Electronics idiot needs LED replacement help, for a good cau

[Joe Bob]

Hi,

I'm in need of bit of help changing a small booklight from operating
with three white LEDs to one yellow or one orange LED bulb. Assuming,
it can be done, for the moment.

Why you ask. Because I have a friend who has E.P. (erythropoietic
porphyria). E.P. is an extremely rare blood disorder which makes my
friend allergic to light. Basically a vampire condition.

Whites, blues, and green bulbs are out of the question! She can
tolerate yellow and red (590nm and above) in small amounts, think 15 to
30 minutes a day, one bulb. Yes, it's that bad.

Thus, my friend has been living in almost complete darkness in a
basement now for about 8 years and she needs some help.

Trust me, I would have absolutely no reason to ever attempt this task
unless there weren't a good reason to do this. My friend is a good
person and she doesn't have anyone else who can even come close to
pulling off this task, as simple as I know the task is for someone who
is experienced.

So, back to the problem at hand. Please be patient with me, this is not
something I know proper terminologies for, etc.

She has provided me with a small booklight of her choice that has inside
it two 3 volt flat quarter sized batteries in it, and a circuit board.

I've placed some photos of the circuit board on a website of mine:

http://www.airporttools.com/led/index.html

I hope the photos help in some manner. I know they are poor, but I was
running out of light, so I had to hurry.

On the circuit board there is a switch that the user depresses once for
full brightness, twice to dim the light a bit, and three times turns the
circuit off.

Upon inspection the circuit board has three white LEDs attached. It
also has what I believe to be a capacitor whose specifications read 22uf
and 10 volts.

I do not see a resister on this circuit board which is a bit confusing
to me. Shouldn't there be one on the board?

Perhaps there is something on the board that I'm not aware of or perhaps
it's invisible BTW, I believe I know what a standard resister looks
like. I've purchased them before for a much larger lamp I built for her
previously.

Perhaps the switch is acting as a resister?

Moving on, at the battery leads to the circuit board I can measure 6
volts of current.

When the lamp is on full brightness I can measure across the leads of
one of the LEDs 3.4 volts. 3.4 volts, from what I read, may be fairly
standard for a white LED?

Eyeballing the circuit board, the three white LEDs appear to be
connected in parallel.

Anyway, basically I want to remove the three white LEDs and insert on
the board just one of the new orange LEDs that I purchased several years
ago for a similar project I did for her.

These loose orange LEDs were purchased from theledlight.com and they are
5mm 2200. I believe each orange LED bulb is basically a 2 volt bulb.

Please note that I have done enough soldering work that I believe I can
remove un-needed components and I can certainly solder on needed
components. I've done it before manually, just not on this small of a
scale. For the moment I'm going to assume I can do this task.

Now for my questions:

- Should I just give up and create a new circuit board?

If yes, then I probably can build this circuit myself. Although, I'd
rather keep the current circuit board since it already fits the
booklight's housing properly and it already has a switch built onto it
that works in the housing as well.

Assuming the answer to the above question is no:

- What role does the capacitor that's currently on the board have?

- Will I need to place a resister on the board to achieve what I want to do?

- What type of resister? Ohms and voltage please, thank you very much.

If, for example, the ohms come out to be 47 and the voltage is 1/8 volt,
can I use a 47 ohm 1/4 volt resister instead? I may have problems
obtaining a 1/8 volt resister. I can probably find 1/4 or 1/2 volt
resisters in my local stores. Otherwise, I'm may need to order some
resisters from a website?

- The measurement of 3.4 volts that I get between the leads of the first
white LED is what is throwing me off. Without a resister I don't
understand how the voltage is going from 6 at the circuit board leads to
3.4 across the first LED.

Is this what the capacitor is doing?

Or are the LEDs, in parallel, themselves creating this situation?

Or is it the switch?

Or something else?

- Can I remove the three white LEDs, then place the orange LED in the
holes where the second white LED currently sits, then place a jumper
wire between the holes where the first white LED currently sits, and
then place a jumper wire between the holes where the third white LED
currently sits?

My guess is that I cannot since 3.4 volts is what I believe to be too
much for my orange LED.

Or am I totally off base?

If there is anyone out there who would be willing to answer my questions
I would truly appreciate it. And I can assure you that your efforts
would be going ultimately towards someone who deserves a break in life.

I understand that your time is valuable, therefore, thank you so very
much for listening to me.

Regards,

Greg

 

[Ecnerwal]

In article <d8OdnR7vq532njTanZ2dnUVZ_tyknZ2d@giganews.com>,
Joe Bob <JoeBob@Bob.com> wrote:

- Can I remove the three white LEDs, then place the orange LED in the
holes where the second white LED currently sits, then place a jumper
wire between the holes where the first white LED currently sits, and
then place a jumper wire between the holes where the third white LED
currently sits?

From what I can see of your pictures, the "black blob" opposite the
switch is what's doing the work (a custom chip mounted directly to the
board and encapsulated with black glop), along with the small transistor
on the backside (in the trace with the middle LED - 3-terminal device).
With any luck, it's doing a current limited approach, in which case it
would get the voltage right without needing an added resistor. Worst
case, you try it, you fry one orange LED, you report back. You have what
sounds like "a supply" of loose orange LEDs, so try swapping one in. DO
NOT short out the spaces where the other two are. Best bet would be to
replace all three, but suit yourself.

--
Cats, coffee, chocolate...vices to live by

 

[Joe Bob]

Ecnerwal wrote:

In article <d8OdnR7vq532njTanZ2dnUVZ_tyknZ2d@giganews.com>,
Joe Bob <JoeBob@Bob.com> wrote:

- Can I remove the three white LEDs, then place the orange LED in the
holes where the second white LED currently sits, then place a jumper
wire between the holes where the first white LED currently sits, and
then place a jumper wire between the holes where the third white LED
currently sits?

From what I can see of your pictures, the "black blob" opposite the
switch is what's doing the work (a custom chip mounted directly to the
board and encapsulated with black glop), along with the small transistor
on the backside (in the trace with the middle LED - 3-terminal device).
With any luck, it's doing a current limited approach, in which case it
would get the voltage right without needing an added resistor. Worst
case, you try it, you fry one orange LED, you report back. You have what
sounds like "a supply" of loose orange LEDs, so try swapping one in. DO
NOT short out the spaces where the other two are. Best bet would be to
replace all three, but suit yourself.

Wow, I'll try that, if it works that would make the whole process
incredibly simple.

I'll get back to you soon.

And yes, I do have a supply of loose LEDs. About 25 of them. At least
I'm realistic.

Thanks.

Greg

 

[DJ Delorie]

It looks like there's a blob on the other side of the circuit board.
If so, I'm guessing it's some sort of constant-current switching power
supply, with the switch changing the current setting or duty cycle.
If that's true, it may "just work" with orange LEDs because it will
change the voltage to make the current correct.

OTOH it looks like there's a switching transistor on the back, so it
might be a variable-duty-cycle control chip.

If you can afford to waste one LED (or a generic red one) to find out,
just wire it in instead of the three white ones, get ready to check
the voltage drop, and turn the light on. You may burn out the LED due
to overcurrent, but you should be able to measure the voltage drop by
then. If it's 2v, it's constant current. If it's 3.4v, it's constant
voltage (see below).

You could also replace the LEDs with a 1k resistor. 3.4v means
constant voltage, 6v means constant current.

If it's not constant current, you may be able to replace those three
LEDs in series, with four new ones as two parallel strings of two in
series:

---->|--->|----
-----| |-----
---->|--->|----

That doubles the voltage drop from 2v to 4v, but may increase the
current-per-led depending on how they do current limiting. Might have
to go with six LEDs to keep the current per LED correct, you'll have
to try it and see. You might also have to switch to red LEDs (1.8v
instead of 2v) to get closer to the "original" voltage.

 

[Joe Bob]

From what I can see of your pictures, the "black blob" opposite the
switch is what's doing the work (a custom chip mounted directly to the
board and encapsulated with black glop), along with the small transistor
on the backside (in the trace with the middle LED - 3-terminal device).
With any luck, it's doing a current limited approach, in which case it
would get the voltage right without needing an added resistor. Worst
case, you try it, you fry one orange LED, you report back. You have what
sounds like "a supply" of loose orange LEDs, so try swapping one in. DO
NOT short out the spaces where the other two are. Best bet would be to
replace all three, but suit yourself.

Hi,

Tried your advice, took out all of the LEDs which were in the board and
I simply put in one of my orange LEDs. And the voltage across the
connection was 2.1 volts. Perfect!

On behalf of my vampire friend and myself thank you so very much!

Have a great day.

Regards,

Greg

 

[Joe Bob]

DJ Delorie wrote:

It looks like there's a blob on the other side of the circuit board.
If so, ....

Hi,

Another person who replied indicated that I should try simply putting in
one of the orange LEDs. I did that and the voltage across the
connections went down to 2.1 which I believe is exactly what I need.

Unless, of course, if you or they can think of a problem. I'm assuming
I will not have the orange LED burn out too terribly premature.

Thank you so very much for your excellent reply and have a great day.

Regards,

Greg

 

[Lord Garth]

"Joe Bob" <JoeBob@Bob.com> wrote in message
news:mo6dnQGR5IdntjfanZ2dnUVZ_hOdnZ2d@giganews.com...

DJ Delorie wrote:
It looks like there's a blob on the other side of the circuit board.
If so, ....

Hi,

Another person who replied indicated that I should try simply putting in
one of the orange LEDs. I did that and the voltage across the connections
went down to 2.1 which I believe is exactly what I need.

Unless, of course, if you or they can think of a problem. I'm assuming I
will not have the orange LED burn out too terribly premature.

Thank you so very much for your excellent reply and have a great day.

Regards,

Greg

I have a feeling that you measured the voltage across the LED. What you
were reading is the forward voltage of the LED. The reading will always
be exactly what is required to turn on the LED. Did you measure the current
through the LED to verify that it is within an acceptable limit? If not,
you will
degrade or destroy the orange LEDs.

 

[DJ Delorie]

"Lord Garth" <lgarth@tantalus.net> writes:

I have a feeling that you measured the voltage across the LED. What
you were reading is the forward voltage of the LED. The reading
will always be exactly what is required to turn on the LED. Did you
measure the current through the LED to verify that it is within an
acceptable limit? If not, you will degrade or destroy the orange
LEDs.

There are two easy tests:

1. No black smoke.

2. Light output from LED seems about right.

Also, keep in mind that with three LEDs in parallel, they'll each get
1/3 of the available current. You'd almost need to run wires from the
PCB to a protoboard, plug in three LEDs, and measure voltage and
current.

 

[redbelly]

On Feb 6, 5:00 pm, Joe Bob <Joe...@Bob.com> wrote:

DJ Delorie wrote:
It looks like there's a blob on the other side of the circuit board.
If so, ....

Hi,

Another person who replied indicated that I should try simply putting in
one of the orange LEDs. I did that and the voltage across the
connections went down to 2.1 which I believe is exactly what I need.

Unless, of course, if you or they can think of a problem. I'm assuming
I will not have the orange LED burn out too terribly premature.

Thank you so very much for your excellent reply and have a great day.

Regards,

Greg

Greg,

Sounds like you at least have something that works now.

One more thing to try: how are things when you use just 1 coin cell
battery rather than 2? The lower voltage of the yellow LED compared
to white, plus the fact you're using just 1 LED instead of 3, means
two things:

1 battery might be sufficient
2 batteries might overdrive a single yellow LED, and shorten it's
lifetime.

Regards,

Mark

p.s Your friend might even appreciate the dimmer light from using just
1 battery.

 

[redbelly]

Greg, I've answered a few of your questions. Read on ...

-- Mark

On Feb 5, 7:54 pm, Joe Bob <Joe...@Bob.com> wrote:

Hi,

I'm in need of bit of help changing a small booklight from operating
with three white LEDs to one yellow or one orange LED bulb. Assuming,
it can be done, for the moment.

Why you ask. Because I have a friend who has E.P. (erythropoietic
porphyria). E.P. is an extremely rare blood disorder which makes my
friend allergic to light. Basically a vampire condition.

Whites, blues, and green bulbs are out of the question! She can
tolerate yellow and red (590nm and above) in small amounts, think 15 to
30 minutes a day, one bulb. Yes, it's that bad.

Thus, my friend has been living in almost complete darkness in a
basement now for about 8 years and she needs some help.

Trust me, I would have absolutely no reason to ever attempt this task
unless there weren't a good reason to do this. My friend is a good
person and she doesn't have anyone else who can even come close to
pulling off this task, as simple as I know the task is for someone who
is experienced.

So, back to the problem at hand. Please be patient with me, this is not
something I know proper terminologies for, etc.

She has provided me with a small booklight of her choice that has inside
it two 3 volt flat quarter sized batteries in it, and a circuit board.

I've placed some photos of the circuit board on a website of mine:

http://www.airporttools.com/led/index.html

I hope the photos help in some manner. I know they are poor, but I was
running out of light, so I had to hurry.

On the circuit board there is a switch that the user depresses once for
full brightness, twice to dim the light a bit, and three times turns the
circuit off.

Upon inspection the circuit board has three white LEDs attached. It
also has what I believe to be a capacitor whose specifications read 22uf
and 10 volts.

I do not see a resister on this circuit board which is a bit confusing
to me. Shouldn't there be one on the board?

Perhaps there is something on the board that I'm not aware of or perhaps
it's invisible BTW, I believe I know what a standard resister looks
like. I've purchased them before for a much larger lamp I built for her
previously.

Perhaps the switch is acting as a resister?

Moving on, at the battery leads to the circuit board I can measure 6
volts of current.

When the lamp is on full brightness I can measure across the leads of
one of the LEDs 3.4 volts. 3.4 volts, from what I read, may be fairly
standard for a white LED?

Eyeballing the circuit board, the three white LEDs appear to be
connected in parallel.

Anyway, basically I want to remove the three white LEDs and insert on
the board just one of the new orange LEDs that I purchased several years
ago for a similar project I did for her.

These loose orange LEDs were purchased from theledlight.com and they are
5mm 2200. I believe each orange LED bulb is basically a 2 volt bulb.

Please note that I have done enough soldering work that I believe I can
remove un-needed components and I can certainly solder on needed
components. I've done it before manually, just not on this small of a
scale. For the moment I'm going to assume I can do this task.

Now for my questions:

- Should I just give up and create a new circuit board?

If yes, then I probably can build this circuit myself. Although, I'd
rather keep the current circuit board since it already fits the
booklight's housing properly and it already has a switch built onto it
that works in the housing as well.

Assuming the answer to the above question is no:

- What role does the capacitor that's currently on the board have?

I'm guessing it's purpose is what's called a "switch debouncer".
Without getting too technical, when you push a switch it's contacts
actually switch on and off quickly several times, over 10 milliseconds
or so. The capacitor helps the circuit register these fast multiple
on/off's as a single click.

For more info you can google "switch debouncing" or "switch debouncer"

- Will I need to place a resister on the board to achieve what I want to do?

It's fairly common for manufacturers to rely on the internal
resistance of the coin cell batteries, and save money by not having an
additional resistance in the circuit. The idea is to sell something
cheap that basically works, not to sell something with optimal
performance that is expensive.

- What type of resister? Ohms and voltage please, thank you very much.

If, for example, the ohms come out to be 47 and the voltage is 1/8 volt,
can I use a 47 ohm 1/4 volt resister instead? I may have problems
obtaining a 1/8 volt resister. I can probably find 1/4 or 1/2 volt
resisters in my local stores. Otherwise, I'm may need to order some
resisters from a website?

The 1/8, 1/4, 1/2 etc. ratings on resistors are for Watts, not Volts.
You might try a google search and read up on the distinction between:
voltage (measured in Volts)
current (measured in Amps, or more often milliamps for LED's)
and power (measured in Watts)

- The measurement of 3.4 volts that I get between the leads of the first
white LED is what is throwing me off. Without a resister I don't
understand how the voltage is going from 6 at the circuit board leads to
3.4 across the first LED.

There actually is a resistor, it's part of the batteries. The proper
term is "internal resistance".

Is this what the capacitor is doing?

Or are the LEDs, in parallel, themselves creating this situation?

Or is it the switch?

Or something else?

Yes, it's in the batteries.

- Can I remove the three white LEDs, then place the orange LED in the
holes where the second white LED currently sits, then place a jumper
wire between the holes where the first white LED currently sits, and
then place a jumper wire between the holes where the third white LED
currently sits?

My guess is that I cannot since 3.4 volts is what I believe to be too
much for my orange LED.

Or am I totally off base?

If there is anyone out there who would be willing to answer my questions
I would truly appreciate it. And I can assure you that your efforts
would be going ultimately towards someone who deserves a break in life.

I understand that your time is valuable, therefore, thank you so very
much for listening to me.

Regards,

Greg

 

[Don Klipstein]

In <LawrenceSMITH-2950CA.20061205022008@news.verizon.net>, Ecnerwal wrote:

In article <d8OdnR7vq532njTanZ2dnUVZ_tyknZ2d@giganews.com>,
Joe Bob <JoeBob@Bob.com> wrote:

- Can I remove the three white LEDs, then place the orange LED in the
holes where the second white LED currently sits, then place a jumper
wire between the holes where the first white LED currently sits, and
then place a jumper wire between the holes where the third white LED
currently sits?

From what I can see of your pictures, the "black blob" opposite the
switch is what's doing the work (a custom chip mounted directly to the
board and encapsulated with black glop), along with the small transistor
on the backside (in the trace with the middle LED - 3-terminal device).
With any luck, it's doing a current limited approach, in which case it
would get the voltage right without needing an added resistor. Worst
case, you try it, you fry one orange LED, you report back. You have what
sounds like "a supply" of loose orange LEDs, so try swapping one in. DO
NOT short out the spaces where the other two are. Best bet would be to
replace all three, but suit yourself.

I suspect the one orange LED would get triple current, maybe more due to
less voltage drop.

I will pull some "educated assumptions out of a hat and assume:

1. The white LEDs are getting 30 mA each for a total of 90 mA.

2. LED current is limited by ohmic resistance. The dimming could be from
this little black blob pulsing them rapidly at some duty cycle. I have
seen that done before.

So 90 mA is experiencing a 2.6 volt drop. This means there is about 29
ohms somewhere.

Now, an orange LED may want 30 mA and probably has a voltage drop of
about 2.2 volts at that current. With 6 volts in, you need to have
something drop 3.8 volts while passing .03 amp. With 29 ohms already
there, you need another 98 - 100 is close enough.

As for resistor wattage - .03 squared times 100 is .09 watt. A 1/4 watt
resistor should be fine with room for error.

Even if my assumption 1 is way off, I think that will make only a minor
difference if you have a 100 ohm 1/4 watt resistor in series with
the LED. You will have to splice it in somehow.

- Don Klipstein (don@misty.com)

 

[Don Klipstein]

In <08406897-57f1-40aa-9d25-65629dd2b01b@v17g2000hsa.googlegroups.com>,
redbelly wrote:

Greg,

Sounds like you at least have something that works now.

One more thing to try: how are things when you use just 1 coin cell
battery rather than 2? The lower voltage of the yellow LED compared
to white, plus the fact you're using just 1 LED instead of 3, means
two things:

1 battery might be sufficient
2 batteries might overdrive a single yellow LED, and shorten it's
lifetime.

p.s Your friend might even appreciate the dimmer light from using just
1 battery.

Coin cells? I missed that - that explains it. Those tend to have a
fair amount of resistance.

Keyfob flashlights with one red, orange or yellow LED tend to have 1
coin cell and no resistor in my experience.

- Don Klipstein (don@misty.com)

 

[redbelly]

On Feb 7, 6:23 pm, d...@manx.misty.com (Don Klipstein) wrote:

In <08406897-57f1-40aa-9d25-65629dd2b...@v17g2000hsa.googlegroups.com>,

redbelly wrote:
Greg,

Sounds like you at least have something that works now.

One more thing to try: how are things when you use just 1 coin cell
battery rather than 2? The lower voltage of the yellow LED compared
to white, plus the fact you're using just 1 LED instead of 3, means
two things:

1 battery might be sufficient
2 batteries might overdrive a single yellow LED, and shorten it's
lifetime.
p.s Your friend might even appreciate the dimmer light from using just
1 battery.

Coin cells? I missed that - that explains it. Those tend to have a
fair amount of resistance.

Yep, he mentioned "two 3 volt flat quarter sized batteries" in the
original post. Easy enough to miss that, though.

Keyfob flashlights with one red, orange or yellow LED tend to have 1
coin cell and no resistor in my experience.

- Don Klipstein (d...@misty.com)

And my white version uses two 2016's on a single white LED. Fresh
batteries push about 50 mA through it.

Mark