Resistor Error

Hint: If the rating of the resistor is lower than 2.9W and there is a short circuit downstream of it, then no wonder.

Another hint: 200 ohms in series with what?

Are you trying to power your lower voltage instrument with just a resistance power drop. Why did you put this unknown wattage resistor in series with anything? What does this have to do with calibration? Do you have a calibration procedure? If you do have a calibration procedure, what does it specifically say?

If you cannot or do not answer these questions I recommend that you stop what you are doing before you damage equipment, property or yourself. You might even try hiring an electronics technician.

Ronron,

I agree with the previous posts, maybe you have a short circuit present in the electrical loop.

If you do the following calculus using P = V. I

P = V.I = 24V . 4mA = 0.096 Watts
P = V.I = 24V . 20mA = 0.48 Watts

The power dissipated is very lower and if the resistor is hot, one possible causes is that you have high current flow trough the electrical loop, caused by some error in the connections in the power supply or transmitter, or the electronic board of the transmitter is bad.

Remember that the impedances is used by two reason:

1.- Convert mA to Volts (4mA = 1Vdc, 20mA = 5Vdc) , using 250 ohms resistor

2.- Get a minimun loop impedance necessary for Hart protocol communication, 250 ohms minimun

Please, Can you show us the electrical drawing used by you?

Include the power supply and transmitter model and brands

Regards
JP

you sure you have the correct values for the components you are using?.. and yes CD will help, with a expansion on the experiment you are trying and the results you are hopefully looking for.

You may have the right resistor value for the testing, but maybe you need one with a higher wattage......

Or you have discovered a short circuit in the equipment.....

I've 24Vdc 10A power supply connected to the instrument to do calibration,

connected to the instrument ....What instrument?

to do calibration ........What are you trying to calibrate: the instrument; the power supply, or, something else?

If this "instrument" needs a 24vdc power supply why are you putting a resistor in series?

If you are trying to drop the supply volts, then get the right volts from a variable PSU.

If you are trying to simulate 4-20 mA you're way off the mark.

If you cannot calculate wattage from basic principles, then ease off and cosult someone.

The usual reason for such a resistor is to reduce the possibility of maximum smoke by keeping the current to a reasonable level.

In this case, simple Ohm's law predicts a maximum current of:- 120ma. even with a short circuit.....but it will dissipate almost 3 watts, which implies that a 5 watt ceramic resistor should be used.......I assume that the OP used a smaller wattage version.....

Please, if you are doing calibration, get a calibrator! Better yet, get someone who knows how.