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How to power LED directly from mains for pennies per year
#1
Sometimes you need to power an LED that is in electrical box and does not have any metal parts exposed. It is fairly safe to power it from mains directly like this:

   

The above circuit is for North American 110VAC / 60Hz. For european 220V / 50Hz use 56nF capacitor instead of 100nF for same ~4mA current.

How does this work ?

The 100nF capacitor has 26.5kOhm impedance at 60Hz, so AC curent will be 110VAC / (26.5 + 1.8) = approx 4mA. Capacitor does not convert electric current into heat like resistor but collects it electric charge on AC ramp-up and returns it back to the grid on the ramp-down - this is called reactive power. There is two types - capacitive and inductive, and they work opposite to each other, when capacitor is charging, the inductor is discharging and vice versa. The power actually consumed, converted into light, heat, motion or other form of energy is called real power. Combination of reactive and real power is called apparent power and in circuit above it would be 110V * 4mA = 0.44W. If you were to replace capacitor above with 26.5k, the real power will be same as apparent power and circuit will not be returning anything back to electric grid.

In Canada and most of the world residential customers are only charged for real power consumed, the meters are smart enough to measure reactive and real separately. Also many electric devices in your home are slightly inductive in nature (furnace fan, fridge motor, heaters, microwave, electric oven, etc), the capacitive reactive power will offset some of inductive reactive power and may not even make it back to the grid. The point is, you will only pay for real power spent by the LED and resistor (2V + 1800*0.004A) * 4mA = 37mW. At average cost $0.20/kWh this circuit will costs $0.06/yr to run if LED is ON all the time.

NOTE: if you use small wall adapter, it has much higher no load current, can be 0.3W - 3W depending on energy efficiency compliance. Each 1W costs approx $1.70/yr, so this quickly adds up.

Restroom Occupancy Indicator

The light switch is outside washroom, but when door is closed, it's not obvious, this has led to few awkward situations. Below is my LED occupancy indicator.

NOTE: at first I connected this circuit without resistor, since the capacitor has already impedance, but resistor has an important task - you flick the switch when 110VAC is not crossing zero (almost always) the discharged capacitor acts as momentary short and current can go to several Amps. This is what actually happened and my circuit without resistor worked 2-3 times and then LED died. After that I added 1.8k resistor which limits transient current to maximum 155V / 1800 = 86mA. These LEDs should be able to withstand 100mA pulse up to 10ms and since one half-periode of 60Hz is 8.3ms, this protection works well within specification without spending lot of power into heat. Technically, most of 37mW still goes into heat on 1.8k resistor but choice of 4mA (modern LEDs are quite brighter than they used to be) instead of traditional 20mA is a big power reduction already.

This is out of scope of this tutorial, but if you wanted occupied/available indicator, you can use dual LED like below. Keep in mind this may not work with other types of light bulb than incandescent.

   


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#2
Waited till tonight to really read this post Roman.
AC circuits still confuses me, and this thread helps out very much, in understanding it.
Good idea for led light with switch.
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#3
Hi Roman.  We have those puck lights underneath our kitchen cupboards shinning light on our counter.  A bulb went out, and my wife told me they cost $8 a piece.  No way I am paying that for a bulb and was thinking to put leds instead, save money on cost and power.

I do not have the double pole capacitors with the amount of capacitance your circuit calls for on hand.  Also would there be an annoying ac flicker associated with your circuit, considering this would be on constantly at night, lighting the kitchen.
What about this bridge rectifying circuit, to smooth out the dc for my (2 white, 1 Blue ) Leds. 

Also the room I have for PCB to fit in puck light is about 1 square inch,  I do not think I can fit this circuit onto such a small board.

Also, is it wise to test 120V ac circuit on bread board?

Any advice ?  

Thanks.

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#4
That circuit you designed will explode the electrolytic cap and R1 and zener will burn. Electrolytic cap cannot be reversed, it will release gas inside and build up internal pressure, it's a pretty decent explosion, people have lost fingers. You doing this in LTSpice, if your simulation works, check the current through R1 and zener.

Btw. your puck light already had all necessary power circuitry included, probably only LED burned up. Open it up and see if you can fix it by replacing discrete white LED. Make sure you fully disconnect it from mains when disassembling, don't just flick the switch.

Also, I do not see any flicker on my indicator, it's 60Hz, should not be visible. With your bridge, it will be 120Hz, so if you adjust C1 to 0.5uF (0.47), you can remove C2 and zener.

Another idea is just use old power adapter, like not-needed phone charger or such.
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#5
Hey Roman. Wouldn't, the bridge prevent the reverse current going through the Caps?
The puck is not Led dc, but incandescent ac.

I like your Idea to replace with dc wall adapter. I might do that.
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#6
Bridge steers the current on the load side, if you look at the line side, each side has a + and - of a diode on it therefore line current flows 2 ways, load has 2x + or 2x - and its current flows one way only.
--------------- ---- --- -- -  -
If things weren't meant to be modified, they would not come with wires attached.
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#7
(2016-04-17, 10:49 PM)darrin Wrote: Hey Roman.  Wouldn't, the bridge prevent the reverse current going through the Caps?
The puck is not Led dc, but incandescent ac.

I like your Idea to replace with dc wall adapter.  I might do that.


If you want to play with caps, see schematic below. Your value 47uF is way too high, so you may need to recalculate. The math you need:

current through the LEDs ... I=20mA
input AC voltage ... V=110V
voltage frequency ... Freq = 60Hz

for simplicity disregard LED forward voltage, impedance needed to limit current will be ... Z  = V / I = 110V / 0.02A = 5500 ohm

Capacitive impedance is Z = 1 / (2 * PI * Freq * C), then C = 1 / (2 * PI * Freq * Z) = 1 / (377 * 5500) = 482nF

Btw. some people swear on connecting electrolytic capacitors back to back without diodes, apparently electrolytic capacitor develops certain rectifying effect that simulates a diode. People use it in speaker filters, but I would not use that on power circuit connected to mains. Keep in mind the resulting capacance is half of two caps, assuming they are the same.


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#8
if you are just looking for some cheap mains low voltage, the usb chargers are hard to beat for the price as long as you don't need more than 5v
--------------- ---- --- -- -  -
If things weren't meant to be modified, they would not come with wires attached.
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#9
(2016-04-17, 10:56 PM)jon Wrote: Bridge steers the current on the load side, if you look at the line side, each side has a  + and -  of a diode on it therefore line current flows 2 ways, load has 2x + or 2x - and its current flows one way only.

Thanks Jon, I can see now what you are talking about.   I was trying to avoid having to use a double pole capacitor, because I do not have any at the recommended capacitance.  I guess I can not.  Not I assume with few components anyway.

(2016-04-18, 12:01 PM)roman Wrote:
(2016-04-17, 10:49 PM)darrin Wrote: Hey Roman.  Wouldn't, the bridge prevent the reverse current going through the Caps?
The puck is not Led dc, but incandescent ac.

I like your Idea to replace with dc wall adapter.  I might do that.


If you want to play with caps, see schematic below. Your value 47uF is way too high, so you may need to recalculate. The math you need:

current through the LEDs ... I=20mA
input AC voltage ... V=110V
voltage frequency ... Freq = 60Hz

for simplicity disregard LED forward voltage, impedance needed to limit current will be ... Z  = V / I = 110V / 0.02A = 5500 ohm

Capacitive impedance is Z = 1 / (2 * PI * Freq * C), then C = 1 / (2 * PI * Freq * Z) = 1 / (377 * 5500) = 482nF

Btw. some people swear on connecting electrolytic capacitors back to back without diodes, apparently electrolytic capacitor develops certain rectifying effect that simulates a diode. People use it in speaker filters, but I would not use that on power circuit connected to mains. Keep in mind the resulting capacance is half of two caps, assuming they are the same.

Thanks Roman for the equations and anayalasis.  Very much to think about.  AC has so much to it.  I will use this advice for future ac circuits.  Reactance and Imputance still confusing .  I understand it then I do not.   Maybe a learning thread on it would help many trying to understand ac and how the components work in an ac circuit.

(2016-04-18, 04:39 PM)jon Wrote: if you are just looking for some cheap mains low voltage, the usb chargers are hard to beat for the price as long as you don't need more than 5v

Good idea Jon.  I will go through my stuff and see what I can find.  Just have to make it look good for my wife.
On vacation for 10 of 11 days, my wife has a bunch of projects for me  to do.  This will be one of them.  More on my side of the projects I want to do. Smile
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