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Recenetly I was looking over the datasheet of an Analog Devices AD5547 DAC, and noticed the following in the application schematics (page 15): enter image description here

Now what interests me here is U2. The purpose of U2 is to provide a negative reference voltage. As R1A and RcomA should be equal, and the positive terminal of U2 is at ground, in order to cancel out the currents it must supply -2.5 V at the output.

However, the positive supply of U2 is connected to ground. This is a rail-to-rail input amplifier, so this is within specifications. However, we can also see on the left that a +5 V positive supply availabe.

Apart from power consumption (it is likely the opamp will draw more current when it has a larger supply voltage), is there any reason why we would want to have the inputs sit against the rails? (perhaps so we are only using one of the two input pairs internally or such?)

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    \$\begingroup\$ Although a R2R output on an opamp absolutely cannot drive beyond the rails, and in fact can only approach them, any amp, not just R2R input types, may sometimes be specificied with an input common mode range that exceeds one or both rail (LM324 goes below ground). However, the 8628 does specify exactly and not beyond the rails for input voltages. \$\endgroup\$ – Neil_UK Jan 22 '18 at 11:35
  • \$\begingroup\$ You may also not want the noise contribution of the positive rail supply in this case as you are generating a 'reference' voltage rail. If this was very important I would scrutinise the internal schematic or test the Vin operation at a fraction above the +V rail to investigate behaviour that might occur in the limiting condition. Presumably latch-up does not regularly occur or this situation would not have been proposed. The 2.5V reference supply could also have been used in the IC could handle 7.5V but not 10V total supply. A +1V rail would have been perfect. \$\endgroup\$ – KalleMP Jan 22 '18 at 18:15
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The AD8628 is a 5 volt op-amp and has a maximum supply rail of 6 volts hence why it can't run from +/- 5 volts. They use it here because of its phenomenally low input offset voltage of typically 1 uV. It also has an input voltage range that includes both power rails. It sounds like a good choice to me.

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  • \$\begingroup\$ I skimmed over the datasheet and in my haste missed that it is a 5 V opamp (and feel silly now). Thanks for the response! \$\endgroup\$ – Joren Vaes Jan 22 '18 at 11:28
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    \$\begingroup\$ @JorenVaes Don't feel silly - we all miss things when looking at op-amps and sims don't help spot these problems either. \$\endgroup\$ – Andy aka Jan 22 '18 at 11:52
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Andy already answered for this specific case, but in general running an op-amp with wider rails than you need to can be problematic.

  1. There are a number of electrical characteristics which are dependent on the supply voltage which may or may not be an issue with a particular application.

  2. By limiting the rail you also limit the extent of the output. If you never want the output to be positive... using +5V for the top rail would be a bad idea.

  3. As you indicated, using wider rails means more power lost. Although that may be a small change in quiescent current, if your load is normally being held at 0.1V with 20mA of current, on a 15V supply, the op-amp will need to dissipate 0.3W.

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    \$\begingroup\$ +1 for "By limiting the rail you also limit the extent of the output." The output can't go positive at all with this setup. \$\endgroup\$ – Transistor Jan 22 '18 at 16:41
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    \$\begingroup\$ Thanks @Trevor_G for this addition. I especially like your second point, and hadn't thought of that yet. I think that is a very good thing to keep in mind when designing systems. I recently saw a case where a opamp was driving a control pin of a 5 V supply chip. The opamp was connected to a +12 V rail. When due to a fault the resistor at the input of the opamp failed, the opamp railed out and damaged the chip it was driving. Had the designers used a 5V rail, this would not have happened and the repair would have been a simple resistor instead of a far more expensive analog IC. \$\endgroup\$ – Joren Vaes Jan 23 '18 at 7:12

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