Electronic – How to measure a negative voltage with a ADC

important note:
this answer was posted to solve the problem for -20V to +20V input, because that was what was asked. It's a clever method but doesn't work if the input voltage limit stays between the rails.

You'll have to scale the voltage with a resistor divider so that you get a voltage between -2.5V and +2.5V, and add 2.5V. (I'm presuming a 5V power supply for your PIC).

The following calculation looks long, but that's only because I explain every step in detail. In reality it's so easy that you can do it in your head in no time.

First this:

R1 is the resistor between \$V_{IN}\$ and \$V_{OUT}\$,
R2 is the resistor between \$+5V\$ and \$V_{OUT}\$, and
R3 is the resistor between \$V_{OUT}\$ and \$GND\$.

How many unknowns do we have? Three, R1, R2 and R3. Not quite, we can choose one value freely, and the other two are dependent on that one. Let's choose R3 = 1k. The mathematical way to find the other values is to create a set of two simultaneous equations from two (\$V_{IN}\$, \$V_{OUT}\$) pairs, and solve for the unknown resistor values. Any (\$V_{IN}\$, \$V_{OUT}\$) pairs will do, but we'll see that we can tremendously simplify things by carefully choosing those pairs, namely the extreme values: (\$+20V\$, \$+5V\$) and (\$-20V\$, \$0V\$).

First case: \$V_{IN} = +20V\$, \$V_{OUT}=+5V\$
Note that (and this is the key to the solution!) both ends of R2 see \$+5V\$, so there's no voltage drop, and therefore no current through R2. That means that \$I_{R1}\$ has to be the same as \$I_{R3}\$ (KCL).
\$I_{R3}=\dfrac{+5V-0V}{1k\Omega}=5mA=I_{R1}\$.
We know the current through R1, and also the voltage over it, so we can calculate its resistance: \$R1=\dfrac{+20V-5V}{5mA}=3k\Omega\$.
Found our first unknown!

Second case: \$V_{IN} = -20V\$, \$V_{OUT}=0V\$
The same thing as with R2 happens now with R3: no voltage drop, so no current. Again according to KCL, now \$I_{R1}\$ = \$I_{R2}\$.
\$I_{R1}=\dfrac{-20V-0V}{3k\Omega}=6.67mA=I_{R2}\$.
We know the current through R2, and also the voltage over it, so we can calculate its resistance: \$R2=\dfrac{+5V-0V}{6.67mA}=0.75k\Omega\$.
Found our second unknown!

So a solution is: \$R1 = 3k\Omega, R2 = 0.75k\Omega, R3 = 1k\Omega\$.

Like I said it's only the ratio between these values which is important, so I might as well pick \$R1 = 12k\Omega, R2 = 3k\Omega, R3 = 4k\Omega\$.
We can check this solution against another (\$V_{IN}\$, \$V_{OUT}\$) pair, e.g. (\$0V\$, \$2.5V\$). R1 and R3 are now parallel (they both have +2.5V-0V over them), so when we calculate their combined value we find \$0.75k\Omega\$, exactly the value of R2, and the value we needed to get \$+2.5V\$ from \$+5V\$! So our solution is indeed correct. [QC stamp goes here]

The last thing to do is to connect \$V_{OUT}\$ to the PIC's ADC. ADCs often have rather low input resistances, so this may disturb our carefully calculated equilibrium. Nothing to worry about, however, we simply have to increase R3 so that \$R3 // R_{ADC} = 1k\Omega\$. Suppose \$R_{ADC} = 5k\Omega\$, then \$\dfrac{1}{1k\Omega}=\dfrac{1}{R3}+\dfrac{1}{R_{ADC}}=\dfrac{1}{R3}+\dfrac{1}{5k\Omega}\$. From this we find \$R3=1.25k\Omega\$.

edit
OK, that was clever and very simple, even if I say so myself. ;-) But why wouldn't this work if the input voltage stays between the rails? In the above situations we always had a resistor which had no current flowing through it, so that, following KCL, the current coming into the \$V_{OUT}\$ node via one resistor would leave via the other one. That meant that one voltage had to be higher than \$V_{OUT}\$, and the other lower. If both voltages are lower there would only flow current away from that node, and KCL forbids that.

Here's a single supply inverting opamp configuration that will do what you want. You will need an opamp capable of output drive to it's lower rail (You will probably want to include a small capacitor across R2 to limit bandwidth, since you don't need much for thermocouple readings)
R3/R2 may need to be increased in order not to load thermocouple depending on type - EDIT, just noticed the output is coming from the AD595, so it's probably low impedance (not checked datasheet) and fine as is:

R3/R2 simply divide the input voltage by 2. R1 and R5 present 400mV to the positive input. Since the opamp tries to keep the two inputs equal, it creates a level shift. For example, when there is -1.2V at the input, to keep the inverting input at 400mV, there needs to be 1.2V at the output. We can now see R3/R2 as a voltage divider with -1.2V at one end and +1.2V at the other, we get 2.4V across R3+R2, so the voltage across R3 is:

2.4V * (R3 / (R2 + R3)) = 2.4V * (10kΩ / 15kΩ) = 1.6V and so:

-1.2V + 1.6V = 400mV

You can run the calculations for the other input voltages and see how it works across the range (remembering there is always 400mV at the inverting input, and effectively no current flows into the input)

Another way to look at it given the above is, say we have -0.6V at the input. We know there must be +0.4V at the other side of R3, so the current flowing through R3 is:

(0.4V - -0.6V) / 10kΩ = 0.1mA

Now we know none of this current flows into the inverting input, so it must flow through R2:

5kΩ * 0.1mA = 0.5V

0.4V + 0.5V = 0.9V at the output

Simulation:

If you need it non-inverting, you can easily do this in firmware or add a simple inverting buffer after this.

ZIGBEE ADC

Just had a look at the Zigbee datasheet and it seems the Vref is fixed at 1.2V (although there is Vref pin, I couldn't find any mention of how to use it in the analog IO section), so you have to work with this unless you use an external (possibly higher resolution) ADC and feed the data to the Zigbee. It's a 10-bit ADC, so 1.2V / 1024 = ~1.17mV LSB, which won't be so bad with with filtering (which use a low cutoff since you have a slowly changing signal from the thermocouple)
Bear in mind the ADC595 has an calibration error of around +-1°C (or +-3%deg;C depending on which variant you are using) so absolute accuracy will not be excellent, but you could go for a higher resolution as mention if you wanted to.
So read the ADC595 datasheet advice thoroughly, pay attention to the PCB layout (if possible a 4-layer with solid ground plane), keep any digital signals away from the analog as best you can and use plenty of decoupling and all should be well.