1
\$\begingroup\$

I am using an amplifier to bias a signal for an ADC. The signal range is -1.25 to 1.25 volts. I have the 2.5V precision reference for the ADC through a voltage divider to the inverting input of the amplifier.

Is it ok if the amplifier is powered from a single supply 0-3.3V to have the -1.25V appear on the input pin since it is biased to 1.25 volts?

This does not apply to any particular part, I have been looking at datasheets for the last hour and every single one doesn't want the minimum voltage on the input pins to be far enough below the V- for this application.

If it is safe I don't have to put in a dual supply.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
  • 3
    \$\begingroup\$ No, this is not safe; it will destroy your amplifier most likely. There's a diode for ESD protection between each input and the negative supply rail. \$\endgroup\$ – Hearth Jan 25 at 16:10
  • \$\begingroup\$ First, what @Hearth said. Second, that's not how you use an op-amp anyway. If you did have a negative rail below -1.25V, the output would just go as low as it could and stick there. \$\endgroup\$ – TimWescott Jan 25 at 16:21
  • \$\begingroup\$ Google something along the lines of "offset a signal with an op-amp", or "op-amp differential amplifier". \$\endgroup\$ – TimWescott Jan 25 at 16:23
2
\$\begingroup\$

Here's what you're actually doing, taking into account the ESD protection diodes that the amplifier very probably has:

schematic

simulate this circuit – Schematic created using CircuitLab

I removed the resistors you had going off to nowhere.

Notice how D2 is forward-biased by V3. Not only that, but it's forward-biased by about twice its forward voltage. This diode is going to release the magic smoke, and the heat will probably damage other parts of the IC as well.

Secondly, this is simply not the correct way to do what you're trying to do. You want a differential amplifier. What you have is configured as a comparator that is just going to constantly output the lowest voltage it can. Or it would, if it didn't break itself, which it will.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.