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To figure this out, you need to know the expected forward voltage and current of the LED. Lets assume 2ma and 2V. (Actually, 2ma is small, but I intend to make a point.)
By Kirchoff's Voltage Law, you know that the signed sum of the voltage drops going around a series circuit must add up to zero. This means that the voltage across the resistor must be 228 volts. (-230 + 228 + 2 = 0)
By Kirchoff's Current Law, you know that the signed sum of the currents entering a node is zero. As a consequence, you also know that the current at every point in a series circuit is the same. Therefore, the current through the resistor is also 2ma.
By Ohm's Law, you know that resistance is voltage divided by current, so you know that the resistor is 228V divided by 2ma, which is 114K. The nearest standard value in the E12 scale is 100K. Recalculate the current for 100K, and you get about 2.25ma. (You could also use 120K, and I'll let you run the calculations yourself.)
Don't stop here. There are some issues...
By the power law, you know that power is voltage times current, so you know that the power dissipated by the resistor is 228V times 2.25ma, which is 513mW. I would put a one watt resistor in there.
However, consider this. 2ma is a low current LED. Some of them pull 25ma. The power in the resistor in that case is about 6.5W, which is getting pretty high.
Secondly, you need to consider the reverse breakdown voltage on the LED. I assume that when you said 230V, you meant AC, not DC, which means that there is going to be 230V (actually, a peak value of 325) across that LED for one half the line cycle. You need to check the datasheet and make sure the LED can handle that. If not, you need to put an ordinary signal diode, such as a 1N4148, in parallel with the LED, in the reverse direction, so that it clamps the reverse voltage at about 0.7 volts. (Don't worry about the reverse breakdown on the 1N4148, because the LED will protect it, on opposite half line cycles.)
Last, but not least, you need to consider the safety of the operator. 230V is a high voltage, and LED's are not the most rugged thing around. If the LED breaks, you need to consider if its internal wiring could come into contact with the operator. I would certainly demand a UL listed device in this application.
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∙ 13y ago
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The 440 volts listed on the cap is the maximum allowable voltage the capacitor can handle. You could actually use a 370 volt cap on 230 volts. ANSWER; 230 volts AC can it actually be 644 volts peak to peak . It is 44ov because it must be rectified and sees only 324 volt peak which is withing the 440 volt capacitor handling voltage
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You are working with two different values. Watts is the product of amps times volts. W = A x V, as you can see if there is a value missing then there can be no answer given.
What is 230 volts divided by 500000 oohms?
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