I'm new to here and I did already some of the research but cannot really find an answer. I'm not very familiar with op-amps.

I have wired a LM358 op-amp as per the below example schematic in the MCP4725 DAC datasheet. The problem is that I'm getting only +2.7 V all the way to -2.2 V (I guess because I used 22 kΩ resistor instead of 20 kΩ) on the 12-bit DAC resolution.

However, I used bipolar supply with +/-15 V, and I was expecting the output voltage to be somewhere close to the 15 V both ways, or at least 10 V. Can somebody please explain what am I missing in order to reach my voltage output? Is the op-amp chosen wrong? If so, can you please recommend a device to use?

Thank you very much in advance for any help.

Example 8-3 - Digitally-Controlled Bipolar Voltage Source - from MCP4725 datasheet

  • R1 and R4 are 22kΩ
  • R2 and R3 are 10kΩ
  • 1
    \$\begingroup\$ The MCP4725 is operated with Vdd = + 5V. It is output Vout could be not higher than 5 V. The first formula gives the value 5(4095/4096) V = 4.9988 V. You have to calculate the values of R1, R2, R3 and R4 to get an output of -10 to + 10 V. You need an amplification of 4 to get -10 to + 10 V from 0 to 4.9988 V. \$\endgroup\$ – Uwe Dec 10 '18 at 12:11
  • \$\begingroup\$ I see now, the datasheet stated gain is 0.5 based on 10k/20k resistors. Am i right? \$\endgroup\$ – Kris Dec 10 '18 at 12:14

The circuit is bogus.

The opamp and its four resistors have the general form of a difference amplifier. Since VOUT can never be greater than VDD, it's simply impossible for the output of the opamp to ever be greater than VDD, no matter what you do with the resistor values.

At a minimum, you'd have to add a resistor between the opamp "–" input and ground, in order to create the required offset. If you make this resistor equal to R1, then the effective reference voltage will be VDD/2 — the middle of the range of VOUT. This means that the differential input voltage (between VOUT and VDD/2) can range between -2.5V and +2.5V.

Then, R3 should be half this value. In order to get a ±10V output range (from a differential input of ±2.5V) you need a gain of 4, so R2 and R4 should be 4× the value of R3 (i.e., 2× the value of R1).


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Any chance you could post a diagram of what you just said? Do i understand that i keep everything as per my above diagram, and than add that resistor to "-" on the opamp? I would appreciate if you could also add the values to reach the +/- 10v. Thanks \$\endgroup\$ – Kris Dec 11 '18 at 13:21
  • \$\begingroup\$ See edit above. If you use 22k instead of 20k, you'll get an output that goes from -11V to +11V. R3 can be two 10k resistors in parallel. \$\endgroup\$ – Dave Tweed Dec 11 '18 at 13:42

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