i have a blown resistor on a pcb its either 68R 3watts or 68K 3watts.
It drops the voltage from 230v to 200v ac its then converted to dc via the diodes. can anyone help by giving me the formula to workout the correct size for the resistorhttp://forum.allaboutcircuits.com/images/smilies/confused.gif

 

P = E^2/R

30^2 / 68 = about 13 Watts, so that won't be it.

30^2 / 68000 = about 13 milliWatts, so a 3 Watt resistor would be overkill.

Are you sure it isn't 680 Ohm at 3 Watt?

 

thanks thingmaker3
the resistor has 68 then the bottom half of a "K" or "R" printed on it.?

 

Without being sure of the load resistance, or the current that the series draw, we canndot figure the resistor value. So, what is the resistance of the rest of the circuit that the resistor would feed. Is the purely resistive? Or try to measure the AC current drawn at 200V AC, if you can get that voltage.

 

If that resistor is blown, seems like there's a really good chance that one (or more) of your rectifier diodes are shorted.

You might want to investigate the cause of the resistor getting blown before just replacing it, and having it blow again.

Your rectifier diodes will give you clues as to how much current the designer expected to be flowing through the circuit. That will help you narrow down your resistance range.

680 Ohms across 30 volts yields about 1.32 watts. (I don't know how I came up with 2.87 watts before!)

You might consider using MOVs (Metal Oxide Varistors) across the input power to protect against power surges if it doesn't already have them.

 

thanks for replies
the diodes are ok, the circuit supplies a 200vdc coil. the supply is 230vac hench the voltdrop resistor and diodes(and capp) as stated the original resistor has 68 and K or R and is rated at 3w

 

OK, so what is the load resistance? If there are coils/transformers in the circuit, that would complicate matters.

You know, you could always start out trying a 68K resistor as a replacement. Very small chance that you would fry anything with a resistor that's too large. That way, you could measure the voltage across the new resistor, and by using Ohm's Law determine the correct value for the resistor.

 

That means you are dropping more than thirty volts! Peaks of 200 V come from an RMS of about 141 Vac.

230-141 = 89

89^2 = 7921

7921 / 68 = about 116 Watts, so it still isn't a 68 Ohm resistor.

7921 / 68000 = about 116 milliWatts, so a three watt resistor is still overkill.

If it were "6.8K," the numbers would make sense. Are you sure there is no decimal point?

 

I'm becoming more and more convinced that 6.8K was the original value.

I'm also starting to think that your coil may have some of it's windings shorted.

You COULD feed it with a 24 VAC transformer (like those used for doorbells and air conditioning controls), and experiment with some different values of resistance. From what I think, and Thingmaker might agree as well, is that across a 6.8K resistor with the 24VAC input, you'll measure a bit over 3 VAC.

If you measure a great deal more than that, you could have shorted windings in your coil, which would place quite a bit more voltage drop across the 6.8k resistor, which would cause it to burn out in a hurry.