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?

"Dave.H" <the1930s@googlemail.com> wrote in message
news:227bac2d-4517-4cd5-8880-d8eaca9b7c4a@v17g2000hsa.googlegroups.com...
> 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

On Feb 10, 5:10 pm, "BobW" <nimby_NEEDS...@roadrunner.com> wrote:
> "Dave.H" <the19...@googlemail.com> wrote in message
>
> news:227bac2d-4517-4cd5-8880-d8eaca9b7c4a@v17g2000hsa.googlegroups.com...
>
> > 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

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

On Sat, 09 Feb 2008 22:10:38 -0800, BobW wrote:

> "Dave.H" <the1930s@googlemail.com> wrote in message
> news:227bac2d-4517-4cd5-8880-d8eaca9b7c4a@v17g2000hsa.googlegroups.com...
>> 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.

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.

"Dave.H" <the1930s@googlemail.com> wrote in message
news:dfcc9dd0-cbb2-4fa1-af88-a7fd4b1bacbe@i29g2000prf.googlegroups.com...
> On Feb 10, 5:10 pm, "BobW" <nimby_NEEDS...@roadrunner.com> wrote:
>> "Dave.H" <the19...@googlemail.com> wrote in message
>>
>> news:227bac2d-4517-4cd5-8880-d8eaca9b7c4a@v17g2000hsa.googlegroups.com...
>>
>> > 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
>
> 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

"Stephen J. Rush" <sjrush@comcast.net> wrote in message
news:LYydndhov9TcBDPanZ2dnUVZ_ofinZ2d@comcast.com...
> On Sat, 09 Feb 2008 22:10:38 -0800, BobW wrote:
>
>> "Dave.H" <the1930s@googlemail.com> wrote in message
>> news:227bac2d-4517-4cd5-8880-d8eaca9b7c4a@v17g2000hsa.googlegroups.com...
>>> 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.
>
> 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

> 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.

On Feb 10, 5:31 pm, "BobW" <nimby_NEEDS...@roadrunner.com> wrote:
> "Dave.H" <the19...@googlemail.com> wrote in message
>
> news:dfcc9dd0-cbb2-4fa1-af88-a7fd4b1bacbe@i29g2000prf.googlegroups.com...
>
>
>
> > On Feb 10, 5:10 pm, "BobW" <nimby_NEEDS...@roadrunner.com> wrote:
> >> "Dave.H" <the19...@googlemail.com> wrote in message
>
> >>news:227bac2d-4517-4cd5-8880-d8eaca9b7c4a@v17g2000hsa.googlegroups.com...
>
> >> > 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
>
> > 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

I can't understand it.

"Dave.H" <the1930s@googlemail.com> wrote in message
news:9c175092-49b4-422a-b435-ca0d8b53dacd@c23g2000hsa.googlegroups.com...
> On Feb 10, 5:31 pm, "BobW" <nimby_NEEDS...@roadrunner.com> wrote:
>> "Dave.H" <the19...@googlemail.com> wrote in message
>>
>> news:dfcc9dd0-cbb2-4fa1-af88-a7fd4b1bacbe@i29g2000prf.googlegroups.com...
>>
>>
>>
>> > On Feb 10, 5:10 pm, "BobW" <nimby_NEEDS...@roadrunner.com> wrote:
>> >> "Dave.H" <the19...@googlemail.com> wrote in message
>>
>> >>news:227bac2d-4517-4cd5-8880-d8eaca9b7c4a@v17g2000hsa.googlegroups.com...
>>
>> >> > 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
>>
>> > 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
>
> I can't understand it.


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

Bob

On Feb 10, 6:21 pm, "BobW" <nimby_NEEDS...@roadrunner.com> wrote:
> "Dave.H" <the19...@googlemail.com> wrote in message
>
> news:9c175092-49b4-422a-b435-ca0d8b53dacd@c23g2000hsa.googlegroups.com...
>
>
>
> > On Feb 10, 5:31 pm, "BobW" <nimby_NEEDS...@roadrunner.com> wrote:
> >> "Dave.H" <the19...@googlemail.com> wrote in message
>
> >>news:dfcc9dd0-cbb2-4fa1-af88-a7fd4b1bacbe@i29g2000prf.googlegroups.com...
>
> >> > On Feb 10, 5:10 pm, "BobW" <nimby_NEEDS...@roadrunner.com> wrote:
> >> >> "Dave.H" <the19...@googlemail.com> wrote in message
>
> >> >>news:227bac2d-4517-4cd5-8880-d8eaca9b7c4a@v17g2000hsa.googlegroups.com...
>
> >> >> > 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
>
> >> > 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
>
> > 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.

On Sat, 9 Feb 2008 23:44:48 -0800 (PST), "Dave.H"
<the1930s@googlemail.com> wrote:

>
>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.
--

On Feb 11, 12:39 am, default <defa...@defaulter.net> wrote:
> On Sat, 9 Feb 2008 23:44:48 -0800 (PST), "Dave.H"
>
> <the19...@googlemail.com> wrote:
>
> >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 http://www.dse.com.au

On Sun, 10 Feb 2008 06:20:10 -0800 (PST), "Dave.H"
<the1930s@googlemail.com> wrote:

>On Feb 11, 12:39 am, default <defa...@defaulter.net> wrote:
>> On Sat, 9 Feb 2008 23:44:48 -0800 (PST), "Dave.H"
>>
>> <the19...@googlemail.com> wrote:
>>
>> >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 http://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.
--

On Feb 11, 1:37 am, default <defa...@defaulter.net> wrote:
> On Sun, 10 Feb 2008 06:20:10 -0800 (PST), "Dave.H"
>
>
>
> <the19...@googlemail.com> wrote:
> >On Feb 11, 12:39 am, default <defa...@defaulter.net> wrote:
> >> On Sat, 9 Feb 2008 23:44:48 -0800 (PST), "Dave.H"
>
> >> <the19...@googlemail.com> wrote:
>
> >> >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 atwww.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.
> --

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.

> 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.

On Sun, 10 Feb 2008 06:57:59 -0800 (PST), "Dave.H"
<the1930s@googlemail.com> wrote:

>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.
--

On Feb 10, 2:00 am, "Dave.H" <the19...@googlemail.com> wrote:
> On Feb 10, 5:31 pm, "BobW" <nimby_NEEDS...@roadrunner.com> wrote:
>
>
>
> > "Dave.H" <the19...@googlemail.com> wrote in message
>
> >news:dfcc9dd0-cbb2-4fa1-af88-a7fd4b1bacbe@i29g2000prf.googlegroups.com...
>
> > > On Feb 10, 5:10 pm, "BobW" <nimby_NEEDS...@roadrunner.com> wrote:
> > >> "Dave.H" <the19...@googlemail.com> wrote in message
>
> > >>news:227bac2d-4517-4cd5-8880-d8eaca9b7c4a@v17g2000hsa.googlegroups.com...
>
> > >> > 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
>
> > > 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
>
> 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

On Feb 10, 1:31 am, "BobW" <nimby_NEEDS...@roadrunner.com> wrote:
> "Dave.H" <the19...@googlemail.com> wrote in message
>
> news:dfcc9dd0-cbb2-4fa1-af88-a7fd4b1bacbe@i29g2000prf.googlegroups.com...
>
>
>
> > On Feb 10, 5:10 pm, "BobW" <nimby_NEEDS...@roadrunner.com> wrote:
> >> "Dave.H" <the19...@googlemail.com> wrote in message
>
> >>news:227bac2d-4517-4cd5-8880-d8eaca9b7c4a@v17g2000hsa.googlegroups.com...
>
> >> > 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
>
> > 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

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

On Feb 11, 2:16 am, redbelly <redbell...@yahoo.com> wrote:
> On Feb 10, 2:00 am, "Dave.H" <the19...@googlemail.com> wrote:
>
>
>
> > On Feb 10, 5:31 pm, "BobW" <nimby_NEEDS...@roadrunner.com> wrote:
>
> > > "Dave.H" <the19...@googlemail.com> wrote in message
>
> > >news:dfcc9dd0-cbb2-4fa1-af88-a7fd4b1bacbe@i29g2000prf.googlegroups.com...
>
> > > > On Feb 10, 5:10 pm, "BobW" <nimby_NEEDS...@roadrunner.com> wrote:
> > > >> "Dave.H" <the19...@googlemail.com> wrote in message
>
> > > >>news:227bac2d-4517-4cd5-8880-d8eaca9b7c4a@v17g2000hsa.googlegroups.com...
>
> > > >> > 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
>
> > > > 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
>
> > 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

"Dave.H" <the1930s@googlemail.com> wrote in message
news:227bac2d-4517-4cd5-8880-d8eaca9b7c4a@v17g2000hsa.googlegroups.com...
> 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 http://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

"Dave.H" <the1930s@googlemail.com> wrote in message
news:16a907c1-6844-4ed3-80a6-76c00536c7a4@s8g2000prg.googlegroups.com...
>
>> 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.
>

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

"Dave.H" <the1930s@googlemail.com> wrote in message
news:16a907c1-6844-4ed3-80a6-76c00536c7a4@s8g2000prg.googlegroups.com...
>
>
> 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

On Feb 11, 8:55 pm, "Steve" <n...@thiswontwork.com> wrote:
> "Dave.H" <the19...@googlemail.com> wrote in message
>
> news:16a907c1-6844-4ed3-80a6-76c00536c7a4@s8g2000prg.googlegroups.com...
>
> > 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

It is 5v.

On Feb 10, 6:57=A0am, "Dave.H" <the19...@googlemail.com> wrote:
\> [red LED]
> 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.

In article <iIQrj.8467$nK5.2348@nlpi069.nbdc.sbc.com>, Bob Monsen wrote:
>"Dave.H" <the1930s@googlemail.com> wrote in message
>news:227bac2d-4517-4cd5-8880-d8eaca9b7c4a@v17g2000hsa.googlegroups.com...
>> 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.

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 (don@misty.com)