Precision Shunt Resistor Sizing

R

Thread Starter

Randy

I plan to use a precision shunt resistor to convert 4-20mA to 1-5VDC. I know I need a 250ohm resistor. How do you determine the wattage the resistor needs to be? Is 1/8 watt sufficient?
 
A

Alan Rimmington

Current x voltage = watts so you need 100mW resistor however your devices may produce more than 20mA, say 25mA so a 1/4W resistor would be a better choice.
 
B

Bouchard, James \[CPCCA\]

Ohm's law P =I2R = 0.02 x 0.02 x 250 = 0.1 watt so 1/8 watt would be barely sufficient I would use at least 1/4 watt. Also you may want to check how the resistance varies with temperature.

James Bouchard
 
The calculation for the wattage should take in to account the loop voltage, usually 24 VDC and the maximum potential current through the 250 ohm resistor.

This calulates to 100 mA possible in the loop, (.1 X 24)= 2.4 watts.

Hope this helps
Regards
 
J

James Johnson

Watts can be calculated by multiplying Current by voltage. We know that your 250 ohm resistor drops 5 volts at 20ma. so..... 5 x .02 =.1 watt. Your 1/8th(.125) watt resistor should be fine.

James Johnson
 
B
Seems to me the worst case would be 20 mA, not 100 mA since the signal is 4-20 mA!

P=I^2 * R = 0.02 *0.02 * 250 = 1/10 Watt maximum dissipation.



Thanks,
Bob Peterson
 
Though calculations show 100 milliwatt as the power dissipated, You will need 1 or 2 watts size resistor as 1/8, 1/4 or 1/2 wattage resistors are not stable with varying temparature. Use a 1 or 2W temp. compensated resistor
regards,
Sekar
 
C
That's a misleading statement. Precision resistors are stable in all sizes. A 1 or 2 watt won't get as hot, but a quality .25 watt or .5 watt should be adequate over the commercial range.

Regards

cww
 
F

Fred Townsend

I agree with Curt but there is another problem. 250 ohms is NOT standard value! In the table of standard electronic parts the closest value would be 249 ohms. One ohm is also a standard value. One solution would be to buy a 1
ohm and a 249 ohm, 1/2, 0.1% metal film resistors and place them in series. 1/8 or 1/4 watt could be used too but 1/2 watt are usually more available and cheaper.

A second solution would be to get an instrumentation 250 ohm shunt. Some instrument companies stocked them at one time. They may still be available or special ordered.

Fred Townsend
 
C
I would typically use the next higher standard value and design the following ground referencing amplifier with about 25k input resistance to the summing junction. Sidesteps the problem and helps with the error budget. It also swamps the input current changes with temperature of the Op Amp.

Regards

cww.
 
F

Fred Townsend

Curt, I suspect this approach could be troublesome. First the summing junction is the inverting input. This means the signal will be inverted unless it is followed by another inverter. Second, the op amp input requires the input signal to be referenced to ground. If one were to successfully insert a ground point there would be poor common mode rejection. An instrumentation op amp would be a better solution. The ap note found at:

http://www.analog.com/UploadedFiles/Technical_Articles/25406877Common.pdf

discusses some of the common mode issues.

Regards,

Fred Townsend
 
C
For the general case you're right. Design of an input that
has the compliance to be anyplace in a current loop is non
trivial. But for direct conversion where one would use a
resistor and a (probably) single ended analog input for a
PLC the point is moot. I simply remove the 1V offset and
ground reference the signal to use the entire span for
better resolution. Since you have gain you can also use
R < 250 ohm and scale.

Regards

cww
 
I want to build an Ammeter for my battery
useing my digital volt meter. my battery is 12v 7Ah so my shunt resistor is 1.7 ohms is this the right way to do it thanks.

WILL
 
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