## Simple Apparatus NFPA 70, 504.10(D) Calculation Method

## Simple Apparatus NFPA 70, 504.10(D) Calculation Method

(OP)

The NEC allows the use of a resistor (i.e. and end of line resistor for a fire alarm circuit) to be installed on terminals in a field mounted device certified as “nonincendive” (a Federal Signal Strobe for example) provided the temperature of the resistor doesn’t exceed the temperature rating of the device (i.e. the “T” Code). This is permitted based on the concept that a resistor is considered a “Simple Apparatus” (see definition below from the NEC).

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

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

(100°C/W)•0.5W=50°C

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

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

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

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

T=(P

Where:

_{o}• R_{th}) x T_{amb}Where:

T = Surface Temperature (of the resistor)

P

R

T

P

_{o}= Output power marked on the associated apparatus (0.7 Amps x 24VDC = 16.8 Watts)R

_{th}= Thermal Resistance of Simple Apparatus (This is where I need the help)T

_{amb}= 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

_{sh}= Θ_{sh}•PΔT

_{sh}=50°CWhere:

ΔT

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

0.5W = Resistor Size in Watts (½ Watt)

_{sh}= Increase in Resistor Temperature100°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

_{sh}= R_{th}...and could be plugged into the equation 504.10(D)???

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

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."

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

All you need is the Θsh of the resistor and the maximum available voltage. The voltage and the resistor's resistance value will define the power dissipation the resistor will be faced with. That power dissipation will result in a temperature. That temperature must remain 1) below the ignition temperature of the environment it's in and 2) below the maximum temperature the resistor is designed to function at.

The Θsh describes how fast the resistor can shed heat into the surrounding air from the power that it's dissipating.

Resistors are sold first by the power they can dissipate and secondly by their resistance value. You never want to EVER drive them near their stated power rating. This is because of the standard old rule of 'chemical reaction rates double for every 10 degrees hotter they occur in'. A resistor's major aging function is chemical oxidation. No one wants their resistors checking-out after only a year or so. The second reason is not because of the resistor but because of the things around it, most notably the solder joints holding it to a circuit. If you run a resistor hot it will oxidize the solder holding it to a board causing eventual trace delamination and an open circuit. (It will even char the board black.)

In your example a hot resistor would oxidize the screw terminals eventually causing lots of problems.

You need the resistor's Θsh gotten from the above equation if you have no better way or the better way, from the exact resistor's data sheet which should have it listed.

Then, using the maximum available voltage that the resistor could see you use:

Power = Voltage

^{2}/ResistanceThen Θsh x Power = The temperature

the resistor will see. That's the temp it will increase to above the ambient it's immersed in.RISEKeeping in mind my aforementioned operating power restriction comment above you typically want to keep a resistor running at about a 1/3 of its power rating or less. Ideally nothing more than cozy warm to your finger as compared to instantly scalding, print smoking, temperatures listed as their typical maximum allowed temperatures.

If you've sized the resistor correctly you'll never be in a situation where it would be a fire hazard since you "could leave your finger on it all day".

But the final say is:

Tmax = Tamb + (Θsh x Voltage

^{2}/Resistance)Where Tmax must be below whatever NFPA states it can be.

A good design would have you aiming for something like 1/2 that.

Keith Cress

kcress - http://www.flaminsystems.com

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

_{thj}= 5.2°C/WTTFN (ta ta for now)

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## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

_{sh}I found a datasheet for a typical resistor we use Xicom Carbon Film. I see on page 2, that for a 11Ω to 99kΩ resistor the TC is 0~-450ppm/°C.

For a 10kΩ resistor, would the TC be (99,000-11)•10,000=45.45°C/W, or would this be the Θ

_{sh}?I looked at the link provided by IRstuff, and I couldn't figure out how they came to a Thermal Resistance value of 5.2°C/W

Is "Resistance" the value of the Resistor or can I calculate power as V•I since I know the current draw of the device (24VDC•0.7A = 16.8 Watts), not the resistance.

Would temperature rise then be 45.45•16.8 = 763.56 Temp rise?...that doesn't seem right.

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."

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

They don't "come up" with the value by calculation. It's either a measured value or one determined by the geometry. Again, it has absolutely nothing to do with the temperature coefficient, which is a measure of the ELECTRICAL RESISTANCE change as a function of temperature

TTFN (ta ta for now)

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## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

You need the

degrees (of some sort) verses watts number. That's the only option. The tempco numbers do absolutely nothing for you. Nada. Have a drink on them too!First page see "Thermal Resistance"

Here they call it "Thermal Impedance"

You will have a number in the 50 to 150C (or K) degrees per watt area.

This number is a function of the device's physical geometry. A 1/2Watt thru-hole cylindrical resistor made by Vishay will have the same R

_{th}as a Stackpole or a Panasonic of the same wattage rating or gemoetry. This is back to the TI equation.If you know the value of the resistor and the Θsh or as Vishay likes to call it R

_{th}, and you know the current that is going to be passing thru the resistor, sure, you can use Power = I^{2}x R to get the resistor's power dissipation.No, V x I tells you about the power consumption of the entire circuit not the specific resistor you're trying to prove safe. Unless your circuit is just a single resistor?? I was under the impression you're using this resistor at the end of a chain of other energy using/providing devices, in which case you really need to know the voltage across the resistor to do it right.

Keith Cress

kcress - http://www.flaminsystems.com

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

TTFN (ta ta for now)

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

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## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

Keith Cress

kcress - http://www.flaminsystems.com

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

The circuit connects one or more devices in parallel. With hazardous location rated devices the current is typically too high to connect more than 1 device on a circuit due to the module providing the power being limited to 2 Amps and the field device (strobe) having an inrush of 1.4 amps, but an average of 0.7 amps. Regardless how many devices will function on the circuit, the resistor will always be at the last device on the circuit.

The end devices are polarized and the circuit reverses polarity when it needs to turn on the devices. Under normal conditions the load is around 0.150 Amps. When the circuit is in "alarm", the load is what ever the current draw of the device is, or in my example 0.7 Amps.

The resistor is there so the circuit can measure the voltage drop during normal (non-alarm) and verify that the wiring is intact to the last device (i.e. supervised fire alarm circuit).

I'm not sure what the technical term is for the style of resistor you linked to (VISHAY). The VISHAY datasheet seems to call this a "Leaded Resistor"?.

I'll digest this info this afternoon and provide my math for review.

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."

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

Yes, those are "Leaded Resistors", though as you can imagine there are scads of different sizes of them.

You say, "normally 0.15A". What value is the resistor you usually use?

Keith Cress

kcress - http://www.flaminsystems.com

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

Truth be told, I've never paid attention to the size of the resistor in terms of Watts. Normally the manufacture will specify 1/2 Watt. The resistor value varies between manufacturers. Notifier is usually 4.7K. We're using a product made by Detector Electronics and they need a 10K resistor. Normally the circuit will take a resistor that upto 50% of the design requirements, so in a pinch we could get away with something close to 7K, or 12K. This is just the tolerance incorporated into the circuit by the manufacture.

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."

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

This takes us right back to what I said first. The only way to know what the resistor is doing is to measure the voltage across it. Alternatively the maximum voltage across it is the system voltage which I believe you've stated as 24V.

24V

^{2}/10,000Ω = 58mWThat's the power being dissipated by the resistor.

If say, the resistor has an R

_{th}of 100K/W then the resistor is going to heat up:58mW x 100k/W = a whopping 5.8 degrees Kelvin or Celsius above ambient.

Keith Cress

kcress - http://www.flaminsystems.com

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

I can tell you that with the circuit reverses polarity and device is active, the resistor isn't even warm to the touch.

I'm thinking that I should ask the manufacture what the "supervisory" current it at 24 volts to better explain this?

If either of you can think of what I should ask them to better explain this, let me know.

DM

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

You won't feel something that small being 5 degrees warmer than the environment. Think about it, this resistor is going to be maybe 78F and your fingers are probably about 85F. Your fingers are maybe 40 times more massive. You're not going to feel any difference.

Classic Heisenberg! Touching the resistor will dramatically change its temperature.

Keith Cress

kcress - http://www.flaminsystems.com

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

This may be huge from your perspective, but not from mine. My objective is to make sure the resistor doesn't rise above the temperature rating of the device, which at 114.4°F doesn't. NEC uses "T" codes for hazardous locations. Most equipment has a "T" code of "T4", which essentially means that the equipment, under abnormal conditions, would never get higher than 275°F.

From an hazardous location perspective (i.e. an atmosphere that has combustible vapors), where the gas present has an "Auto Ignition" temperature above 275°F, as long as the equipment can't get higher than 275°F, I'm good to go.

NFPA 70, Article 500.8(C)(4) specifies that ambient should be 40°C. The "T" codes are defined in Table 500.8(C) and they go from "T1" (450°C) down to "T6" (85°C). The equipment label will have a "T" code stamped onto a tag. If a device has a T code of "T1", the device can't be installed in an area that has combustible vapors with an auto ignition temperature below 450°C.

Chemical List with Auto Ignition Temperatures

Based on the above, I'm at a "T6", or 85°C. As long as my equipment doesn't have a "T" code higher than this, which there isn't a higher T Code, I'm good to go.

Now let's assume that the math yielded a temperature rise of 96°K above ambient and my equipment has a "T4" rating (135°C).

Question:

- Would increasing the wattage of the resistor from say 1/2 watt to 1 watt bring the temperature down?
- Does the resistor wattage come into play with: 58mW x 100k/W, where a 1/2 watt resistor would be 200k, and a 2 watt resistor would be 50k?
- I always though that higher wattage could dissipate more heat?

Regards,DM

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

2> not in the way you are calculating; you want to use the same resistance with a higher wattage rating to get the lower thermal resistance

3> yes, and?

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

You say "...Kelvin or Celsius...", wouldn't it be Celsius since the thermal characteristics are expressed in °C?

DM

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method

The SI unit for temperature is actually "kelvin" uncapitalized and without "degree," while "degree Celsius" is the correct construction for the centigrade.

I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg

## RE: Simple Apparatus NFPA 70, 504.10(D) Calculation Method