Mesuring current with pic

[Charles Linquis]

In most cases, the best way to measure current is with a Hall-effect sensor. No voltage loss, no heat, perfect isolation. Check out the Allegro devices.

http://www.allegromicro.com/en/Produ...rentsensor.asp

[kblim]

Maybe a low ohmic resistor 0.01ohm 5W is suitable in the application.

The hall effect sensor normally is very expensive in my country.

Let say the total amp flows is 15A,:

Voltage dropped = 15 X 0.01 = 0.15V
Wattage dissipated = 15 X 0.15 = 2.25W

I have been using this with the volt meter to measure the voltage up to 22A, and it is working fine. The heat will not be a issue for the resistor. But it is a bit expensive.

[aratti]

Wattage dissipated = 15 X 0.15 = 2.25W

You mean 15x15x0.15=33.75W

Al.

[rmteo]

P = I*V = 15 x .15 = 2.25W

[amgen]

or,
p=1^2 R
15X15X.01=2.25

[rmteo]

or V*V/R
0.15*0.15/0.01 = 2.25W

[Darrel Taylor]

Somebody needs a new Calculator.


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Change R1 to 0.01 ohm, and see what happens.
Note the output voltage to the PIC.
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