HELP to make LEDs work with AC current

[Mark Main]

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.

 

[[email protected]]

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.

From Google:

http://www.marcspages.co.uk/tech/6103.htm
http://www.turbokeu.com/myprojects/acled.htm

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

 

[[email protected]]

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.

Even simpler (no bridge rectifier):
http://www.elecfree.com/electronic/mains-voltage-indicator-with-a-led/

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

Michael

 

[Jan Panteltje]

Even simpler (no bridge rectifier):
http://www.elecfree.com/electronic/mains-voltage-indicator-with-a-led/

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

Michael

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.

 

[Rich Grise]

[email protected] 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.

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

 

[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.

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 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

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

 

[Phil Allison]

From Google:

http://www.marcspages.co.uk/tech/6103.htm
http://www.turbokeu.com/myprojects/acled.htm

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?

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

He has a 25 volt transformer !!!!


Piss the hell off - you ridiculous shithead.




....... Phil

 

[Phil Allison]

"Rich Grise"

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.

** 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.

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

** Correct.




...... Phil

 

[Mark Main]

"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

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.

http://www.electro-tech-online.com/...hat/16808-how-drive-led-ac240v.html#post98018
http://www.electro-tech-online.com/...75d1122412021-how-drive-led-ac240v-ac_399.gif

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.

 

[ehsjr]

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.

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

 

[neon]

LEDs are diode in essence depends on what diodes you buy depends on the forward drop and forward current. usualy 2v to 4v @10-20ma. you must know these parameters then it becomes simple no matter what the voltage source 1000v or 10v a certain amount of current must flow thru the diode to be visible . there are safefty precautions it is called power dissipation you may not exceed this parameter or puff goes the device. for ac just add a diode of the 1n4000 series to pevent reverse breakdown at 120v 240v ac. So now to limit the current add a resistor in series to limit the current. suppose 120ac add 48 2.5v LEDS in series then add a resistor of 500 ohms 1/2w too many LEDS? then 120/20leds 2.5v leds=50v then the resistor becomes 120-50=70v same current 0.010ma then R= 70/.010=7k or 6.8k now we need a bigger resistor and more power like 2w dissipation. any of the above LEDS voltage current can be used. good luck

 

[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- Hide quoted text -

- Show quoted text -

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

 

[Don Klipstein]

[snip]

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.

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

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 ([email protected])