CMOS logic directly powered from mains

you can get a single ic offline buck,

-Lasse

If you're talking about what I think you're talking about, I think it is a bad idea. But then they'll do anything to save a buck, or a yen, or a yuan, or a... you get the idea.

I've dissected two.

1K C1 1/2w 470nF D1 |----+---+---[R4]--. | | | |+ | LED1 | .-' | --- C2 V ~> | ZD1 ^ .-[LDR]-+ --- 100uF --- | | | | | 16V | [R2] | | | | V D2 |470K | | |/ | --- | | +-----| Q1 | | | | | |>. | | | | [R3] | | | | | |10K | | |

They both seem to use a transistor as a zener, although the second one has a zener. What's "LDR"?

And a reversed diode?

Should read "as a shunt".

Come to think of it maybe the word should be crowbar.

light dependent resistor. LED turns off when room is lighted. I'm a bit surprised this is useful since the wattage is so low you could leave them on for years and not pay $1 in electricity.

--

Rick C

As most night light use a capacitor psu it also makes no difference if the led is on or off. But most people think the power consumption is less if the led is off.

--
Dipl.-Inform(FH) Peter Heitzer, peter.heitzer@rz.uni-regensburg.de

Is that true? The capacitive voltage dropper dissipates very little power. When the transistor shorts it also dissipates very little power. So the power consumption is much less when the LED is off... or at least "less".

--

Rick C

As I understand it, the capacitor works as a current limiting resistor.

Most cap psu have a rectifier bridge after the cap and a zener diode to limit the voltage. So the power consumption stays constant independend of the load in parallel to the zener.

--
Dipl.-Inform(FH) Peter Heitzer, peter.heitzer@rz.uni-regensburg.de

In that case the short draws more current when the LED is off.

the

But, into a short circuit that current is 90 degrees out of phase with the voltage waveform. The phase shifts, just a bit, when an LED load is added. According to your home's electric meter, it IS using less power when it shunts that load.

Power consumption is constant, whether the led is on or off.

Perhaps it's an attempt to prolong led life...

RL

How do you figure? The difference should be trivial but even 2x Iled is trivial.

I *highly* doubt that. It's what people want (keeps them from asking questions that the manufacturer doesn't want to take time answering).

Power consumption is primarily determined by the capacitor value, frequency and line voltage. In North America, such a device would draw

20mA AC, with a leading power factor.

There is little reasoning behind cost decisions of made-in-China electronics, so both speculation and doubt is likely a waste of time.

RL

Apparent power [VA] may be nearly constant, but the true power [W] drops due to the changing power factor.

What power factor?

Marketing trumps everything. People think they're wasting money if the nightlights are on during the day (it's illogical but I don't even like it much).

Rick covered that already. Thanks Rick!

No ordinary diodes, just a bunch of series LEDs and resistors. I upped the series resistor to use Christmas lights for "moonlighting" a basement on very low power. Worked great until the LEDs failed. They still work but have very high leakage. Looks like repeated reverse breakdown to me. Two strings, same failure.

Two strings back-to-back might protect them better against reverse voltage spikes... back-to-backing every LED would be even better.

Cheers, James Arthur

if the

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Agreed. Shunting the LED lowers the power-dissipating voltage drop from (~4V at whatever current) to ~1v, same current.

Cheers, James Arthur

ITYM inverse parallel. "Back to back" usually means inverse series, as in "back to back zeners".

Anyway, LED Christmas lights are fuuuuugggglllyyy. But they come out everywhere. :(

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs 
Principal Consultant 
ElectroOptical Innovations LLC 
Optics, Electro-optics, Photonics, Analog Electronics 

160 North State Road #203 
Briarcliff Manor NY 10510 

hobbs at electrooptical dot net 
http://electrooptical.net