LED Reverse Polarity Protection

[Dave.H]

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?

 

[BobW]

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?

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

 

[Dave.H]

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

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

 

[Stephen J. Rush]

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.

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.

 

[BobW]

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

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

 

[BobW]

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.

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

 

[Dave.H]

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.

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.H]

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

I can't understand it.

 

[BobW]

I can't understand it.

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

Bob

 

[Dave.H]

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.

 

[default]

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.
--

 

[Dave.H]

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 www.dse.com.au

 

[default]

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

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.
--

 

[Dave.H]

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.
--

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.

 

[default]

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.
--

 

[Dave.H]

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:

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

 

[redbelly]

I can't understand it.

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

 

[redbelly]

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

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

 

[Dave.H]

I can't understand it.

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

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

 

[Bob Monsen]

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?

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