I would like to design a device with opamp LM324 operating at roughly >5V. The device has to manipulate a signal on +/2.5V (Vpp:5V, DC balanced).

If I want it to work without a split supply, can I use a 2.5V offset to bias the signal? (I understand that the Vdd of opamp should be at least 7V for fear of saturation).

Furthermore, is the diode necessary? In previous trials I found that when LM324 sinks current, output goes to saturation. Or did I just do something wrong?

Schematic of the Opamp with 2.5V offset. Vss at GND. ps forget about the pinouts number. I just randomly pick a opamp symbol.

  • \$\begingroup\$ Sorry for the confusion. I will run LM324 at Vdd=9V. Schematics updated. \$\endgroup\$ Jan 30, 2015 at 3:56
  • \$\begingroup\$ In particular, I would like to ask is it true that the opamp "sense" the already-offseted signal at V+? \$\endgroup\$ Jan 30, 2015 at 4:03
  • \$\begingroup\$ So if I remove the dioide, what will happen if the signal is +/-9V? Shall it become a hystersis (if square wave)? \$\endgroup\$ Jan 30, 2015 at 6:39
  • \$\begingroup\$ If I put the particular question other way, I'm asking is the summation done independent of / before the opamp, so that the negative voltages will not be chopped by the opamp? \$\endgroup\$ Jan 30, 2015 at 6:41

2 Answers 2


If what happens when the input is open is important, then you can follow the answer from WhatRoughBeast (also see comment below). If not, you can simplify things a bit:

Get rid of D1, R1, R2 Change R3 to 20K and connect to the +5V reference voltage Change R5 to 20K

That will have an output of 0V with an input of -2.5V and +5V with an input of 2.5V.

The difference is the the output voltage with the input disconnected will be about 7.5V vs. 5.0V with WRB's circuit.

If that's important, you can still simplify it a bit:

Get rid of D1, replace R3 with a short and increase R1, R2 to 20K each. Leave R5 and R6 at 10K. This will be equivalent (0V out for -2.5V in, 5V out for 2.5V in, 5V out for input open), but will draw less current from the reference input (and have one fewer component).

  • \$\begingroup\$ Does the dioide do harm? Or simply unnecessary? \$\endgroup\$ Jan 31, 2015 at 2:06
  • \$\begingroup\$ Great harm- it won't work at all- it will rail at the positive supply (minus a bit) because there is no path for the bias current flowing out of the input. Even if it worked it would add a lot of inaccuracy due to the diode drop. \$\endgroup\$ Jan 31, 2015 at 2:16

You're very close.

First, as you suspected, get rid of the diode.

Second, change R3 to 9.5k, or R4 to 10.5k. Your 2.5 volt reference will have a Thevenin equivalent resistance of 500 ohms, so you need to compensate in order to keep the gain right.

A few (hopefully minor) points.

Your 5 volts must be stable and noise-free. Any drift or noise will cause drift or noise on the output.

It's probably not important, but be aware that the LM324 will not actually pull all the way to ground. You might expect a worst-case "ground saturation" of about 20 mV.

I'm not sure what your specific question means. The op amp will act to keep the voltage between the + and - inputs as close to each other as possible, within 2 or 3 mV (which is the offset voltage).


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