Measuring negative voltages

[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

[Charles Linquis]

I measure negative voltages accurately without any external active components.

All you have to do is connect a resistor from Vref to the A/D input pin, and another resistor from that A/D pin to the voltage to be measured.

For example: To measure -12V, I use a National 4.096V reference on RA3 and a 2K resistor from that pin to the A/D input I want to use. From there, I use a 7.5K resistor to the voltage to be measured.

The code then loops 19 times and adds the results of the conversions (for averaging). The result is then subtracted from 15360. The answer is in tens of millivolts.

I have used this technique for years with good results.

[Ioannis]

Hi Charles. Yeap, this is the superposition theorem. I really forgot about it. Too many years ago...

Clever though.

Ioannis

[sammy]

Hi Charles:
That is exacly what i was looking for. do you have any formulas as to how to calculated different resistor values and to know how much offset you have and what not.
thank you
sam