# LED Resistor help

Discussion in 'FAQs' started by Play-Doh, Mar 6, 2008.

1. ### Play-DohMember

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Hey folks

Need some help from yall.

I bought some LEDs from ebay, and these will be used to show switch locations on my my layout.

Im powering these from the AC terminals (since they are off the switch machines and snap relays)

My transformer rates the AC at 18.5 V. These LEDs are rated on the listing

50 pcs 5mm Yellow LED 5000mcd & Free Resistor - eBay (item 140152612512 end time Mar-25-08 19:29:08 PDT)

The site says they come with some resistors of different voltage, but everything is in DC.

I need to know what kind of resistors to use for AC. Do these work for AC? Will the resistors listed work? I tried the resistor calculator, and it was all for DC.

Any help?

Thanks!
2. ### beamishHO & Steam Engineer

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The smallest resistor you should be able to use is a 1k ohm (1000 ohm) . personnaly i would go with a 3.3K ohm resistor that way you have less current going through the LED and it should last longer.

I think the ones pictured are 460 Ohm but am not too sure as the colors are hard to read in that pic.

The resistor calculators should work for both dc and ac as long as you know the voltage.
3. ### akremedyNew Member

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Play-Doh, the fundamental problem is that LEDs are diodes (Light Emmitting Diode) - traditionally, they respond to current flowing in one direction only, from source to ground. DC (Direct Current) is predictable...coming from one direction only, so naturally fits the equation. AC (Alternating Current) is cyclical, alternating +/- on each conductor, so in theory, one could expect that the LED would only be lit 1/2 the time - whenever the current was entering the diode from the "right" direction.

If you're committed to using an AC source, you'll need to look into some means of rectifying the power between source and your LED panel to supply the proper type of voltage coming into the panel. Bridge rectifiers can be obtained from outfits like Mouser Electronics - Electronic Component Distributor or Digi-Key Corporation - USA Home Page for a small amount...you'll need to search around for the equation necessary to determine which is the best rectifier given your source voltage and downstream requirements.

Hope that helps a little,
4. ### bclemensMember

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Play Doh, don’t worry. These will work fine with your AC source. Here’s how to figure it:
LEDs are diodes- they will rectify the current themselves, so no external rectifier is needed.
According to the specifications provided by your seller, these LEDs have a forward current rating of 20 mA (.020 amps) The forward-bias voltage drop will be about 2 volts.

Put a resistor in series with the LED that drops the remaining voltage (18.5 – 2) 16.5 volts and provides 20 mA.

Ohm’s law says R = V / I

R = 16.5 / .02

You will need 825 ohms for maximum brightness. Try 1000 ohms (1k) and it’ll work fine.

Oh- and be sure you wire the LEDs in series with the resistor, and wire all LED/Resistor pairs in parallel with each other. In other words, each LED/resistor pair will have a separate wire from one side of your AC terminal to the other.

Cheers!
Bruce
(10 years as an electronics instructor at Ozarks Technical Community College)
5. ### Play-DohMember

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I thankyou all for your help! Once again the great folks at the gauge come to my rescue, you are a class above the rest!

TJ
6. ### PitchwifeDreamer

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Play-Doh, I took the picture of the resistors and played with it on my graphics program and the best I could pull out was the color code yellow violet black which translates into 47 ohm resistors. My recommendation would be to follow Bruce's advice and use 1K or 1000 ohm resistors. Go buy them at any parts house or RS. They are cheap and that way you are using a known value.
7. ### jleslie48Member

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This is the exact math I used to wire my LEDS, Only I had a 12V DC power supply to tap off of.

great lesson on V=IR. thanks.
8. ### dbogotNew Member

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bclemens is correct, except I had a question in the AC v. DC case. I think this is correct, but please let me know if it's not!!!

18.5 VAC is an RMS measurement (at least if a modern DMM was used). Therefore, that value is only .7071 of the PEAK voltage that occurs at the top (and bottom) of the sine wave each cycle.

If the circuit is 18.5 VAC, I believe the peak instantaneous voltage would be 26.16 volts.

Therefore, to be absolutely safe, you would need a current limiting (minimum) resistor of at least 1,308 ohms... As a practical matter, since that peak current flow would occur only for a very short time, it probably wouldn't harm anything if it was a bit higher than 20ma, but why not err to the side of caution, right?

Also note that you can use a potentiometer (variable resistor), but be sure to wire it in series with a fixed resistor, to ensure that the current is limited to less than 20 ma even if the potentiometer is turned to 0 ohms.

Another idea is to test for ideal brightness using a potentiometer, measure the total resistance once the brightness is set just right, then use a regular resistor (or combination of resistors) to achieve a permanent brightness level.

Good Luck!
9. ### bclemensMember

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You are right, dbogot. Most meters give voltage in RMS, which is .707 of the peak value. But if we use RMS as a standard we'll be fine because the voltage and current spend so little time above the RMS value. Indeed the very reason RMS was adopted was to provide a DC equivalent standard to apply to AC circuits. The RMS value applied to AC systems will provide the component values that would work in an equivalent DC system.

So if your meter tells you that your voltage is 18VAC, and it is a RMS meter (like most are) you can safely ignore the fact that the actual peak value is 1.414 times higher. You can spec your components as if the supply was 18VDC.

Cheers!
Bruce
10. ### ezdaysOut AZ way

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Just so you know some practical applications on just how much an LED can take, we had a control system that would pulse each LED in the system through a matrix and when a anode and cathode where pulsed simultaneously, it would light. The puses were brief and occurred once every tenth of a second, so in order for the LED's to not flicker and appear as being on steady, we hit them with a 50 mA pulse. Some of those systems are still running after almost 20 years without loosing any LED's.

The point here is I wouldn't worry about peak voltages when running an LED on AC. The extra current there will help remove any flicker caused by the AC half-wave pulses.
11. ### KentByGN, NP, SP&S

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OK, so 18VAC is the equivalent of 18VDC, but sense the duty cycle of the AC would be 1/2 the DC, would the led be putting out 1/2 the light and using 1/2 the power?

Kent
12. ### bclemensMember

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Now we're getting technical

If the 18VAC that is listed on the power supply has no other terms listed with it (such as 18VAC Peak) then it is most likely an RMS value. That means that in reality the 60 Hz AC sine wave that is coming out of the terminal is peaking at 25.5 volts positive on the first half of the cycle and 25.5 volts negative on the second half of the cycle. So 18VAC is not equivalent to 18VDC. The work we can get out of 25.5VAC (Peak) is equivalent to 18VDC.

Peak voltage = RMS / .707

This voltage repeats 60 times every second in the U.S. Other countries may run at 50 Hz.

The LED is only conducting for the half-cycle that it is forward biased. When it is conducting current, light comes out. The more current the more light, until there is too much current and the diode junction breaks down and the LED is destroyed. That is why we need to limit the current to acceptable levels with a resistor.

The LED is like a check-valve in a pipe. Water can only flow one way through it. On the other half-cycle, the LED stops the current flow and for that half-cycle all the source voltage is dropped across the LED.

So yes, the LED does only give out light for 1/2 of the cycle, 60 times a second. So because of persistence of vision, it will seem to glow continuously but to your eye look about half as bright as a LED that was connected to an 18VDC source. The AC source provides a 50% duty cycle through the LED while a DC source would provide a 100% duty cycle.

To compensate, one can lower the resistance value and overdrive the LED. Like ezdays pointed out, they are very resiliant and tolerant of short-term current surges.

Cheers!
Bruce
13. ### ezdaysOut AZ way

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Yeah, but hit them too hard for too long and wow, you can actually blow their heads off.
14. ### PitchwifeDreamer

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And a fine explosion it is when it happens. :twisted: :twisted: :twisted:
15. ### kf4jqdActive Member

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DO NOT use direct AC to feed those LED's! I have used Atlas snap on relays to isolate the LED's. Use a seperate power supply! I use an old notebook pc's power supply. Then Use a 7805 regulator to drop the 13.8vdc to 5vdc. Then I use a 1000 Ohm (1K) resistor. This will make the LED last for many many years to come.

The 7805 regulator is a very easy device to use. It has 3 wires coming out of it. The first is the input-middle is ground or negative- the output is 5vdc.

One last thing. AC can cause the LED to flicker. AC also burns out LED's, even if the voltage is correct.

Andy

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16. ### jleslie48Member

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Excellent implementation! I love the reuse of those old power supplies. I've got about a dozen of them from all the old kids toys, broken cordless phones, modems, et al. The regulator is the key.
17. ### kf4jqdActive Member

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Save those walwart power suppolies! A 9dv at 300mA will power alot of LED's! Here's a link for a novice to build a simple 5vdc power supply! Simple 5V power supply

Andy
18. ### jleslie48Member

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OK, just to clarify, I believe these are the radio shack parts that are necessary for the project:

RadioShack.com - Cables, Parts & Connectors: Component parts: Transistors & analog ICs: +5V Fixed-Voltage Regulator 7805
RadioShack.com - Cables, Parts & Connectors: Component parts: Capacitors & resistors: 10µF 35V 20% Radial-lead Electrolytic Capacitor
the 100nF part (see below)
RadioShack.com - Cables, Parts & Connectors: Component parts: Capacitors & resistors: 100µF 35V 20% Radial-lead Electrolytic Capacitor

and a mounting board:
RadioShack.com - Cables, Parts & Connectors: Component parts: Breadboards & IC sockets: Pre-Punched IC-Spacing Perfboard

Now my limited education in bread boarding tells me two warning signals:

1) the uF capacitors have one lead longer than the other.
- that means the longer side is positive (just like the leds) which means it
makes a difference which you wire them up.

2) I suck and u,n,p
- on the 100nF 100 nano ferrid capacitors, I always mess up in converting from micro to nano, to pico, to ...
so it is this part:
RadioShack.com - Cables, Parts & Connectors: Component parts: Capacitors & resistors: 0.1µF 50V Hi-Q Ceramic Disc Capacitor Pk/2
or this one:
RadioShack.com - Cables, Parts & Connectors: Component parts: Capacitors & resistors: 0.001µF 500V 20% Hi-Q Ceramic Disc Capacitor

and the last but not least, part are the LED's themselves with there corresponding resistors:

RadioShack.com - Cables, Parts & Connectors: Component parts: Capacitors & resistors: 1K ohm 1/2W 5% Carbon Film Resistor pk/5

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