Would reducing the reference voltage of an ADC have any effect on accuracy?

Question

Regarding the following information taken from a paper:

The paragraph is telling about what happens when the ADC Vref is reduced.

There is the statement from the above quote:

Note that if you reduce the reference voltage to 0.8V, the LSB would then represent 100mV, allowing you to measure a smaller range of voltages (0 to 0.8V) with greater accuracy.

Isn't it wrong to say "with greater accuracy"? Shouldn't it be "with greater precision" instead?

(I'm asking because if I don't clarify this point, I will be misunderstanding all the rest)

Answer 1

I think precision means more numbers, like: 1.23 V vs 1.2300 V, the latter has more precision. However, that says nothing about the true value of the voltage. It is possible that my inaccurate meter says 1.2300 V while the actual voltage is 1.220000 V

More accuracy means that the value I get is closer to the real value. So my accurate meter would show: 1.221 V while the actual value is 1.220000 V.

So 1.221 V has greater accuracy (but less precision)

while

1.2300 V has greater precision (but less accuracy).

In the ADC example the amount of numbers (different reading) stays the same: 8 readings. So precision remains unaffected whatever the reference voltage is. Accuracy does increase though as the reference voltage is decreased because the LSB intervals become smaller as Vref decreases. That means that the value of the error between the actually measured voltage and the value which the ADC outputs (quantization error) will become smaller.

Also: instead "precision" engineers more often use "resolution".

Answer 2

Isn't it wrong to say "with greater accuracy"? Shouldn't it be "with greater precision" instead?

Oh, good. Someone who knows the difference.

To some extent, yes, with greater accuracy. But probably not by the ratio that you reduced the reference voltage, and to a diminishing amount as you do so. Some of the error sources in an ADC are in the front end, and basically reflect back as a voltage on the input. But -- mostly for SAR ADCs -- some of the error sources in an ADC are in the conversion itself.

A vastly simplified description of the operation of a SAR converter is that it makes educated guesses at the answer, applies them to a DAC, and compares the resulting analog signal to the input. Nonlinearity (incremental and integral) of an ADC is almost entirely from the built-in DAC, and reducing the reference voltage should reduce the magnitude of those errors proportionally.

Answer 3

Reducing the reference voltage simply means that the size of the "steps" between digital values is reduced. This increases the resolution of the converter at the cost of range, but if the measurement range is similarly constrained, then the increased resolution is a benefit.

I would not expect changing the reference voltage to have an impact on accuracy, except that in some cases when you set the reference voltage at or near the supply voltage, then the linearity at the top end of the range may suffer in some cases (this is probably more an issue with DACs than ADCs though).

Accuracy in ADCs or DACs has 3 components: offset, linearity and noise. Offset is, effectively, the difference between zero volts and whatever voltage actually results in a zero reading. Linearity is the consistency of the step size across the entire digital range. Noise is how much change you can expect with the same input read multiple times in succession. I wouldn't expect changing the reference voltage to impact any of these except linearity, as I said before.

Answer 4

With smaller Vref, the ADC's comparator will have smaller voltages to use in making decisions, thus the errors (the differential non-linearity) will change.

You will have the same # bits out, but the integral linearity and the differential linearity become unpredictable IMHO.

Additionally, the random noise and the power-supply-rejection become more important. If the successive-approximation is performed by charge-splitting etc, then the size of the capacitors (binary weighted?) matters. Using

Vnoise = sqrt (K * T /C)

you'll compute the TOTAL integrated random noise of a 10pF capacitor is 20 microVolts RMS, or about 130 microVolts PeakPeak at the 1PPM level. As the caps become smaller, say 0.1pF, the PeakPeak noise now is sqrt(10p/0.1p) or sqrt(100) or 10X more random noise, which is 1,300 microVolts.

Answer 5

Typically, the term "accuracy" refers to systematic part of the error, while "precision" refers to the random part:

Reducing the ADC range (by lowering the reference voltage) will increase the resolution and reduce the quantization noise (which is not systematic*), thus improving the absolute precision. However, this improvement will only be visible if the quantization noise is dominant: if another kind of significant random noise (e.g. from AC on the power supply line) is present, the improvement in precision will not be noticeable.

Accuracy (aka trueness) may also get better depending on the internal properties of the ADC. Non-linearity and gain errors are (usually) proportional to the reference voltage, so those absolute errors will typically be reduced, while the offset error may often not change. The overall change in accuracy will again depend on which of those errors is dominant.

According to Wikipedia, "accuracy" can be used to describe a combination of both random and systematic errors, so when both precision and trueness are improved, it's not wrong to say that the measurement has better accuracy.

_{(*) - quantization error actually depends on the signal, but it's a very useful assumption needed for the additive noise model, which is automatically holds when this error is relatively small. When the quantization noise is large, artificial random noise (dither) is often applied to make the additive noise model work.}

Answer 6

I believe that you simply don't have all of the information. It is certainly true that resolution and accuracy are two very different things. The absolute resolution or voltage resolution is directly a function of the converter's reference voltage. With \$N\$ bits, the voltage resolution is \$V_{ref}/2^N\$. The word precision is not a good choice in this discussion; its meaning is less well defined and generally refers more to the repeatability of a measurement.

However, I think what is missing from the one slide you show is that the accuracy of an ADC converter is typically specified as some number of bits rather than as an absolute voltage. The voltage equivalent of a "bit" is the amount of voltage change that would cause a change in one in the LSB of the converted value. This is sometimes called just \$V_{LSB}\$. So, if you change the reference voltage then the value of \$V_{LSB}\$ will also change and the relative accuracy of the converter remains the same. However, lower values of the reference voltage result in smaller values of \$V_{LSB}\$, so decreasing the reference voltage leads to a smaller absolute accuracy (in volts).