Mesuring current with pic

[mitchf14]

I found this circuit to measure current with a pic, it seems like a good circuit that will work, using adcin reading the voltage at the high point of the 0.1 ohm resistor. Hence knowing the voltage and the resistance you can calculate the current that is all fine and dandy, my only concern is that @ 12 v running trough a 0.1 ohm a 5 w resistor would no be at all sufficient. I would need a big heat sink and even then is there something i am not seeing, or is the circuit wrong.

[aratti]

You have a lamp of 50 W in series with the resistor of 0.1 ohm 5 W.

50W/12V=4.2 A this is the current that flow through the resistor.

4.2 * 4.2 * 0.1 = 1.74 W is the power dissipated by the resistor 5 W is more than dimensioned.

Al.

[mitchf14]

That makes lots of sense but what will happen when the 50 w bulb is off, they only load then will be the 0.1 ohm resistor. The power will be over 5 w and burn the resistor out.

[mackrackit]

If the bulb is off then the circuit is broken. No current to flow.

[rmteo]

Never mind. Was not thinking straight.

[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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[amgen]

Mr DT,
what a neat utility that is !!
you can see all the relations between R's, V's, W's etc.
How about lessons on creating that type of object ?
don
amgen

[Darrel Taylor]

Thanks Don!

I'm still looking for those lessons myself.
Been trying to learn Flash for over a year now, and it's still feels like I'm struggling.

Maybe another year of practice will help.
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