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Current sensing resistor vs. resistors in parallel

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Jack.Straw

Member
Hello. I'm building a circuit that calls for a current sensing resistor. Let's say it calls for .25 ohm, could I just use four 1 ohm resistors wired in parallel?
 

MikeMl

Well-Known Member
Most Helpful Member
Yes. You need to calculate how much power the sensing resistor(s) will dissipate, too.
For example, if your 0.25Ω resistor has 5A flowing through it, then the power dissipation would be P=(I^2)*R=25*0.25=6.25W, so it would take four 1Ω 2W resistors...
 

Jack.Straw

Member
Thanks Mike. I'm building a battery charger (getting help on this thread). The power supply i'm currently considering is a switching 1A 12vDC wall-wart. Assuming that means i use 1A as my value for "I" (please correct me if i'm wrong, i'm a newb), then I would use four 1Ω 1/4w resistors?
 
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Chippie

Member
The power supply i'm currently considering is a switching 1A 12vDC wall-wart. Assuming that means i use 1A as my value for "I" (please correct me if i'm wrong, i'm a newb), then I would use four 1Ω 1/4w resistors?

The power unit is capable of supplying 1 Amp...No more!

If this is the load current then 1 Amp would flow through each resistor dissipating 1*1*.25 watts.....then you are right....


If the current of your load is lower then the dissipation would drop...If it were 1/2 an Amp then it would be 1/2*1/2 *.25 watts or 0.0625watts.....
 
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Jack.Straw

Member
Does it matter as long as the wattage is over the needed amount? I mean, say the load is 600mA instead of 1A... could i still use 1/4w resistors?
 

solis365

New Member
Yes, if you calculate your resistors to be able to handle 1A of current then they can handle anything under 1A.

Keep in mind that if you pick the power rating of your resistors to match exactly the power you are dissipating, then the resistors will get HOT! Pick a slightly higher power rating for the resistors than the minimum required.

(1A)^2 * 0.25Ω = 0.25W = total power dissipation by the current sense resistor
since you have 4 resistors in parallel, each one takes 1/4 of the current, so you actually have
(0.25A)^2 * 1Ω = 0.0625W per resistor (multiply by 4 and you have your original power dissipation of 0.25W)

So you could even get away with 1/8W resistors and they would still be at double the required power rating. (1/8 = 0.125 = 2 x 0.0625)
 
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