LED Reverse Polarity Protection

You're on the right track. You don't need a high-current diode like a

Anyway, just put the diode across the led backwards. That is, put the anode of the diode to the cathode of the led, and the cathode of the diode to the anode of the led. That way, if your flashlight batteries are installed backwards, the diode will conduct the current (I=6V/200ohms) and the reverse voltage across the led will be limited to under 1V.

Bob

Thanks, do I put the resistor before or after the diode?

A high-brightness white LED might well draw more current than a small-signal diode can handle, so a 1N4001 (50V, 1A) diode might be better. Any of the 1N400x series will work; they're all cheap and easy to find, but the 4001 is the cheapest. I'd put the diode in series with the LED and adjust the value of the ballast resistor to allow for the 0.6 volt forward drop of the diode. A better solution would be to look up one of the switching regulators usually used with high-power LEDs. With a simple resistive ballast and 6 volts, the resistor will dissipate at least as much power as the LED.

One end of the resistor to one of the battery leads. The other end of the resistor to one end of the paralleled diode/led. The other end of the paralleled diode/led to the other battery lead.

I'm not too good at ascii art, but here's an attempt:

Batt+ ----resistor----|-----cathode(diode)anode-------| |-----anode(led)cathode----------| | Batt- ---------------------------------------------------------|

Does this make sense?

Bob

There's a 220ohm resistor in series. It don't matter what type of led is installed. The max reverse current is determined by the battery voltage, the series resistor, and the voltage drop across the protection diode (note the absence of the led's effect on this).

Bob

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.

I can't understand it.

Well, just leave it as it is and don't install the batteries backwards.

Bob

I would do that, but the battery is a lantern type and it's fairly easy to put it in the wrong way. I don't want to fork out $2 for an LED every time someone installs the battery the wrong way.

Forking out two bucks for a led implies a higher powered led than an ordinary 5 mm led which goes for around ten cents (or you buy them at Radio Shack)

Also a 6 volt lantern battery is a bit overkill for a single 5mm / 20 milliamp led.

Do you have some specifications on the actual LED you plan to use?

The schematic seems correct to me. And I'd use the 1N4000 since it is more robust.

Look up Cat #Z4024 at

Output 16000mcd typ, 2.0V 20mA Cat No. Z4024 Category: Round LEDs

LED 5mm Int Brit RED 16,000mcd

You're using a red led in a lantern?

I looked at the dse NZ site since the AU was down.

I think you could do better price wise . . . they don't show any technical specs. It is easy to get high mcd numbers simply by limiting the size of the beam (so beam angle counts - and they don't say).

I pay 12 cents or so for 5mm 13,000 mcd 9 degree leds.

If you want a killer lamp replacement more suited to a lantern battery, check out some Cree one watt emitters

These are not in the same category as your standard 5mm leds:

These high power leds are usually specified in lumens output not mcd (which is how bright a single spot of light is) and have beam spreads of ~90-140 degrees - and use reflectors if you need a concentrated spot light.

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.

Ah, that makes sense.

I believe we discussed that in another thread a few weeks back, but now I'm opting to go for a lower power one.

Is this better? Use a fixed-width font (like Courier) to view:

Batt+ ---resistor---|---cathode(diode)anode---| |----anode(led)cathode----| | Batt- ----------------------------------------|

Or this (circuit should all be on a single line):

B+ ---R---[LED-diode parallel combo]--- B-

B+ is battery + terminal B- is battery - terminal R is the resistor

Regards,

Mark

Bob,

Try using a fixed-width font (like Courier) for ascii ciruits. If necessary, use Notepad to make it and then copy-and-paste into your post. Even if you post it in another font, if it was created in fixed- width then others can convert the font to fixed-width and read it.

We have a general agreement in here that ascii circuits are to be done in fixed-width fonts, that way everybody can view it.

Mark

Thanks, that makes it much easier. I've decided to use 1n4007 diode, I would use a lower voltage one but Dick Smith only sells the 1N4007, still, cheap enough @ AU$0.09

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.

v+ ------ >| ------ RRRR ----- LED >| ----- GND is OK

GND ------ >| ------ RRRR ----- LED >| ----- V+ doesn't put any current across the LED in the reverse direction

You may want to adjust the 220 ohm resistor to account for the 0.7V drop of the diode. So, use a slightly smaller resistor.

One problem with this circuit is that it'll dim as the battery discharges. A better way to go is to use a 'constant current source'. Here is one: (view in courier font)

VCC --->|---o------. 1N4001 | | | | | | .-. V ->

| | \\| | 2N2222 |----o