How does this work? Its black magic. LED working from 120v AC

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jleslie48

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I was looking for the smallest simplest way to power an LED from USA household current which is 120 volts AC. At the dollar store I found an LED night light and opened it up expecting to see a tiny transformer or something to power the led but instead I found this: <see attached picture>.

I am dumb founded how to explain how this works, and when I checked online for led's powered by 120v ac nothing comes close to this circuit I found. Mean while, it works perfectly. What is going on here?

 
thank you picbits, but this begs the question why is this not done all the time? why do LED light bulbs have transformers to convert the 120v AC to DC ?
 
Also, why would this not be documented on any website I visit as a CHAD way to get an led hooked up to 120v ac? It seems like a great way to put a little LED on a light switch.
 
Just a guess but for a low current white LED, using a transformer or a capacitor dropper just isn't needed. A resistor is simple but only for very low currents.
 
It is a much different problem to light a LED that consumes a few mW than powering a cluster of LED that require ~15W
 
You're looking at Ohms Law at work.

The diode is rectifying the AC coming in. The 15K resistor is limiting the current to about 8 ma. (120V / 15000 = 0.008 = 8ma)

The led is dropping about 1.4 volts, so the majority of the voltage is dropped across the resistor.
 

The led is quite weak, so I could believe that only 8ma is getting through.

so if I replace the 15k resistor with a 6k resistor I'll get 20ma to the led?
and If I put 6 1w leds in parallel and a 20ohm resistor that would work? that doesn't sound right.
 
Less power lost in the resistor if more voltage is across the LEDs.

I don,t know your LEDs, say 1w, 2v, 0.5a.
6@Parrell=2v, 3a, resistor=118v x 3a
6 series= 12v, 0.5a. Resistor=103v x 0.5a
Look at the power loss in the resistor. There is about 6:1 difference in lost power.

You always need a resistor.
 
Don't forget Kirchoff's laws.

In a series circuit, the sum of the voltage drops is equal to the applied voltage.

Therefore:

The current is the same at any point in the series circuit.

jleslie48 said:
ok, so are you trying to tell me that if I put 6 1W led's in SERIES, I can get them to run on 120v AC?
Click to expand...
You would have to change the resistor value. The current will be less and force the voltage drops across each led to be approximately the same.

The leds may actually light, but they will be considerably dimmer.
 
Ok so lets break this down.

question 1)
I want to run an LED (2.1v drop 20ma) that I normally run with 12v, a 560 ohm resistor, directly with 120v AC, this can be done with nothing more than a resistor and a diode?

question 2)
why is this a big secret on the internet and nobody in a google search, or instructables.com mentions that I can do this?
 
It's not a big secret. Look for BigClive's videos on YouTube. It takes apart and analyses many types of LED lights and explains the shortcomings of various designs.
 
That's a 12 meg resistor so the LED is getting 10uA and will be very dim. To get 20mA through your LED would waste around 2W and would get very hot.

Mike
edit, none of the above replies were visible when I posted this!
 
Pommie said:
That's a 12 meg resistor so the LED is getting 10uA and will be very dim. To get 20mA through your LED would waste around 2W and would get very hot.

Mike.
Click to expand...
12k as in kilo not meg correct? What is the ohms of the resistor I have pictured? why is it blue?
 
Nope, meg.

Edit, it could be 15k depending on which way you read the bands.
 
I wouldn't say it's a secret as a simple Google of 120 volt AC LED circuits will get you all of these ideas and schematics ranging from very simple to more complex including the use a a few capacitors. The link should get you some well explained circuits.

Yes to the first question. However, keep in mind that when lighting LEDs from 120 VAC Mains power there is a small potential shock hazard but circuits like used in the link are really, as you found out, pretty common. You also want to keep in mind how much power your current limiting resistor needs to dissipate.

Ron
 
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