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reading analogue voltage on A/D!!!

 
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ambivalent
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reading analogue voltage on A/D!!!
PostPosted: Tue Nov 01, 2005 12:20 pm     Reply with quote

hi guyz, i hope everyone is doing fine.

i want to read analogue voltage on the A/D of pic16f877. the problem is that the voltage varies from -1 --> 6 volts. now i believe that this kind of voltage can not be applied directly to the PIC as 6volts may damage it.
i had the idea of adding 1 volt to the analogue voltage so that i have a range of 0 --> 7 volts, and then half this voltage before applying it to the analogue pin of the PIC, so i would have a range of 0 --> 3.5 volts. and then i would compensate for these manipulations in the converted digital value. Now this solution seems kind of complicated for such a simple issue. can any one please suggest another simpler way of reading an analogue voltage ranging from -1 --> 6 volts.

best regards
Neutone



Joined: 08 Sep 2003
Posts: 839
Location: Houston

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PostPosted: Tue Nov 01, 2005 12:59 pm     Reply with quote

I'm not sure but think a combination voltage divider will do it if the source can drive the resisters in this circuit. Assuming your running on 5 volts.
Code:
                 5V
                 |
                 R2
                 |
                 |
Source --- R1 ---+--- PIC in
                 |
                 |
                 R3
                 |
                GND


R1=R2*5
R2=R3
asmallri



Joined: 12 Aug 2004
Posts: 1634
Location: Perth, Australia

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PostPosted: Tue Nov 01, 2005 5:39 pm     Reply with quote

Deleted - I goofed. I did not see the input was from -1 volt and the divider impedance was too high for the A/D as pointed out by PCM Programmer.
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Last edited by asmallri on Tue Nov 01, 2005 6:43 pm; edited 2 times in total
PCM programmer



Joined: 06 Sep 2003
Posts: 21708

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PostPosted: Tue Nov 01, 2005 6:01 pm     Reply with quote

Quote:
R1=20K

The impedance of the voltage divider circuit which feeds the PIC's
A/D pin must be kept under 10K, per the 16F877 data sheet.

What if he uses a voltage divider which divides the input range by 2
and then biases the bottom resistor at 1.0v so the output is 0 to 3.5v.

Code:

Vin           4.7K
-1.0v  >----/\/\/\/---------->  To PIC A/D pin 
  to                    |
+6.0v                   >
                        < 4.7K
                        >   
                        |
                      +1.0v       



------------
I just realized that my proposed solution is the Thevenin equivalent
to Neutone's circuit. Just a rough check gives approximate values
for his circuit of:

R1 = 4.7K
R2 = 27K
R3 = 6K
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