From NFPA 70, 2016 Edition, Equation 504.10(D)

T=(P[sub]o[/sub] • R[sub]th[/sub]) x T[sub]amb[/sub]

Where:

Where:

T = Surface Temperature (of the resistor)

P[sub]o[/sub] = Output power marked on the associated apparatus (0.7 Amps x 24VDC = 16.8 Watts)

R[sub]th[/sub] = Thermal Resistance of Simple Apparatus (This is where I need the help)

T[sub]amb[/sub] = Ambient Temperature (Lets' assume 104°F [40°C])

P[sub]o[/sub] = Output power marked on the associated apparatus (0.7 Amps x 24VDC = 16.8 Watts)

R[sub]th[/sub] = Thermal Resistance of Simple Apparatus (This is where I need the help)

T[sub]amb[/sub] = Ambient Temperature (Lets' assume 104°F [40°C])

I’m trying to figure out how to do these calculations. Most of what I’ve read says you need to consider the “temperature coefficient” or “TC” of the resistor which apparently is expressed in terms of “ppm / °C”. I looked up a resistor and found that the TC is “500 to 350 ppm / °C”.

I came across some information on a Texas Instrument Web site titled "How to Calculation the effects of resistor self-heating." Equation 3 seems to be what I need

ΔT[sub]sh[/sub] = Θ[sub]sh[/sub]•P

ΔT[sub]sh[/sub]=50°C

Where:

ΔT[sub]sh[/sub] = Increase in Resistor Temperature

100°C/W = Temperature Coefficient of a typical Resistor (100ppm/°C)

0.5W = Resistor Size in Watts (½ Watt)

100°C/W = Temperature Coefficient of a typical Resistor (100ppm/°C)

0.5W = Resistor Size in Watts (½ Watt)

Based on the above, I'm assuming...

ΔT[sub]sh[/sub] = R[sub]th[/sub]

...and could be plugged into the equation 504.10(D)???

...or has someone already plugged this into a spreadsheet somewhere?

Regards,

DM

"Real world Knowledge isn't dropped from a parachute in the sky but rather acquired in tiny increments from a variety of sources including panic and curiosity."