Measuring negative voltages

[Ioannis]

One idea might be to use an op-amp, like LMC662, and raise the negative voltage with the offset fed into the op-amp. But since the span is |-2-12|=14 volts you will need a divider too. So I would use a divide first to down scale the voltage range with a 2 resistor divider and obtain maximum 3 volts when the input will reach 12. That is because the op-amp would have an offset of 2 volts,so the output would be raise to 5Volts.

The divider should be a divide by 4, like 3K from input to common node and 1K down to earth.

The common node is going to feed the +input of the op-amp and the - input of the op-amp would get -2 offset voltage (with a low Z source). The output should be connected to the -input (follower). Now the power supply should be +/- 5 volt at least (sorry...).

Ioannis

[sammy]

thanx for the idea but is there any way you can give me a diagram.
thanx
sam

[Ioannis]

What part you don't understand?

I do not ussually draw schematics on screen...

Ioannis

[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