# Thread: Sensing Resistors vs Current limiting Resistors

1. ## Sensing Resistors vs Current limiting Resistors

Im trying to get my arms around the differences of using a constant current motor driver with current limiting resistors or a current sensing resistor setup with the SLA7062M. Questions to follow.

My motor for the sake of arguement and numbers is 5V @ 1amp. with a 30V power supply.

To construct a constant current board with current limiting resistors I calculate the resistor value as follows:
(30-5) / 1 = 25 I should use a 25 ohm resistor

To calculate resistor power, I calculate current squared multiplied by resistor value

1*25=25 ....so my 25 ohm resistor should be rated at 25 watts.

Am I out of whack here ??? Is this the formula to use???

Lets say I want to use current sensing resistor in the SLA7062M

href="http://www.pminmo.com/SLA7062/DIYSLA7062schematic.pdf"

I have used a formula that makes little sense and have difficulty in finding the basis for the 1.25 value.

Current sensing resistor calculated using 1.25/Amps out=ohms

So in my case 1.25/1=1.25 ohm resistor

Question: can I calculate power as in the previous example using current squared*resistor .
The example in the link shows two resistors in paralell (10,11) (19,20) so I expect the heat generated should be disappated over a couple of ? watt resistors.

Any help or assistance is appreciated.
Whitney

2. Power dissipation is purely a function of the current flowing thru the resistance that it flows thru.

Current limit resistors need to be "big" enough to handle the thermal load induced heat that the current they are dissipating will generate.

Sense resistors are selected to generate a certain voltage for the current that you're programming the resistor to "react" to. I suspect that the 1.25 factor is a current threshold voltage limit that has to be matched/exceeded for the current limiter to kick in.

As far as power dissipation in the sense resistors goes, these may not be so critical as a low ohm resistor won't get that hot so it won't change value all that much. It would not hurt, however to use a decent sized one as well as a aluminum encased precision one (as supplied by DALE).

This way you can heat sink it so as to have a more accurate and consistant current sense funtion - keeping a constant temp across the current sense resistor is critical to accurate current sensing. A review of the resistance coefficient vs temp value will more clearly show/explain why.

3. Couple of points, I set the copper up to allow a person to have an easier choice of sense resistors buy using parallel resistors. You don't have to use parallel resistors. For the sake of discussion, I will refer to one sense resistor, Rsense. If a person uses parallel resistors, consider the parallel electrical equavalent as Rsense.

For the SLA7062 Itrip=Vref/RS i.e. ohms law I=E/R. The voltage at the sense input needs to be less than 2V at Itrip (coil current rating) and no less than .1V (Allegro's recommendation). Above 2V the SLA7062 goes into sleep mode. Say you can get four 1ohm 1W resistors. The parallel value is .5ohm at 2W. At 3A you will develop 1.5V across the sense resistor, but it would require 4.5W total. So that would be a bad choice. So lets say you can get 1/2ohm 1 watt resistors. the electrical equavalent is .25ohm 2 W 3A at .25ohm is 2.25W still a bad choice but close.

For 1A the minimum sense resistance is 1 , max 2 ohm. 2W
For 2A the minimum sense resistance is .5ohm, max 1 ohm. 4W
For 3A the minimum sense resistance is .34ohm, max is .66ohm. 6W

For 3A I would recommend two .5ohm 2W resistors for each coil. (electrical equavalent of one .25ohm 4W resistor) or one .2ohm 2W resistor.
For 2A, one .5ohm 2W resistor for each coil, etc.

Vref voltage should be adjusted to Rsense * Itrip. So if your motor is 1.5A, and Rsense is .2 ohm, Vref should be set to .75V

Postscript. Don't use wirewound resistors.

Now using resistors to limit current in the linear sense (no chopper or electronics current limiting). You are on the right track with your power resistor calculation, but it's just a tiny bit more complicated depending on full step/half step/wave drive......

4. Got It!! I need more study of the data sheets I suspect.

Thanks again,
Whitney