). The absolute maximum 15.5V supply voltage is a bit closer than one might like to 12V.

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  • \$\begingroup\$ Here is the vane and recommended circuit: nrgsystems.com/assets/resources/anVaneIn.pdf My problem is choosing an opamp and powering the vane with 10V voltage regulator or 10V voltage reference? My power supply will be 12V \$\endgroup\$
    – user1245
    Sep 8 '17 at 12:41
  • \$\begingroup\$ I want to use this reference uk.rs-online.com/web/p/voltage-references/5343043 for the vane but they dont mention if I can use it with 12V input \$\endgroup\$
    – user1245
    Sep 8 '17 at 12:52
  • \$\begingroup\$ How about this opamp I have docs-europe.electrocomponents.com/webdocs/10ed/…? Which one would be better? \$\endgroup\$
    – user1245
    Sep 8 '17 at 13:19
  • \$\begingroup\$ @user134429 The op-amp looks like gross overkill to me. I didn't look at the output drive- the first one you picked is probably better and it does not look any more robust. Your source is just a 10K pot- rated linearity is 1%. There will be friction (it's not got magnetic levitation or air bearings). Why do you think you need or want microvolt accuracy for such a crude device? \$\endgroup\$ Sep 8 '17 at 15:17
  • \$\begingroup\$ Sorry for the late reply, I was outside. I see what you mean but the supply for the vane at least should vary much. Some voltage regulators have %5 uncertainty; voltage references less uncertainty. Never done a similar project before and I cannot decide whether I should use a voltage regulator or a reference for the vane's potentiometer. The opamp will use 12V supply but I want vane to output max 10V. So I want obtain 9 or 10V from 12V power supply for the vane. Voltage regulator or reference? I have this regulator: ti.com/lit/ds/symlink/lm317.pdf Thanks for the answers a lot. \$\endgroup\$
    – user1245
    Sep 8 '17 at 22:45
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The datasheet will tell you what voltages the input and output can support. As expected for an opamp advertized as "Rail-to-Rail Input and Output", the specifications for VCM and VO show that you can go almost to the rails.

Typical CMRR is specified as 82 dB. Any dB calculator shows that −82 dB corresponds to an amplitude ratio of 0.0000794, so a common-mode change of 10 V at the input, attenuated by 82 dB, would result in a 794 µV change at the output. However, worst-case 60 dB corresponds to 10 mV.

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  • \$\begingroup\$ Thanks, I just wanna ask to be sure if I got what you mean. In unity gain buffer configuration Vinverting = Vnoninverting. Does that mean if Vin = Vnoniverting the common mode voltage Vcm is equal to Vin always? And the error due to the CMRR is always Vin*0.0000794? \$\endgroup\$
    – user1245
    Sep 8 '17 at 12:06
  • \$\begingroup\$ A "common mode" change means that both inputs change by the same amount so that the differential input does not change. With a buffer, all changes are common-mode changes. The single CMRR value is a simplification; see figures 19–21. \$\endgroup\$
    – CL.
    Sep 8 '17 at 12:54
  • \$\begingroup\$ So 2V input means 2V common mode voltage right? (In buffer) \$\endgroup\$
    – user1245
    Sep 8 '17 at 12:56
  • \$\begingroup\$ "Common mode" is a property of the change. CMRR specifies how much a change at the inputs affects the output. \$\endgroup\$
    – CL.
    Sep 8 '17 at 13:04
  • \$\begingroup\$ How about this opamp I have docs-europe.electrocomponents.com/webdocs/10ed/…? Which one would be better? \$\endgroup\$
    – user1245
    Sep 8 '17 at 13:19

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