Resistor size for LED? 1

[draymorris]

I want to install LED as indicators on certain parts of my machines. Some of the circuits are 24V DC and others are 120V AC. How to I calculate the resistance needed based on the LED's that I purchase to do the job?

 

[nbucska]

Let's suppose you want 10 mA:
for 24V: 2.4kOhm, 1/2 W
for 120 V AC:
Connect the LED into the diagonal of a diode-bridge
and use a series 12Kohm min. 3W

OR use a series cap to reduce the dissipation:

First connect a diode anti-parallel to the LED
(anode to LED's cathode, cathode to LED's anode)
next connect the cap ( .47 uFarad, 600V ) in series
with it.

Or: connect the LED into a diode bridge
and reduce the cap to .22 uFarad.

If you use cap, MUST use fuse !!!! <nbucska@pcperipherals.com>

 

[draymorris]

No Offense, but I was hoping for some formulas. What if I want to use multiple LED's on one circuit for more light? Any suggestions on LED specs?
Thanks

 

[nbucska]

search <

http://www.dogpile.com>

for LED

V=I*R is the formula you need.
<nbucska@pcperipherals.com>

 

[jimthompson]

The Voltage drop for Red and Amber LEDs is approximately 2 volts, Green and Blue LED's have approximately 3.5 volt drops. Unfortunately the reverse voltage is usually rated at only 5 to 10 volts. Most of the 3 mm and 5 mm LED's can handle 20 ma continuous current.

Since you are using the LED's as indicators, you do not need the maximum current, so the easy way is to use a resistor and a diode in series with the LED. Since that would result in half wave current, assume that the current is double (40 ma) and calculate the resistor value. The voltage drop of the LED's will give you some extra margin.

The modules that I design need the maximum light and I measure the current with a true RMS current meter and size resistor for the maximum allowable current. I have not found a formula that I trust.

 

[Lewish]

Any formula will only get you in the ballpark, as LEDs vary greatly between vendors and even between shades of the same color. Choose a vendor, look at the datasheet to see what the minimum forward voltage across the LED is, subtract that from your supply voltage, use Ohm's law to calculate a resistor that will give a current that is slightly less than the maximum allowable for that LED.

Or you could use a constant current source if your supply is DC. Search the web for how to design a ccs.

If you want to use multiple LEDs, put them in series.

Other caveats in the above posts apply.

 

[SteveHuck]

For the 24v DC use

Resistor = Voltage / Desired Current
R = 24 / I

Wattage = Voltage * Current
P = 24 * I

Make sure you put a resistor of a high enough wattage in the circuit or it will overheat and fail. If you have a LED that requires 20ma then you need
24 / .020 = 1200 ohms or 1.2K
24 * .020 = .480 Watt or 1/2 Watt

There is your formula!! Good luck

 

[draymorris]

Thanks to everyone. This forum is great! I know it takes time to give all this free info - Many Thanks!

 

[jbartos]

Suggestion: Visit

http://www.pti-ic.com/ptipdf/appnote/PT8A320PAN01.pdf

on the more involved LED application within circuitry. The LED needed resistance calculation may become more complicated depending on the application.

 

[lcsjk]

Formulas:
1. Run the LED at about half it's max current rating: If you need more brightness get a high brightness LED. I expect this to be 12mA.
2. Resistors get hot, so make sure the power dissipation is no more than 1/2 the resistor's rating for 120 volts (170vpeak).
3. For 120VAC, use a diode connected in anti-parallel across the LED. THis reduces the RMS voltage to 60 volts, but the peak remains 170V.
Assume the LED forward voltage is 2 volts or read the spec sheet for 1/2 max current.

Now the formulas:
for 24 volts.
The resistor must be sized to limit the current to 12mA rms, 20mA max., where 20 mA is assumed to be the max. operating LED current rating.

R =V/I gives the resistor ohms. V is V24-Vf or V60-Vf, I is 12mA

P = (V)(I) using the above values.

Example: Vf = 2V, I = 12mA, Operating voltage = 24VDC
R = (24-2)V/12mA = 1.83k ohms (use the nearest standard value)
P = (24-2)V * 12mA = 0.264 watts. (Use a 1/2watt in open air or 1watt if the space is confined.)

Example: Vf = 2V, I = 12mA, Operating Voltage = 120VAC.
R = V/I, V= Vrms - Vf = 60-2v = 58V.
R = 58/12mA = 4.83k ohms.
P = 58 * 12mA = .7Watts ( use 2 watts or more)

Vpeak = 170 - 2 = 168V peak
The current must not exceed the absolute maximum device ratings:
Imax = Vpeak / R = (168) / 4.83k =34.7mA.
If this is too close to your device rating, go back and choose the resistor so that the peak current is no more than 80percent of the absolute rating.
Example: Device abs. max rating is 40mA.
.8 x 40mA = 32mA
R = 168/32mA = 5.25k ohms.
Inew = 58/5.25k = 11.05 mA
Pnew = 58 * 11.05 mA = .64 watts




The resistor voltage will be 24V - the forward voltage. V=VfR = V/I

 

[jbartos]

Suggestion: The design margins in resistor wattages indicated in the previous posting are appropriate. Also, the resistor may be preferred to be wirewound.