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I get that each mV is equivalent to a bit in a 12-bit ADC, but why is 4.096V used so often? I've heard of this being from a "bridge". What does this mean and what is the historical significance?

Pugz
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  • Looks like it's the closest power of 2 to 4V – scld Apr 15 '14 at 15:25
  • I always thought it was to make the binary to decimal conversion easy. – kenny Apr 15 '14 at 15:33
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    Just to make it 1mV/count with a 12-bit converter. It's also a number that gives you most of the available range with a 5V supply. If you drop to 2.5V, typically, you'll lose some accuracy, and you can run a 4.096V reference from a 5V supply. – Spehro Pefhany Apr 15 '14 at 15:43

3 Answers3

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It may or may not have historical significance, but if you have a perfect 12 bit DAC/ADC, each A/D count (or LSBit) corresponds to 1 mV, making some math easier.

FWIW, I personally see 2.5v being more common and I use internal bandgap references for non-critical measurements.

Nick Alexeev
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HL-SDK
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  • I agree. That's what I see most as well and use. Analog Devices and everybody else uses 2.5V all the time. I'm just wondering where the 4.096 came from. As I said, I've heard this mentioned right by the word Bridge a few times. – Pugz Apr 15 '14 at 15:31
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    4.096 is a good number for an A/D working off a +5V supply rail. If it's a 12-bit A/D, then one A/D count corresponds to 1mV. It it's a 10-bit A/D, then one A/D count corresponds to 4mV. Convenient round numbers. – Nick Alexeev Apr 15 '14 at 16:04
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4.096 volts appears to have nothing in common with various Standard Cells, such as they Weston Cell. Even using Li, you'll not get much more than 3 volts from a standardizable chemical reaction.

NIST Standard cell calibrations (pdf):

The US. Legal (or NBS) Volt is presently defined in terms of the atomic constants h (the Planck constant) and e (the elementary charge) via the ac Josepson effect[26]. Critical to this is definition is the role of the Josephson junction which may be regarded as a frequency to voltage converter, where the frequency-to-voltage ratio is precisely equal to to the combination of physical constants 2e/h.... The U.S. legal definition of voltage is known to be smaller than the SI unit by about (9+-1) ppm...

There's no obvious conection to 4.096 volts there either, so I strongly suspect the value 4.096 is chosen for its convenience when using 12 bit ADC, rather than any underlying fundamental physical principle.

Wayfaring Stranger
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You run that 4.096 as the reference to your ADC, and as the power to your bridge. Then, if it varies with time or temp, the reading into your ADC stays the same. I've heard it called a radiometric measurement. The 4.096 gives you plenty of range on your 5V ADC, some headroom for your esd diodes and is commonly available.

ArielP
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    If you have a ratiometric circuit, then there is no reason to use a precise voltage reference. – markrages Apr 15 '14 at 15:51
  • I agree with markrages on that based on what I know but my knowledge is limited. Really, maybe I should be asking "what is meant by bridge?" I know what an H-bridge is for a motor driver but is this the same thing? – Pugz Apr 15 '14 at 16:02
  • @Pugz No, he's most likely talking about a Wheatstone bridge which is commonly used for sensing changes in resistance. – helloworld922 Apr 15 '14 at 16:36
  • I know what that is... usually used for RTDs and other resistive sensors. I don't think this is what anyone was talking about because I'm not aware of any sensors or processes that specifically require 4.096V or result in that voltage. – Pugz Apr 15 '14 at 18:06