# Confusion with using non-inverting input to fix common mode voltage

Regarding this paper, and the following diagram:

We can see above that the non-inverting input of U2 is connected to Vref of the DAC. Signal input goes to the inverting input and Vref to the non-inverting input.

An in this pdf it mentions:

So my questions are:

1-) Why and how does the non-inverting input is used to fix the common mode voltage? What is the reason behind fixing it to a value? And how is not possible by the inverting input. I'm not very into the opamp internals hope a easy explanation is possible.

2-) Regarding the first diagram, in similar situation if we do not have access to Vref but Vdd of the DAC, can that be used to connect to the Vcm pin(to the + input of U2) through a voltage divider? And I really don't get what is the reason non-inverting input is connected to Vref but not to any other reference voltage.(?)

The circuit converts the output of the DAC (which goes from 0V to $$\V_{REF}\$$) to a differential signal ($$\V_{OUTP}-V_{OUTN}\$$). So, assuming no knowledge of opamp internals, using the ideal model to understand the two options for $$\V_{CM}\$$ and $$\R_1=R_2=R\$$, for the first set (1) of questions, we have:

Set $$\V_{CM} = V_{REF}\$$. For the two limits of the DAC output, the differential outputs reach the following values:

Set $$\V_{CM} = 0 V\$$. For the two limits of the DAC output, the differential outputs reach the following values:

In summary: by using the common mode voltage in the non-inverting input of the second opamp you can change the differential signal and avoid that one of the outputs reaches negative values.

Regarding the second set of questions, if the common mode voltage varies independently to $$\V_{REF}\$$, this will have a direct effect on the differential voltage, as it is given by:

$$\V_{OUTP} = V_{DAC}\$$

$$\V_{OUTN} = V_{CM} \left(1+\frac{R_2}{R_1}\right)- V_{DAC}\$$

By using a reliable reference like $$\V_{REF}\$$ (instead of the power supply rail, for example) it is possible to reduce the uncertainty of the differential voltage that is measured.

As an example, if the reference voltage is capable of driving a 2R load, the voltage shift could be changed using a voltage divider, relatively to the voltage reference:

• Thank you for your answer, In my case the DAC I have is DAC8554 and it is with this board mikroe.com/dac-8-click with schematics here: download.mikroe.com/documents/add-on-boards/click/dac_8_click/… In this board Vref for the DAC8554 comes from DAC60501 as you see in the schematic. My problem is the board pinouts does not have access to Vref but only Vdd(Vcc) Do you think I can use Vcc of the board instead of Vref. (For the board Vcc I can use a 3.3V voltage regulator or reference). I would be very glad if there's a workaround. I'm stuck. Apr 21 at 20:26
• electronics.stackexchange.com/questions/616908/… Apr 22 at 18:55