Connecting 200 LEDs to AC (120V)?

[miso@sushi.com]

Hi All,

I am planning to remove the light fixture from one of our rooms and put
200 LEDs instead. LEDs are going to be:

Reverse Voltage:5.0 V
DC Forward Voltage: Typical: 3.4 V Max: 3.8V
DC Forward Current:20mA

How can I connect them all to a AC outlet (110V or 120V? in Canada).

I believe I need to put a "bridge rectifier" to prevent blinking and
they should be mixture of parallel and series.


Thanks if advance,

Homer

I see no current regulation in your design.

Looking at failed commercial LED arrays in traffic lights, I gather
they do a combination of serial and parallel. If I were to build one,
I'd do strings of LEDs that are current regulated. While you can get
effectively brighter output by pulsing LEDs, I'd rather have DC. [Less
RF noise.]

Dig up a 48V power supply. Current regulation can be done with an
opamp, voltage regulator, and NPN BJTs. [NPN E to resistor. Resistor to
ground. Intersection to op amp negative input. Op amp positive input to
reference. Assume BJT matching is close enough that only one have to be
in feedback loop. ] LED string to plus rail of supply. Other end to C
of NPN.

 

[Bob Scott]

bericht
Oops. 200 LEDs require 8 strings of 25.

I think I'd forgo rectifier and the resistor. Limit the current with an
inductor instead, like a fluorescent lamp.
Hmm, a LED ballast. Better efficiency with no resistor losses. Wire the LEDS
back to back in strings of 20 back-to-back pairs in series.

But can it be done in Canada? I don't know.

Regards,
Bob

 

[Homer J Simpson]

Transformers are not rated to go open circuit on 110 VAC either.

Explain the hazard in an open circuit transformer.

 

[redbelly]

Four strings of 25 LEDs in series each with a 1k2/1W series resistor. These
strings in parallel and connected to the rectifier.

petrus bitbyter

I think you're forgetting that 120 Volts RMS means a 170 Volt amplitude
(the maximum voltage in the sine wave's cycle). The above circuit will
result in a too-high peak forward current of 66 mA (assuming 3.6 V per
LED), and a too-high peak reverse voltage of 6.8 V

The most efficient way would use a rectifier and capacitor to supply a
(roughly) DC voltage to work with. But if you were to do this using AC
with just current-limiting resistors and the LED's, you might try using
THREE of the following:

A string of 34 series LED's
In parallel with that, another 34 series LED's with polarity
opposite to the first string.
And, in series with those two parallel strings, a current limiting
resistor of 2.4k (or whatever the closest standard resistor is), 1 Watt
or greater.

Wiring the LED's with opposite polarity would mean that some LED's are
on during each half-cycle of the mains supply, and also that the peak
reverse voltage would be under 5V.

Three of these circuits would use a total of 204 LED's.

Again, if you want more efficient lighting, it would be much better to
use a rectifier to convert to DC.

Mark

 

[Alan theTech]

I think I'd forgo rectifier and the resistor. Limit the current with an
inductor instead, like a fluorescent lamp.
Hmm, a LED ballast. Better efficiency with no resistor losses. Wire the LEDS
back to back in strings of 20 back-to-back pairs in series.

But can it be done in Canada? I don't know.

Regards,
Bob

There's a product on the market. I used to work for an electronics
company that manufactures an LED light stick that is a direct
replacement for an incandescent bulb in exit signs 110 or 220 Vac
50/60 Hz. It has 16 LED's and was the size of an average cigar.
I have several at home that I use for night lights. They have a 30
year life span, the company guarantees them for 25.

 

[LinkWaves]

I think you are right. Just buy the LED bulbs

LinkWaves Corp
29980 Technology Drive, Suite 6
Murrieta, CA 92563
http://www.linkwaves.com/productsmain.asp
909-725-9143
We buy sell all Cisco, Discount up to 80% off on New Used Cisco.

 

[nospam]

Explain the hazard in an open circuit transformer.

Explain the hazard in an open circuit LED.
--

 

[Homer J Simpson]

Explain the hazard in an open circuit LED.

120 VAC across a device designed for 2 or 3 volts DC. You don't see a hazard
there? You might want to check with a fire marshal or electrical inspector.

 

[Donald]

120 VAC across a device designed for 2 or 3 volts DC. You don't see a hazard
there? You might want to check with a fire marshal or electrical inspector.

WOW, someone needs to learn what "open circuit" means.

I lost track of where this started.


But, it sure has gone over the edge !!

 

[Homer J Simpson]

WOW, someone needs to learn what "open circuit" means.

What do you think it means?

 

[jasen]

I think I'd forgo rectifier and the resistor. Limit the current with an
inductor instead, like a fluorescent lamp.

That'd work, but you'd need 8 of them...

Hmm, a LED ballast. Better efficiency with no resistor losses. Wire the LEDS
back to back in strings of 20 back-to-back pairs in series.

a bridge rectifier is more efficient.

Bye.
Jasen

 

[petrus bitbyter]

I think you're forgetting that 120 Volts RMS means a 170 Volt amplitude
(the maximum voltage in the sine wave's cycle). The above circuit will
result in a too-high peak forward current of 66 mA (assuming 3.6 V per
LED), and a too-high peak reverse voltage of 6.8 V

The most efficient way would use a rectifier and capacitor to supply a
(roughly) DC voltage to work with. But if you were to do this using AC
with just current-limiting resistors and the LED's, you might try using
THREE of the following:

A string of 34 series LED's
In parallel with that, another 34 series LED's with polarity
opposite to the first string.
And, in series with those two parallel strings, a current limiting
resistor of 2.4k (or whatever the closest standard resistor is), 1 Watt
or greater.

Wiring the LED's with opposite polarity would mean that some LED's are
on during each half-cycle of the mains supply, and also that the peak
reverse voltage would be under 5V.

Three of these circuits would use a total of 204 LED's.

Again, if you want more efficient lighting, it would be much better to
use a rectifier to convert to DC.

Mark

Mark,

I did not pay much attention to the peak current but it has to be taken into
account. Most LEDs that are designed for 20mA continuous DC current can
handle peak currents of a multiple of that 20mA. Whether 66mA is too high or
not should be concluded from the datasheet. When in doubt you can always
raise the series resistor to 1k5, 1k8 or beyond.

No reverse voltage accross the LEDs while using a bridge rectifier. (Both
the OP and I mentioned it.)

The bridge rectifier and series resistor method is the most simple approach
that I can imagine. That's to say without becoming hopelessly inefficient.
When using a smoothing capacitor and currentregulators you can do - let's
say - 5 strings of 40 LEDs. Each with it's own current regulator. If you
want to be energy aware you can even use switching regulators. But that's
not simple anymore.

petrus bitbyter

 

[Michael A. Terrell]

That'd work, but you'd need 8 of them...


a bridge rectifier is more efficient.

Bye.
Jasen

Really? How? It adds more loss to the circuit, lowering efficiency.


--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

 

[Homer]

Thanks all for your comments. I will try your solutions. Also I got
some idea from following links (by putting 170 Volt and 200 LEDs into
the first link)

http://led.linear1.org/led.wiz
http://ourworld.compuserve.com/homepages/Bill_Bowden/page10.htm

 

[miso@sushi.com]

Thanks all for your comments. I will try your solutions. Also I got
some idea from following links (by putting 170 Volt and 200 LEDs into
the first link)

http://led.linear1.org/led.wiz
http://ourworld.compuserve.com/homepages/Bill_Bowden/page10.htm

I really don't like hooking up to the AC without transformer isolation.

 

[Homer J Simpson]

I really don't like hooking up to the AC without transformer isolation.

The transformer will add a substantial safety factor to the lamp assembly.
Without it there may be insurance liability problems. A reduced secondary
voltage is also highly desirable.

 

[a7yvm109gf5d1@netzero.com]

Thanks all for your comments. I will try your solutions. Also I got
some idea from following links (by putting 170 Volt and 200 LEDs into
the first link)

You really haven't paid any attention at all, have you? Only an idiot
would waste time and effort to build something so unsafe when you can
just buy the finished product.

 

[Jeff Findley]

You really haven't paid any attention at all, have you? Only an idiot
would waste time and effort to build something so unsafe when you can
just buy the finished product.

They're called LED Christmas lights. I bought three sets on clearance at
Target in Jan/Feb for a really cheap price (one set is blue, one is white,
and the last is multi-colored). From what I can tell, the LED's are just
wired in series from one side of the 120 volts AC to the other. They're
their own half wave rectifier. I believe the plug has the "standard"
Christmas light fuse for safety.

I actually bought them because, on clearance, they're the cheapest source
for fairly high intensity LED's I've seen. They're as easy to remove from
their plastic sockets as the little, low voltage, incandescent Christmas
lights. ;-)

Jeff

 

[Homer J Simpson]

They're called LED Christmas lights. I bought three sets on clearance at
Target in Jan/Feb for a really cheap price (one set is blue, one is white,
and the last is multi-colored). From what I can tell, the LED's are just
wired in series from one side of the 120 volts AC to the other. They're
their own half wave rectifier. I believe the plug has the "standard"
Christmas light fuse for safety.

As purchased, they presumably have a UL label, which means they are fit for
use as is. Alter the design and that no longer applies.

 

[redbelly]

Mark,

I did not pay much attention to the peak current but it has to be taken into
account. Most LEDs that are designed for 20mA continuous DC current can
handle peak currents of a multiple of that 20mA. Whether 66mA is too high or
not should be concluded from the datasheet. When in doubt you can always
raise the series resistor to 1k5, 1k8 or beyond.

Hello Petrus,

Yes, the 20 mA can be exceeded for pulsed operation. Not all
datasheets address this. One sheet I looked at spec'ed a pulse current
for 10 us pulses, which would not apply here since the pulses are
several ms long.

No reverse voltage accross the LEDs while using a bridge rectifier. (Both
the OP and I mentioned it.)

Okay, I missed that.

The bridge rectifier and series resistor method is the most simple approach
that I can imagine. That's to say without becoming hopelessly inefficient.
When using a smoothing capacitor and currentregulators you can do - let's
say - 5 strings of 40 LEDs. Each with it's own current regulator. If you
want to be energy aware you can even use switching regulators. But that's
not simple anymore.

Efficiency should only be a minor concern. The LED's themselves will
consume about 14W at most. Even with a 50% efficient drive circuit,
you're still using under 30W. Agree that going to switchers is more
trouble than it's worth for wall-outlet power. If it were to be a
battery driven configuration, you'd likely want to do anything possible
to extend battery life.

Mark