# Resistor Temperature

1. Jun 22, 2007

### emlombardo

I am working on getting an MSHA approval for a device and I would like to estimate what the surface temperature of a resistor will be over time.

Is there a way to calculate the surface temperature (temp change per unit time perhaps) of a known resistor if a simple DC voltage is applied?

Any advice would be greatly appreciated.

2. Jun 22, 2007

### Staff: Mentor

Hmmm, I don't see a thermal resistance listed for a jellybean resistor on its datasheet:

http://www.panasonic.com/industrial/components/pdf/AOA0000CE14.pdf

The thermal resistance for the resistor is what you need -- and it will need to be tailored to how it is mounted on the PCB, and what the heat sinking characteristics of your PCB and package are.

The easiest way to determine this (and how we typically do it for our products) is to measure it with a thermocouple, especially at the highest rated ambient operating temperature for your procuct. Do you have a (low-mass) thermocouple setup?

For something as small as a low-power resistor, a non-contact IR thermometer may be a better way to do the measurement, but that assumes that you can see the resistor while the product is operating normally (covers closed), and would normally be made at ambient temperature instead of in an oven.

3. Jun 22, 2007

### emlombardo

The problem is that the resistors are part of a strain gauge bridge inside of a welded pressure transducer. I have the resistance values but the samples were sent direct to MSHA so I can't take any actual measurements. We did not manufacture the transducer and the manufacturer is actually the worst that I have ever dealt with.

I was hoping that there was a way to estimate mathematically based on current, etc...

4. Jun 22, 2007

### Staff: Mentor

Well, with heat-sinked components or ICs, you will be given a range of thermal resistance, depending on how the part is mounted. The thermal resistances can range from a fraction of a degree C per Watt to 10's of degrees C per Watt. A good heat sink mounted to a metal enclosure would come in at the low range of C/W, and a DIP IC or through-hole component with just traces leaving the solder pads would come in at the high range.

What is the power dissipation of this resistor, and what package is it in (SMT, axial, etc.)? Is one end connected to a ground plane via, or are both ends just connected to skinny PCB traces?

5. Jun 22, 2007

### emlombardo

I know that they are axial leaded resistors. I think that they are mounted to a PCB inside of the case but I can't get that info from the manufacturer.

If I remember correctly, the testing is done with the element covered in coal dust and if it smolders, it fails. If I am thinking of the wrong test, it may be done in open air and the surface temperature is what is measured.

Thanks for your help, it's nice to be able to have questions like this answered. I am the only EE that my company employs so I don't usually have anyone to ask if I get stuck.

6. Jun 23, 2007

### NoTime

If it's a welded black box then it seems to me the best you could do is calculate the temp rise in the device as a whole.
So you could probably calculate for cube in free air (or coal dust) disa[ating the input power.

7. Jun 23, 2007

### xez

Well this doesn't directly solve your problem, but it may
avoid the issue depending on the regulartory specification /
test requirements.

Is there a limit to the input *power* that you apply to the
"black box" strain gauge bridge? For instance, if you
only periodically read the bridge, say 100 microsecond
measurements conducted once per second, and in between
measurements the module/bridge is deactivated
from excitation power then that'd only be 1/10,000th of
the maximum pulse input power that'd be available to
heat the bridge on average.

So even if you used significant momentary excitation
powers, the average power over any reasonable amount
of time might be just miniscule relative to the time
that any component heating could occur over.

I'd guess the module you're using has some rated value
of maximum excitation power, and bridge resistance,
so from that and the duty cycle of your excitation
power input you could figure out the overall input energy
and power that's being applied to the bridge.

Perhaps the regulatory test / specification specifically
would exempt equipment operating over sufficiently
low input voltage/current/energy levels from thermal
aspects of regulatory investigation. At sufficiently
low levels it becomes conceptually in the same category
as a hearing aid, digital watch, or whatever.

From the overall mass of the welded unit you could
(based on input power applied) calculate the
watts / kilogram or joules / kilogram excitation or
similar for the overall transducer module and perhaps
specify some insignificantly small (assuming that's
the case based on your input power) surface temperature
rise for that as a whole (ignoring its individual parts)
if it's a hermetically welded / sealed unit.

Relative to explosive atmosphere certification I'd assume
that either some kind of minimum
voltage / power / energy INPUT requirement might be
relevant, as well as some kind of certification of hermiticity
and quality of the module's enviromental seal; if the
module wasn't environmentally sealed, and the input
power/voltage wasn't limited, even if the normal
component surface temperatures were sufficiently low
by design, a fault (cracked wire / resistor) could lead to
a very high temperature spot or arc/spark which clearly
wouldn't be safe in an explosive atmosphere.

So given other constraints of the system / design and
specification you MAY in such fashions find exemptions
that would obviate the necessity for you to know / specify
what goes on INSIDE the black box sealed module.

8. Jun 23, 2007

### xez

Oh, this is a bit of a stretch, but if you could figure
out the temperature coefficient of resistance of the resistors
in the bridge, you could just MEASURE the change in
resistance of the bridge relative to known environmental
temperatures (soak it in an oven until it's stabilized),
and thereby be able to calculate the approximate rise
in bridge resistor temperature based on an increased
excitation level.

That all assumes that the bridge isn't temperature
compensated and the temperature coefficient of
resistance behaves in a simple and sufficiently measurable
manner over the possible range of operating temperatures.

You'd probably need a couple sample units,
a precision and calibrated multimeter / ohmmeter,
a test oven that'll go from 0 to 100C or something like that,
and the ability to measure / test the device while it's
in the oven at a stable known ambient temperature.

9. Jun 25, 2007

### emlombardo

@NoTime:
Testing the entire device would be ideal and a very practical solution but their testing is done on each individual resistor in the circuit, not on the circuit as a whole. This means that the entire 12V from the batteries will be placed across each resistor for the testing.

10. Jun 25, 2007

### emlombardo

Xez:

The idea of only powering the transducer when sampling is a good one that I had not thought about. However, the circuit is analog and connected to an analog meter. It was designed in 1977 and the transducer replacement is the first change to the electrical system. Also, if I made any changes to the circuit board it would also have to be retested and that could open a can or worms that I don't even want to think about.

There is an exception for testing if the resistor does not have the potential of exceeding 2/3 of its rated power. However, I was not here when the transducer was designed and they gave no thought to this at the time. Of course, it would require huge resistors because of the testing method; you have to assume that you will have a fault condition that will place the entire battery supply voltage across your resistor (in series with a current limiting resistor).

The resistance values used to offset this bridge are very small, 5-10 ohm, so they will allow a large current flow if placed across the entire supply voltage. In fact, with a 10 ohm current limiting resistor and a 5 ohm resistor in the circuit, the current draw @12V would be .8A. The only thing that might save me is a 1/8A resistor in the circuit. It will be used for the testing so if it blows before the resistor gets hot the component will pass. If they slowly ramp up the voltage, the resistor will be allowed to pass up to 1/8A and will be dissipating .5W. It is only a 1/8W resistor so I don't think it will go over well.

I am going to send the components for testing and see what happens. In the meantime, I will research adding a second, lower power fuse to the system.

11. Jun 25, 2007

### emlombardo

I thought this was a stupid question but I am glad I asked. It is a fascinating topic and I would have liked to evaluate it more while I still had access to my university lab.

Thanks for the help everyone!

12. Jun 26, 2007

### NoTime

The resistor has to survive a fault condition?
Fuses or Overcurrent Foldback don't count?
Given the test condition you state then I would think the circuit does not count. All you would need is the resistor size and type (with some possible consideration for lead dress).