I have an input signal with an amplitude of a few mV that I want to filter, amplify and then feed to an ADC ("Analog Signal").

The input signal (Coming though C1) is lifted at 2.5V and the range of the ADC is 0-1.25V. The ADC has a reference voltage pin of 1.25V (it's the LTC2470).

Until now I was using the Adjustable resistor R4 to center the output signal of the opamp by hand. But this solution is not elegant. I would like to remove R3, R4 and the +5V input and manage to center the output without external intervention.

How would you proceed to center the output signal of the opamp to about 1.25/2?

ADC input

  • \$\begingroup\$ Second op-amp. First amp to scale the signal, second one to set offset. \$\endgroup\$
    – JimmyB
    Commented Aug 9, 2016 at 20:59
  • \$\begingroup\$ Why not just replace R3+R4 with a 1.333 M resistor (or 1M 1% + 332K 1%)? \$\endgroup\$ Commented Aug 9, 2016 at 21:00
  • \$\begingroup\$ Change the 2.5V on pin three to 1.25/2, and change the ground on C4 to be that voltage as well. This is still in the common mode input voltage range of your op amp. \$\endgroup\$ Commented Aug 9, 2016 at 21:20
  • \$\begingroup\$ Is your input an AC signal? What frequency range? Does the input signal (to the left side of C1) have a DC offset? Do you need to measure this DC offset with the ADC? \$\endgroup\$
    – FiddyOhm
    Commented Aug 9, 2016 at 21:24
  • \$\begingroup\$ @ScottSeidman: If Vin+ in not in the middle of the supply voltage, doesn't it reduce the available range for my signal? \$\endgroup\$
    – Kevin P
    Commented Aug 9, 2016 at 21:46

1 Answer 1


Okay, so you want to use the 1.25V reference for the offset. This is basically the comment Scott Seidman made in schematic form:


simulate this circuit – Schematic created using CircuitLab

The input common mode voltage is fine- it sits at 625mV fixed which is well within the range. Maximum input voltage is about 9mV peak. The op-amp output can't swing quite down to the negative rail with the load of the ADC input so you would get slightly more range by biasing it up by a smidge, but probably not enough to be worth bothering with given the large Vos uncertainty of that particular op-amp.

  • \$\begingroup\$ Thank you for this very clear solution. I have two more questions: 1) Why do I have to set the ground of C4 to 0.625 as Scott Seidman mentionned? 2) Would you recommend to use that solution, or keep the "summing amp" you suggested before? \$\endgroup\$
    – Kevin P
    Commented Aug 9, 2016 at 22:30
  • \$\begingroup\$ I don't think it's necessary or desirable- I would keep it grounded sans some kind of rail splitter buffer configuration. Maybe Scott could explain. This is probably a slightly better solution than just putting the 1.333M resistor in because the 0.625V derived from the 1.25V is low-pass filtered (so maybe less noise than wherever your 2.5V source is coming from) and based on the ADC reference (so slow variations cancel out). \$\endgroup\$ Commented Aug 9, 2016 at 22:32
  • \$\begingroup\$ Because at high frquency, your output would go to ground, and my understanding is that you would want it to go to 0.625 \$\endgroup\$ Commented Aug 9, 2016 at 23:39

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