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HELP to make LEDs work with AC current

I have a transformer that is reducing the AC supply voltage down to

25.5VAC, which connects to a thermostat that connects to a Honeywell water valve for my home boiler system, which then completes the circuit back to the transformer. I don't see an amperage limit marked on the transformer--it's about baseball size, so it's not small.

Only 1 of the poles of the relay is used and I would like to use the other poles of the relay switch to power a red LED (either a standard LED or a Bi-Polar LED).

This is my home, safety is top of the list (e.g. in my research I've seen some people offer designs that don't use resistors with the LEDs or they connect diodes straight across the power supply, both of which don't seem like they maximize safety).

I want the design to still be safe when one of the parts does eventually fail.

I need help with the design, and I'd especially like to know the formulas that were used in the design because I eventually want to create some LED lighting under my kitchen counter that uses 110V AC... So if you can help me calculate how I alter the design for 110V vs. this current 25.5V design that would be great.

I'd like to know how to do this using standard LEDs and also how to also do it using the bi-polar LEDs.

Thanks for any help... I'm new and learning and have only done a few basic projects so far.

Reply to
Mark Main
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From Google:

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The usual disclaimers apply; use at your own risk; I am NOT an electrical engineer qualified to make assessments re: electrical safety.

Folks, any comments on the safety of those circuits? Electronic current-limiting countermeasures recommended?

Michael

Reply to
mrdarrett

Even simpler (no bridge rectifier):

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I recall seeing a schematic similar to the above, but I just can't find the exact link anymore...

Michael

Reply to
mrdarrett

On a sunny day (Wed, 8 Oct 2008 14:13:54 -0700 (PDT)) it happened snipped-for-privacy@gmail.com wrote in :

The reactance of the capacitor at 50 Hz = 1 / 2.pi.f.C =

1 / (314 . 22.10^-8) = 100000000 / (22 x 314) = 14475 Ohm, say 15 kOhm. The current at 230 V = 230 / 15000 = 15 mA. You need no resistor.

Hopefully I did the math right.

Reply to
Jan Panteltje

Hopefully, yes, but you _do_ need a resistor to limit the inrush of charging current through the cap when you switch it on.

And at 25V, you shoudn't need to do any tricks - say your LED is

1.2V Vf, subtract that from 25V and calculate the resistance at that voltage that gives 10 mA. Also calculate the power, and size the resistor accordingly.

To run it on AC, put a 1N400x (or any GP power diode) in antiparallel with the LED to limit its reverse voltage.

Cheers! Rich

Reply to
Rich Grise

On a sunny day (Wed, 08 Oct 2008 22:14:23 GMT) it happened Rich Grise wrote in :

That is very true. How inductive do you think the mains feed is? Could be all sorts of caps from filters in parallel. So for a peak of 230 x sqrt(2) = 325 V and a max charge current of say 100mA (most LEDs can handle that as pulse) Z would have to be 3250 Ohm. Then at 15mA eff the power would be .7 W, so a 1W resistor of 3k3. So indeed 4k7 / 1W would be fine too.

And, in case no diode in the box, use 2 LEDs in anti parallel.

Reply to
Jan Panteltje

snipped-for-privacy@c22g2000prc.googlegroups.com...

** Both are VERY dangerous and NOT what the OP needs at all.

He has a 25 volt transformer !!!!

Piss the hell off - you ridiculous s*****ad.

...... Phil

Reply to
Phil Allison

"Rich Grise"

** LEDs respond to the average value of DC current flow - which is about 10% less than the rms value when sine waves are involved.

PLUS: a LED is a *diode* and only conducts half of the time with AC feed - so the average current is about 45% of the value found by the simple calc you suggested.

The series resistor's power dissipation WILL have to be calculated using rms values.

** Correct.

..... Phil

Reply to
Phil Allison

ut

ed -

using

I've read the thread and I still can't figure out how to do the math to calculate what the relationship should be between the resistors and the capacitors when trying to drop down from my 25.5V to an LED of say

1.5V.

I've also done some other reading, and in this example (see links below) "Roff" advises connecting the diode in parallel to the LED and have both connected in series to the resistor.

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Right now I'm seeing lots of information, but I can't organize all this info into actual math formulas that help me decide what resistors, capacitors, and diodes I should be selecting; and how to safely arrange them.

I'm especially safety concious when I hook this up to 110V, but even at 25.5V I want to be safe.

Thanks to all who are helping.

Reply to
Mark Main

For the 25V design, you need a 2200 ohm 1/2 watt resistor in series with the LED and a 1N400x diode in antiparallel with it. That is explained below.

A simple approach may be best for you. An LED needs to have the current limited to a (relatively wide) range, and needs to be protected against reverse voltage. Placing a diode in antiparallel with the LED accomplishes the latter. To limit the current you can use a resistor, and since the range is relatively wide, precision is not necessary.

To determine the resistance needed to limit current to 10 mA in a circuit with a 25 volt source, figure 25V/.01A That equals 2500 ohms. A close standard value resistor is

2200 ohms. If you use a 2200 ohm resistor that limits current to no more than 25V/2200ohms which equals 11.36 mA, which would be fine for an LED. Such a resistor would need to dissipate about .284 watts, so you would use a 1/2 watt resistor. That is enough figuring in this case for your needs, but it is not the full story.

Your current will actually be a little lower than the 11.36 mA, because the LED drops the voltage roughly 1.5 volts. So for completeness, you can re-compute using (Vs-Vf)/R where Vs is the source voltage and Vf is the voltage drop of the LED. Then you get (25-1.5)/2200 or ~10.68 mA

The answer to your second question (about how to design the circuit for 110 VAC for under cabinet lights): DON'T. It won't be safe.

Instead, get a wall wart with a safe low level DC output, and compute your resistor using the formula used above: Vs-Vf/I = R where I is current, R is resistance, Vs is the source voltage and Vf is the voltage drop of the LED.

Ed

Reply to
ehsjr

Thank you. That gave me the info that I was seeking. I'll be rigging it up this weekend.

Reply to
Mark Main

[snip]

Or another LED wired the other way around for 2X the light.

--
Paul Hovnanian	paul@hovnanian.com
-----------------------------------------------------------------------
Have gnu, will travel.
Reply to
Paul Hovnanian P.E.

What I like to do is use a bridge rectifier to have both halves of the AC cycle bcoming DC for the LED. A bridge rectifier will protect the LED from reverse voltage while also never giving the LED reverse voltage.

Put the resistor upstream from the bridge rectifier, so that the resistor limits current if the bridge rectifier shorts. And 400 volt bridge rectifiers are cheap.

Have a resistor of sufficiently high power rating that no more than

50-60% of the power rating is actually dissipated into the resistor - and consider worst case input voltage. Measure what the transformer's or wallwart's output voltage is when loaded by nothing but either a voltmeter, or a voltmeter in parallel with a resistor drawing a few mA. That can be well above the nominal output voltage.

I have seen what the failure rate of resistors is when power dissipation gets to about 60-70% of the rating. It is low, but I consider it significant. A few times I have already fixed things where the problem was a resistor failing while dissipating about 60-70% of rated power. In one case, the resistance decreased greatly.

In critical applications, there are "flameproof" resistors. Some are even UL recognized components.

One more thing - you can reduce heat production here: There are now plenty of LEDs that get plenty bright at just a few milliamps.

- Don Klipstein ( snipped-for-privacy@misty.com)

Reply to
Don Klipstein

Or capacitor. This is an often used technique for powering an LED from mains voltage when the option is finding someplace to dump all that I^2R heat. Some budget designs demand absolute minimum parts counts. A bridge rectifier can count as one, but wall warts might be out of the question.

--
Paul Hovnanian     mailto:Paul@Hovnanian.com
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If Mama Cass had just split that ham sandwich with Karen Carpenter,
they\'d both be alive today.
Reply to
Paul Hovnanian P.E.

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