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LED Reverse Polarity Protection

Discussion in 'Electronic Basics' started by Dave.H, Feb 10, 2008.

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  1. Dave.H

    Dave.H Guest

    Hi, I'm considering using a high brightness LED in an old incandescent
    flashlight, in series with a 220 ohm resistor, for use with a 6 volt
    lantern battery, but I also want some type of reverse polarity
    protection for the led, as I'm putting the LED, resistor, etc. in a
    regular flashlight lamp base. I assume this would most likely be done
    with a diode, perhaps a 1N400x type, but how would I wire this up?
     
  2. BobW

    BobW Guest

    You're on the right track. You don't need a high-current diode like a
    1N400x, however. A 1N4148 or 1N914 will work fine because the maximum
    current will always be very low (6V/220ohm = okay for a small signal diode).

    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
     
  3. Dave.H

    Dave.H Guest

    Thanks, do I put the resistor before or after the diode?
     
  4. 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.
     
  5. BobW

    BobW Guest

    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
     
  6. BobW

    BobW Guest

    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
     
  7. Dave.H

    Dave.H Guest

    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.
     
  8. Dave.H

    Dave.H Guest

    I can't understand it.
     
  9. BobW

    BobW Guest


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

    Bob
     
  10. Dave.H

    Dave.H Guest

    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.
     
  11. default

    default Guest

    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.
    --
     
  12. Dave.H

    Dave.H Guest

    Look up Cat #Z4024 at www.dse.com.au
     
  13. default

    default Guest

    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:

    http://www.dealextreme.com/details.dx/sku.1776
    Red CREE LED Emitter (20mm 1.9~2.2V)
    $4.97 free shipping RED

    http://www.dealextreme.com/details.dx/sku.2134
    Cree P4 XR-E 7090 (WD) Emitter on Star (5-Pack)
    $21.39 free shipping WHITE

    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.
    --
     
  14. Dave.H

    Dave.H Guest

    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.
     
  15. default

    default Guest

    Ah, that makes sense.
    --
     
  16. Dave.H

    Dave.H Guest

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

    redbelly Guest

    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
     
  18. redbelly

    redbelly Guest

    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
     
  19. Dave.H

    Dave.H Guest

    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
     
  20. Bob Monsen

    Bob Monsen Guest

    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 ->
    10k | | -
    | | | Your LED
    '-' |
    | |
    | |/
    o----| 2N2222
    | |>
    | |
    \| |
    2N2222 |----o
    <| |
    | |
    | .-.
    | | |
    | | | 33ohms
    | '-'
    | |
    GND ---------o------'
    (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)

    This circuit should keep the light output constant (about 20mA) down to
    about 4V or less, depending on the voltage drop of your LED.

    Regards,
    Bob Monsen
     
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