I need to amplify a voltage, which comes from a DAC and is 0-2.5V and scale it 0-25V in an operational amplifier. After amplified, I need to offset it -1.25V, so the final voltage is -1.25 to 23.75V. In other words, being an operational amplifier, I want to do the y=ax+b operation. with a=10, and b=-1.25V

I thought of 2 different ways of doing this.

  1. The first idea I had is to use a second amplifier as a summing amplifier, with a fixed -1.25V voltage.

  2. The second idea is to use a 1.2V zener in series with the output, along with a biasing resistor to -V.

Why do I want this? I want to set the ADJ pin in an LM338 voltage regulator from an MCU. LM338 out is ADJ+1.25V. If ADJ is -1.25V then the output is 0V.

  • 1
    \$\begingroup\$ -V is a negative supply rail, which is available in your system, right? If so, what's the voltage of this negative supply rail? Just making sure that we understand your setup. By the way, check out this related thread. \$\endgroup\$ Commented Dec 12, 2013 at 1:45
  • \$\begingroup\$ I'm flexible. I can add any -V or +V as needed. That link you posted seems to do exactly what i want! \$\endgroup\$
    – hjf
    Commented Dec 12, 2013 at 2:24
  • \$\begingroup\$ Instead of generating the reference voltage with an analog output, I would use a digital pot like this one for the variable resistor in the reference voltage circuit. \$\endgroup\$
    – tcrosley
    Commented Dec 12, 2013 at 5:04
  • \$\begingroup\$ A DPOT is a good idea, however, the DPOT resistor terminals cannot stray outside the power supply range for the part. This would severely limit the range at which the regulator could be used. \$\endgroup\$ Commented Dec 12, 2013 at 7:44
  • \$\begingroup\$ @alex.forencich Good point. A DPOT would work great for example a variable 0-5v supply, but not for the range the OP wants. \$\endgroup\$
    – tcrosley
    Commented Dec 12, 2013 at 8:27

3 Answers 3


You can use a differential amplifier

enter image description here

As you can see from the shown equation VR is just an offset and doesn't get amplified so set the resistors for a gain of 10 , connect GND to V1 and -1.25 to VR

The reply here shows an alternative solution that doesn't need a 1.25v reference but it needs a negative rail voltage to supply the opamp

  • \$\begingroup\$ This could be achieved simply by using already existed differential amplifier chip with gain of 10, such as INA106 with low source output impedance of both the DAC and reference. If the output impedance of the DAC or the reference happens to be high, then buffer it. \$\endgroup\$
    – Unknown123
    Commented May 19, 2019 at 5:27

Great answers as for the suggested analog circuits. However, the main point is missed here - the LM338 voltage regulator does not just output ADJ+1.25V , it needs the feedback from Vout, and just connecting an arbitrary voltage to ADJ will kill this feedback. Digital potentiometer is OK and will keep the feedback. The LM338 cannot regulate an output voltage which is less than its internal reference voltage (1.25V).


Easist way to do that would be to build a differential amplifier.


Connect your DAC to the positive input and connect the negative input to the power rail. Then calculate the resistor values so that the positive input gets a gain of 10 and the negative input gets a gain of whatever you need to scale it to 2.5 volts. The equations to do so are in the Wikipedia page. I would suggest calculating gain and resistor values for the negative input first, then find the resistor values to get a gain of 10 on the positive input.

Also, I would recommend putting a current limiting resistor immediately at the output of the op amp, but before any of the feedback resistors are connected. This will not affect the accuracy of the circuit, but it will protect the op amp in case something goes wrong with the regulator.


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