# Analog to Digital Converter 12 Volts

Discussion in 'Electronic Basics' started by [email protected], Jul 16, 2005.

1. ### Guest

Dear All,
I need to know if you have or know of any analog to digital
converters that can fully convert 12 volts of input voltage or more. I
recently got an adc0804 only to find out that it can only convert up to
5 volts. Thank you very much for your time and consideration in this
matter.
Sincerely,
Christopher Koeber

3. ### John FieldsGuest

---
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

4. ### John LarkinGuest

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

5. ### JamieGuest

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

6. ### Kitchen ManGuest

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).

7. ### John FieldsGuest

---
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.