Hello forum,

I'm using a PIC's analog input pin with the internal AD converter to measure the voltage from a voltage devider, as in the circuit diagram below.

Now, the PIC's datasheet says max 10K input impedance, but I'm not sure how to calculate this value.

The resistor devider supply can vary from 4V to 10V.

Thanks for any help.

Regards,
Futterama

Attached Images

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The source impedance should be no more than 2.5K, your divider is considerably more than that!. This will restrict how fast you can switch between channels, and may have some bearing on accuracy.

If you can't use a lower value potential divider, you should consider an opamp buffer between the PIC and divider - check my analogue tutorial!.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

But why? The datasheet says 10K. Please explain

Ok, then please tell me how much more, and include a calculation so I can calculate it myself :wink:

Oh, but I can lower the resistor values, but not below 1K total because of the current consumed by the resistors then.

I'm sorry Nigel, I don't find any helpful tutorial on you site, it is a bit confusing finding anything in there :roll:

Regards,
Futterama

 

Sorry, I see the 12F675 states 10k, other PIC's are usually 2.5k, so the 12F675 is somewhat better.

Actually, when you sit down and work it out, it's slightly LESS than 10k, but it looks more at first glance.

You only need ohms law and parallel resistors to work it out - you need to work out what value resistor you would put in parallel with the bottom resistor to give half the voltage - the extra resistor equals the source impedance.

Having seen your other post, I don't see any problems with the existing values, there's no need for speed or accuracy.

The tutorial is at http://www.winpicprog.co.uk/pic_tutorial11.htm and the hardware used is at http://www.winpicprog.co.uk/pic_tuto...ogue_board.htm.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

You'd consider any low impedance DC voltage to be a ground, and then calculate the resistance to the measuring point. So 30k in parallel with 10k = 7.5k.

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I thought what I'd do was I'd pretend I was one of those deaf-mutes.

 

Thanks, both methods works for me, but Oznogs is much easier to work with :wink:

Regards,
Futterama

 

But I don't think it gives an accurate value? - I couldn't be bothered working it out exactly, but it came out between 9K and 10K. Not that it matters much for this application!.

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk

 

Nigel, that's odd, I used an Excel worksheet to do the calculation, and I also ended up with 7.5k with the method you mentioned.

 

Perhaps I messed up? - I'll try it again when I have time, I did it before with a calculator in one hand, a phone in the other, and a piece of paper on the bench :lol:

__________________
PIC programmer software, and PIC Tutorials at:
http://www.winpicprog.co.uk