Re: Running LED from Mains power

In article <p1ol55l50rrt1usq529janj5vht8ipp6dv@xxxxxxx>, Charlie+ wrote:

Reading the thread on battery power with interest...
Is there a minimal and cheap junk box circuit that anyone can recommend for
running a (blue 25 to 40mA) LED at near its maximum output but from mains
power - 230V ac in my case?

Probably easiest is to use a "wall wart" type power supply, such as a
cellphone charger. Add an appropriate resistor. In my experience,
many modern cellphone chargers are good at minimizing power consumption
when lightly loaded or unloaded, so power consumption may be less than
half a watt.

To run an LED more directly from the mains, feed the LED with the output
of a bridge rectifier, put some impedance in series with one of the AC
leads of the bridge rectifier, then connect the AC.

If this impedance is a resistor, to pass 20 mA average current with
average voltage across it being about 205 volts, the resistance would be
10.2 K-ohms. The nearest common value is 10K ohms.
The RMS voltage across the resistor would be close to 225 volts. (230
minus the voltage drop of the LED ad two diodes in the bridge rectifier)
225 squared divided by 10,000 is 5.06 watts. Given tolerances in LED
voltage drop, resistor value and line voltage, the minimum wattage rating
of the resistor should be 10 watts. Keep in mind that a 10 watt resistor,
even with 5 watts going into it, can get egg-frying hot.

To reduce losses, use a capacitor instead of a resistor. If capacitive
reactance is 10K ohms at 50 Hz, capacitance is 1/ (2 * pi * F * Xc) or
..318 microfarad. The nearest common value is .33 microfarad. The
capacitor needs an actual AC voltage rating of at least 250 volts.

(Capacitors rated 400 or even 600 volts DC but lacking an AC voltage
rating, during prolonged use with AC, are likely to go KABLOOEY. I have
had a few 600 volt capacitors and even an 800 volt capacitor blow up from
use with 240 volts AC during my more reckless youth. Some of those
capacitor failures were spectacular. One sprayed the ceiling with hot oil
as it failed, leaving a very tough stain.)

If you use a capacitor, you should have some resistance in series with
it to limit the peak current that occurs when power is applied. If the
resistance is low enough to allow a peak current that is safe for
switches, etc. but unsafe for the LED, you need another capacitor across
the LED to smooth that out to something safe for the LED. The second
capacitor may need to be somewhat large unless a bit of resistance is
added in series with the LED.

If you use a capacitor, you also need to add a resistor to bleed it down
after power is shut off. If it is possible to touch the input connections
to this LED product after shutdown (such as by unplugging it), then the
bleeder resistor will have to discharge the capacitor to a safe voltage in
eyeblink time (.2-.25 second). Worst case peak capacitor voltage is peak
line voltage, about 325 volts. To discharge that down to say, 18 volts,
requires almost 3 time constants (natural logt of 325/18). So, one time
constant should be say, .07 second. Divide that by .00000033 farad, and
the bleeder resistor should be 212 K-ohms. The nearest common value is
220 K-ohms, though to err on the side of safety I would make it 180 or 150
K-ohms. A 150K resistor with 225 volts RMS AC aross it will dissipate
about 1/3 of a watt - use a 1/2 watt one minimum. It may dissipate more
power than a lightly loaded good modern cellphone charger does.

If you use a capacitor, you also need a fuse and anything else to ensure
that the LED product can only fail safely. If you have a second capacitor
across the LED, it should operate safely (or fail safely) if the LED
should fail open. (I have seen them do that, and I have seen solder
joints "crystalize" with age.) The product needs to not become unsafe if
any one component fails open or fails short (it may be a good idea to have
the bleeder resistor be two in parallel in case one fails, breaks, or a
solder joint fails).

You probably want to use a cellphone charger or other "wall wart"
power supply to power your LED.

- Don Klipstein (don@xxxxxxxxx)
.

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