Analog to Digital Converter 12 Volts

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Why not use a voltage divider to scale your input voltage to match the
input voltage range of the ADC?

An ADS7805 will digitize +-10.24 volts or something with just a +5 supply... one of the few ADCs that accept true bipolar inputs that way.

Some series input resistance would probably extend that to +-12 or so.

But for unipolar, just use a resistive voltage divider as others suggest.

John

drive that output to a + input on an op-amp. set the loop back gain to tailor your needs.

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Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

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No. 

You can\'t get a gain of less than 1 from a non-inverting amp:



Ein>-----|+\\
         |  >--+-->Eout
      +--|-/   |
      |        |
      +--[R2]--+
      |
     [R1]
      |
GND>--+

                 Ein (R1 + R2)
         Eout = --------------- 
                      R1

I had the same thought, but see John's answer that the gain of a non-inverting op-amp cannot be less than unity. Use the inverting input, and a voltage-follower to re-convert the polarity. But again, see John's first answer that all this amplifier design is unnecessary if the input voltage can be converted by as simple an implementation as a voltage divider (use large resistance values to minimize current and thus power loss in the divider).

--
Al Brennan

"If you only knew the magnificence of the 3, 6 and 9,
 then you would have a key to the universe." Nicola Tesla

--- Interestingly, for this application something like a 10 or 20 turn pot would be better than fixed resistors for a couple of reasons, the first being that even with something like an 8 bit ADC, to scale the

12V down to 5V and get the full resolution of the ADC, the ratio of the resistances has to be:

Ein | [R1] | +---->Eout | [R2] | 0V

R1 Ein - Eout 7V ---- = ------------ = ---- = 1.4 R2 Eout 5V

Which works out nicely for standard 1% resistors if R1 is 14k and R2 is 10k. However, because of the 1% tolerance, R2 could vary anywhere between 13860 ohms and 14140 ohms, and R1 could vary anywhere between

9900 ohms and 10100 ohms, which means that with Ein equal to precisely 12V Eout could vary anywhere between: Ein R2min 12V * 9900R Eoutmin = --------------- = ---------------- = 4.942V R1max + R2min 14140R + 9900R

and:

Ein R2max 12V * 10100 Eoutmax = --------------- = ---------------- = 5.058V R1min + R2max 13860 + 10100R

That's about a 60mV error each way, which could overrun the ADC on the high end and introduce an error of 3 LSB's (about 1%) on the low end.

With a 10 turn pot in there, 12V would be spread out over 3600 degrees of rotation, so 5° (which is pretty easy to resolve manually) would correspond to an error of

5° * 100 E% = ---------- ~ 0.14% 3600°

_and_ with the pot adjusted to just 255 with 12V on the input, the overrun problem would go away.

Using a 20 turn pot would bring that error down to about 0.07%, which would allow a 10 bit ADC to be used to its full resolution.

The second reason is that the pot's resistive element would likely exhibit the same tempco from end to end while two discrete resistors would not likely exhibit each other's trmpco, and the element would likely be in an environment more nearly isothermal than that of the two discrete resistors. That's important because in a voltage divider what's important is the ratio of the two resistances, and that ratio is more likely to remain constant for a pot going through a thermal excursion than two separate resistors going through the same excursion.

-- John Fields Professional Circuit Designer

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The third reason is that the pot would allow the ADC\'s output(s) to be
adjusted to precisely 255 for a 12V input to the pot regardless of the
ADC\'s input resistance.