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Would this work? (simple circuit diagram) Answered

Well, i'm pretty much clueless about electronics, so I was just wondering whether or not this diagram would work, and if not, why? Thanks =]

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Does this help? (works of 120 without transformer)

whitelednight.jpg

That's a cute trick!

That first capacitor essentially limits how much charge is injected into the rectifier bridge each half cycle, essentially suplying a constant current to the LEDs, regardless of how many you have in series (provided you stay <<110V).

This is the original page at discovercircuits.com, in case you have a hard time reading the circuit diagram above.

Here's an even simpler version:

AC Line Powered LED Pilot Light

And a more complicated version, with a light sensor to shut off during the day:

Modified LED Night Light

it might work, what's the voltage supply?

Haven't figured that out yet, nor that resistor value. There will be more LEDs than that, too. I'm mainly trying to make sure that the wiring configuration would work.

that could work, but it's recomended to have a resistor for each individual led

plus as good hart said each led will only be on 40% of the time, put a bridge rectifier in and it'll be on 80% of the time

half of the sine wave, the other half is negative and the leds won't conduct

So with a transformer, it takes away the need to worry about that, and the LEDs will be on constantly?

many Wall warts (transformers) have the added bonus of a bridge rectifier (set of four diodes or a diode IC that does the same thing) that effectively smooths out the sine wave, while a voltage regulator brings it to a relatively precise voltage, and a few capacitors smooth out the little bumps that are left. this is a basic one:

03263.png

Thank, G. Here's a nice little animation showing how a bridge rectifier works. Note that current flows through the diodes only in the direction of the triangle of the symbol:

Bridge Rectifier

Hey thanks, that is really cool. I hadn't come across that before.

Using Goodhart's little diagram - if you hook up the LEDs to the AC voltage directly, they will only conduct when the voltage is positive. So the current through the LEDs will look like the first image below.

If you had a second set of LEDs, connected in the opposite direction, those would light up during the other half of the AC wave. Now, it turn out that you can do a little trick with 4 diodes (that's the "bridge" Goodhart mentioned), to send that current during the second half of the wave through that same set of LEDs. Now the current will look like in the second image below.

Note that this is not quite "DC" yet, because there is still a huge fluctuation in the current through / voltage across the LEDs. But you can smooth that out easily enough using a capacitor.

half_wave.gifrectified.gif

yeah, listen to that, not to my discription

a sine wave is a signal that goes up and down and up and down and p and down (and in yourcase, 40 times a second). They are all slopoes, so it's not sudden. half of the time the wave is positive and the other half it's negative.

yep, or you could leave out the transformer, have a ton of led's in series, and then put a real diode (like a 1n1004 or something, I forgot the name) in series. Then you need one big fat resistore (like 1 or 2 watt)

series would require so much more voltage though. I don't know what a real diode is. Without a transformer, I won't be able to plug it into the wall, will I?

you could still plug it into the wall (I'd put a fast blow fuse somewhere though). By a real diodes, I mean something that could handle the voltage, like an led reversed biased could break down at 10 volts. a "real" diode (like a 1n4004) could let up to 200 volts pass through, and oppose up to like a 1000 volts (don't hold me to it)

ok, how many volts is from the socket where you live what color leds do you want how many

I don't know I don't know I don't know I've got a friend with a connection for LEDs, so it's based on what he has.

Think of it this way; imagine those LED's all hooked up to a singe connection point (which is literally what is being done anyway) on both sides. The one problem I see is the 40Hz power supply. Half the cycle will be "clipped" by the LED's, or if one or more burn / go out, the circuit could overload.

Remember the D in LED = Diode :-) one way gate unless the breakdown voltage is reached

A Diode will allow current to flow in one direction only (up to a point), a bridge of 4 diodes are used in power supplies to alter AC into DC. Ok, if the power supply is actually DC and not AC as in the drawing, and IF there is enough power (there is a voltage drop across each LED, depending on which are used), it will work, but if just one LED fails it could mean "good bye" to the whole string, and so, too little power and nothing happens, and too much fries a lot of LEDs.

Isn't wall power AC? Bridge? I don't quite get why the LEDs would be fried if one goes out.

wall power is AC, yes....one thing we don't see in the drawing is which way the LED's are "facing". Since power moves through them like any diode, in one direction only, the return wave could wipe the LED out, depending on the power level. The reason I warn about the LED in parallel problem is that, once you figure the proper voltage / amperage for the circuit, and the voltage drop across each LED, when one "drops out", the figures are then skewed. More power will be available to each of the remaining LED's and they could fail in a cascading fashion or all at once.

Ah, I understand now. So that's why a resistor for each one is best? Is there a way to stop power from coming back? Is that why a normal diode was suggested?

Due to the distribution, one resistor will waste much more current (as heat) than several resistors (and you'd need a 5 or ten watt resistor, too.) If the current limiting resistors are the correct value, the reverse polarity won't hurt the LEDs, but they won't be lit for half the cycle (and they won't be fully lit for most of the 'on' cycle, either.) Some one already suggested using a bridge rectifier. That will convert the negative portion of the AC wave to positive. Add a capacitor or two to filter the ripple, and you've got a decent DC power supply. The DC will be somewhat higher voltage than the AC wave form. AC voltages are given as RMS, which is more of an average (VOMs also measure RMS.) Since the peak AC voltage is higher, the DC will be higher.

Don't worry about the resistor or power supply numbers...the diagram thing I used chose those automatically.

I recommend using the "Guru" on http://ledcalc.com. Note that if you use the "parallel leds" calculator on that page, it will put all the leds in parallel, with a single limiting resistor - this is wrong. For the same choice of parameters, the "Guru" calculator will correctly put a resistor in series with each led.

Also, you can put a number of leds in series with a single resistor, and then put a number of these series circuits in parallel.

well I have to go to bed, nighty night (school night)

oh and by the way, does this have to deal with the led contest in any way ;)

ummm.... what's the circles?