Why use amplifier with adc?

I don't know very well how adc works, so i have a question. Most adc have a Vref pin that defines Vin range of adc. Vin range of adc is then divider for 2^n (n=resolution) to obtain LSB. If i resize Vref, new Vin range (smaller) is divided for 2^n, so LSB has smaller value.If i have a sensor, why i shoud use amplifier to drive Vin pin of adc?can i resize Vref as i like (or there is a limit)?

Reply to
idkfaidkfaidkfa
Loading thread data ...

An ADC contains one or more comparators, whose accuracy determines a good part of the overall accuracy of the device.

So you can expect that a 5V, 10-bit ADC will have an absolute accuracy of around 5mV (maybe 1 or 2mV, but probably not much better).

When you reduce Vref you increase the number of counts that the ADC measures, but you keep that basic absolute accuracy, so you don't gain much (if any) precision.

Putting an amplifier ahead of the ADC to make the sensor range match the ADC range keeps the ADC as accurate as possible, and makes the best use of it.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html
Reply to
Tim Wescott

The larger the voltage the lower the noise floor(relative).

A simple extreme case to show how it works: Suppose you scale the signal with the amplifier to 1/10000mV and set the adc. Do you think it would work? You think it would be accurate? Surely there is a minimum voltage that the adc can "work" at?

The same idea exists in communication lines. Suppose you just send your signal along the line with X volts peak. Any noise on the line will distort the signal making it impossible(or at least difficult) to distinguish what is noise and what is the signal. I.e., if the noise floor is Y volts then when the signal is below Y volts it is impossible to tell the signal from noise.

But if we first scale up the signal before transmission then we decrease the effects of the noise on the signal.

Mathematically we might have somethign like s(t) = A*f(t) + e(t)

where e(t) is the noise, A is an amplification factor, and f(t) is the signal pre transmission.

s(t) is the transmitted signal but before de-amplication,

After we de-amplify on the other side we have

B*s(t) = B*A*f(t) + B*e(t)

in general B = 1/A so the final signal is

f(t) + e(t)/A

So the point being that if we first amplify the signal then transmitt and then deamplify we end up lowering the noise floor by that amplification factor.

It's a similar idea with the ADC. The pre-amp increases the signal so any "noise" by the op amp is reduced. Since we don't have to de-amplify because it is digital after that and we can just scale it mathematically without any loss.

Note that in reality it's not quite analogous because noise is usually "external" in communications lines while for ADC's it is a propery of the design and non-ideal components used.

The main point is that by "scaling" the signal first you reduce the effects of any "noise" or errors because they are not scaled. In some sense the ADC has an easier time to distinguish the signal over the noise because you've made the signal larger.

A very simple analogy is a magnifying glass. In some sense your eyes when reading a ruler is like an ADC. If the notches are too close then any "error"(blur, say) can easily throw off the results. But if you first magnify then read it is much easier because the "error" doesn't scale. (although this is not to say that the magnifying glass doesn't introduce it's own error)

Reply to
Jon Slaughter

You can use Vref to scale the overall resolution into less range.

Instead of say having 1000 5mv steps, hook Vref to 3.3 volts and you now have

1000 3.3mv steps - up to 3.3 Volts.
Reply to
plesterf

On another aspect of this same subject, I have a resistive sensor that I can get to put out 3 to 4 volts of range. What type of signal conditioner or amplifier will allow me to change the 3-4 volts into say 0-3.3 volts?

Reply to
plesterf

The best answer, is a Wheatstone bridge. That's because you presumably have to excite the resistive sensor, and if your ADC uses the second leg of a bridge for its reference voltage, the pure resistor ratio is what gets measured (independent, or nearly independent, of the available exciting voltage source noise and drift).

Reply to
whit3rd

ElectronDepot website is not affiliated with any of the manufacturers or service providers discussed here. All logos and trade names are the property of their respective owners.