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LED Reverse Polarity Protection

B

Bob Monsen

Jan 1, 1970
0
Dave.H said:
According to the LED's datasheet (which is a 16,000 mcd red unit) the
power dissipation is 80 mW, forward current is 20 uA, Peak pulse
forward current is 150 mA. The LED draws about 20 mA at 2.0-2.3
volts.

His point is that at 6V, the resistor is dissipating more power (power is
voltage times current) than the LED.

There are special power supplies for driving big LEDs which use a technology
called "switch mode power supply". It uses a trick with inductors to
minimize this power loss in the resistor. However, for 20mA, it is probably
overkill. If you have a big lantern battery, the thing will last for days
pulling 20mA. The incandescent light bulb is probably pulling 5x or more of
that current. If you used a switch mode power supply, it might double the
time before the battery fails to 10x the endurance of the lamp with an
incandescent bulb. However, it is somewhat complicated, and will require
special chips or a purchased power supply to do it.

Regards,
Bob Monsen
 
S

Steve

Jan 1, 1970
0
Dave.H said:
According to the LED's datasheet (which is a 16,000 mcd red unit) the
power dissipation is 80 mW, forward current is 20 uA, Peak pulse
forward current is 150 mA. The LED draws about 20 mA at 2.0-2.3
volts.
Dave, since you have the LED's datasheet on hand, what is the Vr rating?
Many LEDs can handle 6V to 10V across them in reverse-polarity, although
most are about 5V. If you're lucky...

.... Steve
 
D

Dave.H

Jan 1, 1970
0
Dave, since you have the LED's datasheet on hand, what is the Vr rating?
Many LEDs can handle 6V to 10V across them in reverse-polarity, although
most are about 5V. If you're lucky...

... Steve

It is 5v.
 
W

whit3rd

Jan 1, 1970
0
If already got the LED, had it for about 2-3 months now, It's pretty
damn bright, especially with the large flashlight reflector, I mainly
want the flashlight for walking at night where I don't want my night
vision impaired.

The forward voltage of a red LED is compatible with two
cell flashlights (3V fresh, 2V depleted) and it's a waste to
drive it from a lantern battery (6V). Also, light from an
LED is only shed into a hemisphere, not spherically as
from a filament, so most of the flashlight reflector is
not needed.

At 20 mA, two D cells will last about 500 hours; I'd go with
that and not bother with more expensive lantern batteries.
 
D

Don Klipstein

Jan 1, 1970
0
I read the thread with BobW and default. I'm confused as to why can't you
just put your diode in series with the LED, pointing in the same direction?
Then, it'll block voltage in the opposite direction.

Blue, white, non-yellowish-green, violet, UV, purple and pink LEDs are
extremely intolerant of reverse breakdown. The amount of reverse current
through a 1N400x diode is enough to cause actual damage. To make your
LED of such color immune to reverse polarity, add a second diode in
parallel with the LED.

Then again, I never blew one of these LEDs at 9 volts, let alone 6.
(I know, the datasheets only say they're good for up to 5 volts reverse
reverse voltage, and some of the datasheets say avoid reverse bias.)

A reverse parallel diode is good if the LED will have any exposure to
static. I have blown these LEDs with imperceptible amounts of static.

- Don Klipstein ([email protected])
 
D

Don Klipstein

Jan 1, 1970
0
I said:
Blue, white, non-yellowish-green, violet, UV, purple and pink LEDs are
extremely intolerant of reverse breakdown. The amount of reverse current
through a 1N400x diode is enough to cause actual damage. To make your
LED of such color immune to reverse polarity, add a second diode in
parallel with the LED.

Then again, I never blew one of these LEDs at 9 volts, let alone 6.
(I know, the datasheets only say they're good for up to 5 volts reverse
reverse voltage, and some of the datasheets say avoid reverse bias.)

A reverse parallel diode is good if the LED will have any exposure to
static. I have blown these LEDs with imperceptible amounts of static.

I noticed elsewhere in the thread that this LED is a red one. I would
just put a diode in series with it, probably a 1N4148 although I would not
worry too much about a 1N400x one.

- Don Klipstein ([email protected])
 
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