Measuring negative voltages

[HenrikOlsson]

Hi Sam,
Attached is a schematic of an inputstage I've designed and used succesfully with the internal ADC. It's an OP-based rectifier with adjustable gain.

Apply your -12V signal to the input and adjust P1 untill you get your desired output (+5V or +4.096 or whatever). (Do this before you connect the output to the ADC)

The sign output will be high or low dependig on if the input voltage is positive or negative and you can use a digital input to read the sign and 'add' one bit of resoultion to the ADC resulting in +/- 10bit. Since your voltage is always negative you can leave that part out if you don't need it.

There's one drawback though. You need to feed the OP-amp with +/-15V.

Hope it helps.

/Henrik Olsson.

[Ioannis]

On the schetch of Henrik please consider the voltage drop on the output resistor of 1K that is going to be in series with the ADC input resistor of your PIC (about 10K). That will give a ratio of about 1/10th.

Ioannis

[HenrikOlsson]

Ioannis, Sam,
Valid concern that MAY need attention!

If the voltage drop over the output resistor does present a problem it can easily be adjusted for by changing the gain of the amplifier/attenuator with P1.

BUT.....

The datasheet for the 16F877 (which is the PIC I used and I suspect others are similar), table 15-12 says that the recommended impedance of the voltage *source* is max 10k. So the 1k shouldn't make that large impact on the result. I couldn't find any info on the actual input impedance of the ADC. Where did you find that?

Further on, I belive that if the input impedance was 10k the ratio would be ~9/10 with a 1k source impedance.

Thanks!
/Henrik Olsson.

[Ioannis]

Well, actually the input resistance of the ADC will change relative to the Vcc. For 5 volts would be 7K according to the diagram in Fig.11-2 of 16F877 DS and ading the interconnect resistance of 1K, makes for a total 8K. So this is a factor to concern when sampling the input. So, I would suggest to skip the maths and do a longer sampling just to be sure.

Ioannis

[HenrikOlsson]

Ioannis,
You're right, ~7k @ 5V Vdd - I totally missed that graph!
As you say, a longer sampling period may be needed. Trying will tell. Also, the PIC Mid Range Manual sugests a 0.1uF capacitor on the input if the signal isn't changing to quickly, may be worth a try if Sam decides to use the circuit.

Thanks for the heads up on that graph!

/Henrik Olsson.

[Ioannis]

You are welcome! That's the idea of the forum, help each other!

Ioannis

[sammy]

thank you all for all your responses. I think i have all the info that i need and now have to do some experiments.

sam