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I can't figure out why the refence voltage of a DAC is used in the amplifier stage in many circuits.

Here are some examples:

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Why is Vref of the DAC used in the op-amp amplifier stage? Does that have to do with noise elimination or precision? What would happen if another source were used in the amplifier stage instead of the DAC's Vref? I would appreciate a clear didactic explanation since it is very confusing to me.

Edit for an answer:

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Plots

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3 Answers 3

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Most of the time, a DAC will provide an output that ranges from 0V to Vref. The code that you send to your DAC then defines what voltage will be sourced relative to the value of this reference.

$$V_{DAC} = V_{ref} \cdot \frac{DAC_{code}}{2^{N_{bits}}}$$

However, in many applications you want a DAC that can produce voltages below 0V, typically from -Vref to Vref, which can be achieved using an external operational amplifier. OpAmp offset removal

The output of this operational amplifier is:

$$V_{OUT} = V_{ref} \cdot \frac{DAC_{code}}{2^{N_{bits}}} \cdot (1 + \frac{R_{FB}}{R_{INV}}) - V_{ref} \cdot \frac{R_{FB}}{R_{INV}}$$

In case of Rfb=Rinv, the output range range is increased to +/-Vref.

The refeference voltage is taken as the non-inverting input to assure the most precise offset.

For instance:
Let us assume that your Vref is 5V, your DAC has 8bits and Rfb=Rinv. The DAC code then ranges from 0-256 where the code 128 corresponds to a DAC output of exactly 2.5V and the opAmp output is exactly 0V.

However in reality, your reference will be slightly different from 5V, lets say 4.9V, due to manufacturing tolerances, temperature drift and ageing effects. In this case your DAC output at a code of 128 is only 2.45V. But here`s the thing: Since the reference is acting on the opAmp, this imperfection will be "compensated" and the opAmp output is still exactly 0V at a code of 128 regarless of the actual reference voltage value.

You can take some other voltage source than the reference as the non-inverting opamp input, but then this kind of compensation will not work anymore.

EDIT:
The Vref of a converter (DAC or ADC) is typically the most precise and stable value in your circuit. It is usually provided by a voltage reference IC, which has tight tolerances and is rated for drifts. These ICs can cost a lot of money depending on their driving output capabilities, accuracy and drift.

To give you an idea about the numbers: The REF6250 (link to datasheet) is a buffered 5V reference, rated for 18bit converters, that can source and sink up to 3mA of current. It has an initial accuracy of 0.05%, and a temperature drift of max. 3ppm/°C from 0-70°C and costs about 11$ on digikey. So the uncertainty of your reference voltage is just 4.9975V-5.0035V inside this temperature range.

For the circuit as shown above, it just makes sense to use this very stable Vref to get precise offsets and gains at your operational amplifier's output.
Of course you can take other voltage sources to modify the offset of your output, and there are for sure applications/circuits where this is required.
However, if your goal is to achieve high precision and stability, you will end up buying a second (potentially expensive) reference voltage IC for that second source. The uncertainty of this source will be reflected on the circuit's output as well as the uncertainty of the Vref reference, which decreases your total accuracy.

Be aware that there will be other error sources in the circuit above such as the tolerances of the resistors and the output offset of the opAmp. Also the reference IC has specified more drift parameters than I mentioned above.

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  • \$\begingroup\$ Please see my edit in question: also here i.sstatic.net/Cm7pU.png. I simulated what you mentioned. Red output is the ideal output when Vref is common and no drift occurs. Green output is when Vref is common and there is 0.5V drift at Vref. Blue output is when Vref is not common(Vreg is used for the opap) and there is 0.5V drift at Vref. For the common Vref case(green output) things are symmetric around zero and error minimizes at the center. But for the separate case(blue output) the error is minimum at the very beginning and increases with voltage. \$\endgroup\$
    – cm64
    Commented Jan 18, 2023 at 12:47
  • \$\begingroup\$ But the total error is the same. Only one is symmetric around zero. What is the benefit of that? Here are the output plots i.sstatic.net/QmqXP.png \$\endgroup\$
    – cm64
    Commented Jan 18, 2023 at 12:47
  • \$\begingroup\$ @cm64 In the common design you have a guaranteed point at 0V where your circuit is "ideal". The error then scales only with the value of Vref which is something you can easily calibrate. If you take a second source, then you will introduce uncertainties of that second source (tolerances/drifts etc.) in addition to the uncertainties that you already have from your reference. It will just be much harder to calibrate this thing. \$\endgroup\$
    – Mau5
    Commented Jan 18, 2023 at 12:58
  • \$\begingroup\$ @cm64 If you extend your simulation by making Vreg drifting. In addition, you can also make the initial value of Vref and Vreg different from each other e.g. 4.95V and 5.02V. And think about how you would calibrate it if you had to :) \$\endgroup\$
    – Mau5
    Commented Jan 18, 2023 at 13:15
  • \$\begingroup\$ @cm64 look at my edits in this answer. Maybe it helps you clarifying some things \$\endgroup\$
    – Mau5
    Commented Jan 18, 2023 at 22:21
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Why is Vref of the DAC is used in the opmap amplifier stage?

It's used to convert a unipolar output voltage from the DAC to a bipolar output voltage from the op-amp. To obtain a bipolar output from a unipolar signal, an offset needs to be used and, that offset comes from using Vref feeding the inverting input of the op-amp hence, it acts as a negative offset and shifts the unipolar positive output in a negative direction.

What would happen if another source were used at the amplifier stage instead of the DAC's Vref?

For a simple DAC that produces an output voltage between 0 volts and Vref, you want to negatively offset this by Vref/2 so that the new output voltage is bipolar about a centre-point of 0 volts. You may also choose to apply amplification to make the overall output span bigger but, that is unimportant to your question.

So, the new output voltage will span from -Vref/2 to +Vref/2. If Vref drifts up or down a little bit then of course the span increases a little bit but, importantly, the centre-point will remain at 0 volts.

However, if you derived the negative offset from a different supply unrelated to Vref there is no link to ensure that the centre-point is held at 0 volts.

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  • \$\begingroup\$ I read all answers and still didn't get why Vref is used for that. Lets say DAC's Vref after the voltage divider is coupled to inverting input of the opamp stage. And lets say this voltage is 1V. Now if I obtain that 1V not from the DAC's Vref but from Vcc what would go wrong? Is there a particular reason why Vref is used as the source? \$\endgroup\$
    – cm64
    Commented Jan 17, 2023 at 23:05
  • \$\begingroup\$ No problem with using 1 volt from any source you want except for one thing... If you use Vref (and it drifts up 1%), the unipolar DAC output will be 1% higher in magnitude and, to keep the op-amp output exactly symmetrically bipolar, the 1% higher Vref is perfect for doing so. It doesn't mean the bipolar output won't be 1% higher (negatively and positively) but, it does mean it is still centred perfectly at 0 volts (that is important enough to use Vref) @cm64 \$\endgroup\$
    – Andy aka
    Commented Jan 17, 2023 at 23:39
  • \$\begingroup\$ Your last comment is what Im asking about. Can what you say be simulated for clearity? \$\endgroup\$
    – cm64
    Commented Jan 17, 2023 at 23:45
  • \$\begingroup\$ Sure it can be. What sim tool do you have? \$\endgroup\$
    – Andy aka
    Commented Jan 18, 2023 at 1:15
  • \$\begingroup\$ Please see my edit in question: also here i.sstatic.net/Cm7pU.png. I simulated what you mentioned. Red output is the ideal output when Vref is common and no drift occurs. Green output is when Vref is common and there is 0.5V drift at Vref. Blue output is when Vref is not common(Vreg is used for the opap) and there is 0.5V drift at Vref. For the common Vref case(green output) things are symmetric around zero and error minimizes at the center. But for the separate case(blue output) the error is minimum at the very beginning and increases with voltage. \$\endgroup\$
    – cm64
    Commented Jan 18, 2023 at 12:51
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As the LTC2641 datasheet states on its first page:

Both the LTC2641 and LTC2642 feature a reference input range of 2V to VDD. VOUT swings from 0V to VREF. For bipolar operation, the LTC2642 includes matched scaling resistors for use with an external precision op amp (such as the LT1678), generating a ±VREF output swing at RFB.

So, it is meant for generating positive and negative swings in the output voltage from a signal which is only on the positive side of the ground or 0V point, or swinging from 0v to some positive voltage and back to 0V. In other words, getting a bipolar op-amp output from a unipolar DAC output.

The reason why Vref is used as a reference for the op-amp is because the DAC uses Vref as its maximum output voltage point when all data bits are set to 1 or high.
Using the same Vref avoids problems of the two voltages being different to any extent, and also reduces noise and interference, ground loops or any current loops in the circuit from affecting the output. That's why it is important that it is exactly the same voltage/potential as the one the DAC is referring to.
Technically, another voltage reference of exactly the same voltage could be used, but making it an exact match down to a mV or less would extremely difficult, any changes in it due to temperature or any other reasons would not be exactly the same, and using the same Vref also saves on space and additional components.
Whatever the mismatch between DAC's Vref and another Vref, it will show up as a voltage other than zero when DAC wants to output zero (a DC offset), and would also show up as noise, though very small for small mismatches.

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  • \$\begingroup\$ Still unclear why is Vref i used for that. It could be any source. Lets say DAC's Vref after the voltage divider is coupled to inverting input of the opamp stage. And lets say this voltage is 1V. Now if I obtain that 1V not from the DAC's Vref but from Vcc what would go wrong? Is there a particular reason why Vref is used as the source? \$\endgroup\$
    – cm64
    Commented Jan 17, 2023 at 23:06
  • \$\begingroup\$ Because Vref is used as the maximum output voltage of the DAC. It is the voltage DAC outputs when all bits are set to high (1). \$\endgroup\$ Commented Jan 17, 2023 at 23:15
  • \$\begingroup\$ Imagine Vref is 3.3V. Why not using another 3.3V voltage source instead of Vref? Im asking to understand. I still dont get why Vref is used incitingly. \$\endgroup\$
    – cm64
    Commented Jan 17, 2023 at 23:18
  • \$\begingroup\$ Is that because if it drifts or something? \$\endgroup\$
    – cm64
    Commented Jan 17, 2023 at 23:19
  • \$\begingroup\$ Using the same Vref avoids problems of the two voltages being different to any extent, and also reduces noise and interference, ground loops or any current loops in the circuit from affecting the output. It is very important that it is exactly the same voltage/potential. \$\endgroup\$ Commented Jan 17, 2023 at 23:21

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