overheating resistors

Hi, im using the arduino to turn on a relay and complete a curcuit connecting the USB to a curcuit like this one :

The IR's are TSHA203 , 100mA , Vr -5V, 180mW,

Now one thig i dont understand is how 5 * 0.1 = 0.5 W and not 0.18 W, so I assumed that it could work on lower values like 2 volts or 50 mA and these are the maximum ones.

...But i guess i was wrong couse the resistors are overheating.Can you guys point me in the right direction?

Thanks!

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dirtdiver over 13 years ago

P.S. with the Vr=5V , so if that is the maxi Voltage , i wont need a resistor, so is it? ..I cant even find the datasheet
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R_Phoenix over 13 years ago in reply to dirtdiver

Vr is reverse voltage max, and yes - you do need the resitor.

I will assume the voltage drop of the Ir's is about 2 volts, so that leaves 3 volts droped across the resistor, so 3v / 30r = 100ma, 100ma * 3v = 300mw. What size are your resistors? For this they will need to be a half watt (half 500 mw and 1/4 is 250 mw). So I can see why they may be over heating.
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Nate1616 over 13 years ago

I agree with Phoenix in that you would need a half watt resistor. 2 volt drop across the IR is a good estimate since its usually between 1.6V to 2.2V. hope this helps
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dirtdiver over 13 years ago in reply to R_Phoenix

well my resistors are 0.25 W .. I will get 0.5 W tomorrow. thanks for the help!
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Nate1616 over 13 years ago in reply to dirtdiver

did upping the resistor wattage help you dirtdriver?
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dirtdiver over 13 years ago in reply to Nate1616

i just tried it with .5W and they are still verry hot.I just don't see the reason!
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fustini over 13 years ago in reply to dirtdiver

Curious, have you measured the voltage drop across the resistors and LEDs individually?
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R_Phoenix over 13 years ago in reply to dirtdiver

Then you will need to accurately calculate the V drop (measure it with a meter) across the IR, and the Resistor. Each group should be close to the same. Then calculate the current draw of each to get the watts being dissipated by the resistors.

quick calculations say that the v drop on the Ir should be around 1.7v so you should be using a 33 ohm resistor, but I bet a 100 ohm would work fine if you have some handy to try.
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dirtdiver over 13 years ago in reply to R_Phoenix

Each one should be the same? why?
I measured it and its like this:
I gues the USB port isnt exactly 5 volts
so 3.9 /30 = 130mA, 3.9*130=507mW
oh man 7 mW above the max and the resistors are capable of cooking my dinner
its back to the store then..and again thanks for the help!
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R_Phoenix over 13 years ago in reply to dirtdiver

dirtdiver wrote:

Each one should be the same? why?
Each group, so each resitor should meassure 3.9 v drop, assumeing each resistor is identical. And each Ir should messure 1.7, again assuming they are the same values. IN your first drawing you have 3 in parallel.
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