Big Project LED's

LScamper wrote:
Gdetrailer wrote:

"Half Wave rectification without filter caps will result in HALF the voltage at 89V in theory but in practice generally you can count on about half of the AC RMS or about 60V-70V.
Assuming those LEDS are designed from 32V to 36 V or 96V-108V for three in series which should be well high enough."

Not correct but not going into it.

The LEDs will see the peak voltage of about 180V. At that time the current will be at maximum, well over what it is at 127V. The LEDs are sensitive to peak current and will smoke.

Also, dissipating 400W will need a huge heat sink with fans on it. Or maybe water cooling.

Half wave DIODE will only LET ONE HALF of the CYCLE through, PERIOD.

If you don't believe me then perhaps you should break out the O scope...

Putting a STANDARD DIODE in series with the LED string will block the remaining LEDS in series with the blocking diode from ever seeing the PIV. The PIV of the ALL diodes in COMBINATION (because series in this case will ADD to the PIV of the ENTIRE STRING).

The LEDS do not see the PIV at all since the blocking diode will prevent any current flow through the diodes and the PIV of the blocking diode is well above the PIV of the AC.

Now placing the LEDS straight across the mains would be rather spectacular to say the least... Which you could do if you had enough in diode junctions in series but not really a good thing to do..

Where you can get into trouble is if you place a CAPACITOR across the rectifying diode. The filter capacitor not only filters the 60 cycle half wave DC but it will increase the average voltage of the DC..

Now if you used a full wave rectifier then you WILL get BOTH "halves" of the AC giving you a higher voltage unfiltered (120 HZ)DC..

There is a VOLTAGE difference between half wave and full wave rectification.

I have been doing the electronic thing since I was a kid, went to school for it then repaired too many electronic things over the years.