simulate this circuit – Schematic created using CircuitLab

I am taking a -8 volt signal and using a resistor divider to make it -3 volts by using a 16.5k ohm resistor as r1 and 10k as r2. I am feeding the -3v into a single supply opamp (+5 and gnd) in the inverting configuration with unity gain (using two 10k resistors) in order to obtain 3 volts. Will the resistor divider before the unity gain non-inverting opamp effect my unity gain?

Essentially i am taking a signal that goes from -8 to 0 and i want to turn this into 3v to 0 for input into a ADC.

  • \$\begingroup\$ A schematic would really help this question. Click the 'edit' link, then there's a button in the editor toolbar, or press CTRL-M. \$\endgroup\$
    – Phil Frost
    Jun 4, 2013 at 20:05
  • 1
    \$\begingroup\$ What is your R3 doing there in 'series' with the inverting input? That's not the standard opamp inverting amplifier circuit! \$\endgroup\$ Jun 4, 2013 at 21:01
  • \$\begingroup\$ You do realize an LM324 will not achieve what you want in terms of output voltage capabilities? \$\endgroup\$
    – Andy aka
    Jun 5, 2013 at 7:16
  • \$\begingroup\$ if my single supply rail is 5v..and ground...i believe it will drive from 3 volts to zero at output. What spec are you looking at that says otherwise? \$\endgroup\$
    – tsop
    Jun 5, 2013 at 12:26

1 Answer 1


Yes, the divider definitely will affect the gain calculation, because its equivalent source resistance (Thévenin resistance) of 6.23 kΩ is in series with one of your 10 kΩ gain-setting resistors. The overall gain will be -0.2325, rather than the -0.375 that you're looking for.

If you want an overall gain of -3/8, it would be simpler to forget about the voltage divider and just set up your opamp with 20 kΩ input and 7.5 kΩ feedback resistors.

Also, be sure to use an opamp whose common-mode input range includes ground.

  • \$\begingroup\$ i always hear that below unity gain is unstable for opamps..how can i be sure that that will be stable? Thanks \$\endgroup\$
    – tsop
    Jun 4, 2013 at 20:03
  • \$\begingroup\$ Actually, that was already a concern, even with your original circuit. \$\endgroup\$
    – Dave Tweed
    Jun 4, 2013 at 20:44
  • \$\begingroup\$ ya so i searched and looks like i just have to put a cap across the feedback resistor with a rolloff value above the freq of interest \$\endgroup\$
    – tsop
    Jun 5, 2013 at 12:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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