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As title mentioned, why too large common mode voltage can damage a device? Especially an instrument amplifier?

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  • \$\begingroup\$ Why should any voltage damage a device? What is "an instrument amplifier"? Is it something for an electric guitar or keyboard? \$\endgroup\$ – Andy aka Oct 1 '15 at 8:59
  • \$\begingroup\$ It depends on the device. Any device can be damaged given enough voltage/current. It depends on the construction of the device what gets damaged. \$\endgroup\$ – Bimpelrekkie Oct 1 '15 at 9:01
  • \$\begingroup\$ an 'instrumentation' amp is just an arrangement of 3 op-amps - typically in a single package. All the same limits apply to the individual amps, as they would if they were separate - in particular, typically no voltage should exceed the supply voltages. \$\endgroup\$ – Icy Oct 1 '15 at 9:20
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In the diagram below V1 is the common mode voltage

schematic

simulate this circuit – Schematic created using CircuitLab

You would expect VOut to be 0V (in an ideal amplifier) because its inputs are at the same potential (being connected together). In fact the op-amp will probably just catch fire, because the common mode voltage is way higher than the supply voltage and well outside the specification for common mode voltage for (most) real world devices

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  • \$\begingroup\$ Additional info: In this case what will fail are the ESD protection diodes which are on nearly all pins on nearly all ICs. These diodes will go into forward mode as the input voltage becomes higher than the supply voltage. Even if there would be no supply then an internal crowbar circuit will short the supplies when above a certain value (this is also for ESD protection). Optionally the input transistors will also be damaged. \$\endgroup\$ – Bimpelrekkie Oct 1 '15 at 9:25
  • \$\begingroup\$ Will there be a condition that common mode voltage lower than power supply voltage, but maybe too high to damage an op-amp? If yes, then why? \$\endgroup\$ – hongcc Oct 3 '15 at 1:07

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