modifying existing LED cirtuit

[Robbob]

Hello all,

i have a basic question to ask, i tried to find it on the web but with
no luck.
i am trying to use a LED array taken from a lamp sold in a discount
store for 2$.
It is quite simple, 5 whites LEDs (in parallel), a push button switch
and a 10 Ohms (brown, black,black, gold) resistor. It is powered by 3
AAA batteries. Since that makes 4,5 volts, i taught naively that i
could hook it up directly on the 5 volts from a PC power supply. Well
you guessed it, the resistor became real hot real fast!

Now to my question, can i simply replace the resistor with on that is
the proper strength to limit correctly the current to the LEDs?

I've include a rude diagram of the actual array.
http://docs.google.com/leaf?id=0B8sOh0II6JLrMjA1ZmI0YTctZmE3NC00ZGYzLTkzOGUtZDBhNzFjYTg2ZjEw&sort=name&layout=list&num=50

Sorry for the long link i can't find the proper way to do it

Robbob

 

[Jeff Johnson]

"Robbob" <robbob1er@gmail.com> wrote in message
news:76cf2652-55bd-4f1c-b248-6e44bf082f61@t20g2000yqa.googlegroups.com...

Hello all,

i have a basic question to ask, i tried to find it on the web but with
no luck.
i am trying to use a LED array taken from a lamp sold in a discount
store for 2$.
It is quite simple, 5 whites LEDs (in parallel), a push button switch
and a 10 Ohms (brown, black,black, gold) resistor. It is powered by 3
AAA batteries. Since that makes 4,5 volts, i taught naively that i
could hook it up directly on the 5 volts from a PC power supply. Well
you guessed it, the resistor became real hot real fast!

Now to my question, can i simply replace the resistor with on that is
the proper strength to limit correctly the current to the LEDs?

I've include a rude diagram of the actual array.
http://docs.google.com/leaf?id=0B8sOh0II6JLrMjA1ZmI0YTctZmE3NC00ZGYzLTkzOGUtZDBhNzFjYTg2ZjEw&sort=name&layout=list&num=50

Sorry for the long link i can't find the proper way to do it

Yes, you must increase the resistance. Normally one would use a constant
current source but the resistor is suppose to be acting like one. It is a
cheap way that doesn't usually work whenyou alter the parameters.

Easiest way would be to measure the current using the batteries then find a
resistance using the 5v that gives the same current.

So if it was normally drawing 0.25A with the 10Ohm resistor at 4.5V then you
might need 30Ohms or so when you go to 5V to get approximately the same
current.

Basically LED's are diodes and when they are forward biased their internal
resistance drops significantly(almost acting as a short). It seems giving
the extra 0.5V causes the LED's to act as a short(so you have 0.5A through
the resistor).

Also they could be using the internal resistance of the batteries to help
limit the current which the computer supply doesn't have(or which is much
much smaller).

e.g., you might need a 50Ohm resistor in series to get the same current
through the led's.

 

[Jon Kirwan]

On Sun, 15 Aug 2010 15:01:42 -0700 (PDT), Robbob
<robbob1er@gmail.com> wrote:

i have a basic question to ask, i tried to find it on the web but with
no luck.
i am trying to use a LED array taken from a lamp sold in a discount
store for 2$.
It is quite simple, 5 whites LEDs (in parallel), a push button switch
and a 10 Ohms (brown, black,black, gold) resistor. It is powered by 3
AAA batteries. Since that makes 4,5 volts, i taught naively that i
could hook it up directly on the 5 volts from a PC power supply. Well
you guessed it, the resistor became real hot real fast!

Assuming, for a moment, that the AAA batteries weren't being
limited by their own internal R (which should be under 1/2 an
Ohm) and that the R in question is supposed to run at about
1/8th watt and no more at 4.5V, then I'd figure this:

P = I^2 * R, therefore I = sqrt(P/R) or about 100mA.

This is consistent with the 5 LEDs taking 20mA for their
drive. So it seems reasonable from this perspective.

This 100mA via R=10 suggests the R drops about 1V. In short,
the white LEDs operate on about 3.5V, or so.

A change in V from 4.5 to 5.0 would be dV = 5 - 4.5 = 0.5.
dV/R = dI, so this suggests about another 50mA into the 5
LEDs, or about 10mA into each assuming equal distribution
(which won't be the case, but may be okay for rough guessing
right now.)

The new estimate of 150mA, now at a 1.5V across R instead of
the earlier 1.0V, would be about 1/4 watt instead of 1/8
watt. About doubling the power that needs to be dissipated.
I suspect, but didn't look up just now, that the thermal
resistance of an 1/8watt to 1/4watt resistor must be in the
area of 200C/W. 1/4W at 200C/W is about 50C. If ambient is
about 25C, that's 75C... which is getting hot. Beforehand,
it would have been perhaps 50C, which feels less concerning
when you touch it.

Now to my question, can i simply replace the resistor with on that is
the proper strength to limit correctly the current to the LEDs?

Assuming you still want the same brightness, and let's call
that 100mA or so, then you want to drop 1.5V (I'm assuming
that my earlier 3.5V figure was about right for the LEDs) and
that means 1.5V/0.1A or about 15 ohms. Which is a standard
value. And the dissipation should still be about in the same
area, say 150mW. So getting one that is about the same size
as the current one should be okay.

I've include a rude diagram of the actual array.
http://docs.google.com/leaf?id=0B8sOh0II6JLrMjA1ZmI0YTctZmE3NC00ZGYzLTkzOGUtZDBhNzFjYTg2ZjEw&sort=name&layout=list&num=50

Sorry for the long link i can't find the proper way to do it

I can't get access to it. But that's okay.

Jon

 

[Jon Kirwan]

On Sun, 15 Aug 2010 17:30:17 -0500, "Jeff Johnson"
<Jeff_Johnson@Hotmail.com> wrote:

"Robbob" <robbob1er@gmail.com> wrote in message
news:76cf2652-55bd-4f1c-b248-6e44bf082f61@t20g2000yqa.googlegroups.com...
Hello all,

i have a basic question to ask, i tried to find it on the web but with
no luck.
i am trying to use a LED array taken from a lamp sold in a discount
store for 2$.
It is quite simple, 5 whites LEDs (in parallel), a push button switch
and a 10 Ohms (brown, black,black, gold) resistor. It is powered by 3
AAA batteries. Since that makes 4,5 volts, i taught naively that i
could hook it up directly on the 5 volts from a PC power supply. Well
you guessed it, the resistor became real hot real fast!

Now to my question, can i simply replace the resistor with on that is
the proper strength to limit correctly the current to the LEDs?

I've include a rude diagram of the actual array.
http://docs.google.com/leaf?id=0B8sOh0II6JLrMjA1ZmI0YTctZmE3NC00ZGYzLTkzOGUtZDBhNzFjYTg2ZjEw&sort=name&layout=list&num=50

Sorry for the long link i can't find the proper way to do it

Yes, you must increase the resistance. Normally one would use a constant
current source but the resistor is suppose to be acting like one. It is a
cheap way that doesn't usually work whenyou alter the parameters.

Easiest way would be to measure the current using the batteries then find a
resistance using the 5v that gives the same current.

I think that's good.

So if it was normally drawing 0.25A with the 10Ohm resistor at 4.5V then you
might need 30Ohms or so when you go to 5V to get approximately the same
current.

It probably wouldn't be that much current. That would imply
a drop of 2.5V across the R, leaving 2V for the white LEDs.
Which no white LED works at, so far as I know. That's a red
LED voltage.

Basically LED's are diodes and when they are forward biased their internal
resistance drops significantly(almost acting as a short). It seems giving
the extra 0.5V causes the LED's to act as a short(so you have 0.5A through
the resistor).

A change of .5V across 10 ohms would be a change of 50mA, not
500mA.

Also they could be using the internal resistance of the batteries to help
limit the current which the computer supply doesn't have(or which is much
much smaller).

I checked. AAA alkalines are spec'd to about .150-.300 ohms,
when fresh. This is significantly less than the 10 ohms. But
it is correct to keep it in mind, just in case. In this
case, I don't think it is that important, though.

e.g., you might need a 50Ohm resistor in series to get the same current
through the led's.

As I wrote elsewhere, I'd go with 15 ohms (maybe 18) just to
start. But your point about measuring the current at 4.5V
holds, too. And I'd go that way, if an ammeter is available,
and work out the R from there (after also measuring the 5V
supply voltage.)

Jon

 

[Sjouke Burry]

Robbob wrote:

Hello all,

i have a basic question to ask, i tried to find it on the web but with
no luck.
i am trying to use a LED array taken from a lamp sold in a discount
store for 2$.
It is quite simple, 5 whites LEDs (in parallel), a push button switch
and a 10 Ohms (brown, black,black, gold) resistor. It is powered by 3
AAA batteries. Since that makes 4,5 volts, i taught naively that i
could hook it up directly on the 5 volts from a PC power supply. Well
you guessed it, the resistor became real hot real fast!

Now to my question, can i simply replace the resistor with on that is
the proper strength to limit correctly the current to the LEDs?

I've include a rude diagram of the actual array.
http://docs.google.com/leaf?id=0B8sOh0II6JLrMjA1ZmI0YTctZmE3NC00ZGYzLTkzOGUtZDBhNzFjYTg2ZjEw&sort=name&layout=list&num=50

Sorry for the long link i can't find the proper way to do it

Robbob
You should replace it with 5 resistors, one for each led.

Thats how it should have been done in the first place.
First guess, 8 times the original resistor.

 

[Jon Kirwan]

On Mon, 16 Aug 2010 01:37:35 +0200, Sjouke Burry
<burrynulnulfour@ppllaanneett.nnll> wrote:

Robbob wrote:
Hello all,

i have a basic question to ask, i tried to find it on the web but with
no luck.
i am trying to use a LED array taken from a lamp sold in a discount
store for 2$.
It is quite simple, 5 whites LEDs (in parallel), a push button switch
and a 10 Ohms (brown, black,black, gold) resistor. It is powered by 3
AAA batteries. Since that makes 4,5 volts, i taught naively that i
could hook it up directly on the 5 volts from a PC power supply. Well
you guessed it, the resistor became real hot real fast!

Now to my question, can i simply replace the resistor with on that is
the proper strength to limit correctly the current to the LEDs?

I've include a rude diagram of the actual array.
http://docs.google.com/leaf?id=0B8sOh0II6JLrMjA1ZmI0YTctZmE3NC00ZGYzLTkzOGUtZDBhNzFjYTg2ZjEw&sort=name&layout=list&num=50

Sorry for the long link i can't find the proper way to do it

Robbob

You should replace it with 5 resistors, one for each led.

Yes,

Thats how it should have been done in the first place.

Yes.

First guess, 8 times the original resistor.

Figuring 20mA per LED and about 1.5V drop for each, would be
1.5/0.02 = 75 ohms. So yup-ish, yet again.

Jon

 

[pawihte]

Robbob wrote:

Hello all,

i have a basic question to ask, i tried to find it on the web
but with
no luck.
i am trying to use a LED array taken from a lamp sold in a
discount
store for 2$.
It is quite simple, 5 whites LEDs (in parallel), a push button
switch
and a 10 Ohms (brown, black,black, gold) resistor. It is
powered by 3
AAA batteries. Since that makes 4,5 volts, i taught naively
that i
could hook it up directly on the 5 volts from a PC power
supply. Well
you guessed it, the resistor became real hot real fast!

Now to my question, can i simply replace the resistor with on
that is
the proper strength to limit correctly the current to the LEDs?

I've include a rude diagram of the actual array.
http://docs.google.com/leaf?id=0B8sOh0II6JLrMjA1ZmI0YTctZmE3NC00ZGYzLTkzOGUtZDBhNzFjYTg2ZjEw&sort=name&layout=list&num=50

Sorry for the long link i can't find the proper way to do it

Robbob

A 15-ohm resistor will give you about the same brightness as with
the original 10 ohms on a 4.5V supply. A 1/4-watt resistor will
still get fairly hot but will not burn out. A 1/2-watt type will
run cooler. You could also use two 33-ohm 1/4-watt resistors.

 

[Jeff Johnson]

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news1rg66tgp2iimod476eild7vmscajh253a@4ax.com...

On Sun, 15 Aug 2010 17:30:17 -0500, "Jeff Johnson"
Jeff_Johnson@Hotmail.com> wrote:

"Robbob" <robbob1er@gmail.com> wrote in message
news:76cf2652-55bd-4f1c-b248-6e44bf082f61@t20g2000yqa.googlegroups.com...
Hello all,

i have a basic question to ask, i tried to find it on the web but with
no luck.
i am trying to use a LED array taken from a lamp sold in a discount
store for 2$.
It is quite simple, 5 whites LEDs (in parallel), a push button switch
and a 10 Ohms (brown, black,black, gold) resistor. It is powered by 3
AAA batteries. Since that makes 4,5 volts, i taught naively that i
could hook it up directly on the 5 volts from a PC power supply. Well
you guessed it, the resistor became real hot real fast!

Now to my question, can i simply replace the resistor with on that is
the proper strength to limit correctly the current to the LEDs?

I've include a rude diagram of the actual array.
http://docs.google.com/leaf?id=0B8sOh0II6JLrMjA1ZmI0YTctZmE3NC00ZGYzLTkzOGUtZDBhNzFjYTg2ZjEw&sort=name&layout=list&num=50

Sorry for the long link i can't find the proper way to do it

Yes, you must increase the resistance. Normally one would use a constant
current source but the resistor is suppose to be acting like one. It is a
cheap way that doesn't usually work whenyou alter the parameters.

Easiest way would be to measure the current using the batteries then find
a
resistance using the 5v that gives the same current.

I think that's good.

So if it was normally drawing 0.25A with the 10Ohm resistor at 4.5V then
you
might need 30Ohms or so when you go to 5V to get approximately the same
current.

It probably wouldn't be that much current. That would imply
a drop of 2.5V across the R, leaving 2V for the white LEDs.
Which no white LED works at, so far as I know. That's a red
LED voltage.

Basically LED's are diodes and when they are forward biased their internal
resistance drops significantly(almost acting as a short). It seems giving
the extra 0.5V causes the LED's to act as a short(so you have 0.5A through
the resistor).

A change of .5V across 10 ohms would be a change of 50mA, not
500mA.

It wasn't a change but a maximum current assuming 0 led voltage drop. I
don't know the LED voltage drop nor the configuration(parallel or series) so
I didn't take it into consideration.

Also they could be using the internal resistance of the batteries to help
limit the current which the computer supply doesn't have(or which is much
much smaller).

I checked. AAA alkalines are spec'd to about .150-.300 ohms,
when fresh. This is significantly less than the 10 ohms. But
it is correct to keep it in mind, just in case. In this
case, I don't think it is that important, though.

e.g., you might need a 50Ohm resistor in series to get the same current
through the led's.

As I wrote elsewhere, I'd go with 15 ohms (maybe 18) just to
start. But your point about measuring the current at 4.5V
holds, too. And I'd go that way, if an ammeter is available,
and work out the R from there (after also measuring the 5V
supply voltage.)

I just made up random numbers because it depends on too many unknown
factors. If they set the current depending on the knee voltage of the LED's
then the internal resistance could be significantly reduced resulting in a
significant increase in current.

For example, Lets assumine a standard diode(1N4001) instead of LED's. The
knee voltage at room temp is about 0.63V. The bulk resistance is relatively
large here and we can estimate the bulk resistance to be about 10Ohms.

This means if we can supply 0.63V we would get about about 0.063A.

Suppose we add an additional 10Ohms series resistance and some voltage V to
give the same current above,

V - 0.63 - 0.063*10 = 0 ==> V = 2*0.63 = 1.26V

NOW!!! suppose we have the same circuit but now alter V. Suppose V is set at
4.8V, we can now guess that the diode's bulk resistance is significantly
less and current will be

I = (4.8 - 0.8)/10 = 0.4A

In this case the current as increased almost 4x. (The diode's voltage will
increase to about 0.8V)

For a supply of 2V a similar effect with the current being approximately
0.12A.


I'm not saying they did this in their design of the LED's but they may have
chose a point close to the knee and when the extra voltage was added it was
enough to significantly decrease the LED's bulk resistance.

It depends on too many factors to guess at which is why I suggested using a
meter and gave him pretty save values to start with incase he doesn't have a
meter. My initial guess of 30Ohms may hav been too high as even with a short
5/30 << 4.5/10 but it was mainly to point out that the resistance may need
to increase non-linearly with the voltage. (a linear approximate would
suggest 15Ohms which may be lower than required and has the potential to
destroy the LED's depending on how they originally chose the 10Ohms)

When you factor in the batteries resistance(again, which I don't know), the
non-linearity of the bulk resistance, and unknown design practices 30Ohms is
a very reasonable starting choice which should guarantee preventention of
failure due to overcurrent.

If, say, they design the LED's to run at 95% maximum capacity and the 15 Ohm
serial resistance was 5% smaller and the power supply was 5.1V instead of 5V
then it is possible to ruin the LED's or significantly reduce the lifespan.
Doubtfull but possible.

 

[Jeff Johnson]

"Sjouke Burry" <burrynulnulfour@ppllaanneett.nnll> wrote in message
news:4c687a40$0$14120$703f8584@textnews.kpn.nl...

Robbob wrote:
Hello all,

i have a basic question to ask, i tried to find it on the web but with
no luck.
i am trying to use a LED array taken from a lamp sold in a discount
store for 2$.
It is quite simple, 5 whites LEDs (in parallel), a push button switch
and a 10 Ohms (brown, black,black, gold) resistor. It is powered by 3
AAA batteries. Since that makes 4,5 volts, i taught naively that i
could hook it up directly on the 5 volts from a PC power supply. Well
you guessed it, the resistor became real hot real fast!

Now to my question, can i simply replace the resistor with on that is
the proper strength to limit correctly the current to the LEDs?

I've include a rude diagram of the actual array.
http://docs.google.com/leaf?id=0B8sOh0II6JLrMjA1ZmI0YTctZmE3NC00ZGYzLTkzOGUtZDBhNzFjYTg2ZjEw&sort=name&layout=list&num=50

Sorry for the long link i can't find the proper way to do it

Robbob
You should replace it with 5 resistors, one for each led.
Thats how it should have been done in the first place.
First guess, 8 times the original resistor.

IIRC White LED's have a very gradual knee voltage and hence more less likely
to have issues with current sharing. They design the LED's this way to
reduce the cost of using extra resistors. Of course current sharing is still
an issue but depending on the design it may be relatively safe to use.

 

[Jon Kirwan]

On Mon, 16 Aug 2010 12:26:45 -0500, "Jeff Johnson"
<Jeff_Johnson@Hotmail.com> wrote:

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news1rg66tgp2iimod476eild7vmscajh253a@4ax.com...
snip
A change of .5V across 10 ohms would be a change of 50mA, not
500mA.

It wasn't a change but a maximum current assuming 0 led voltage drop. I
don't know the LED voltage drop nor the configuration(parallel or series) so
I didn't take it into consideration.

It has to be included to make any kind of reasoned
suggestion, though. And white LEDs are based upon blue ones,
which operate in the roughly-3.5V-4V range, depending
somewhat on operating current and design.

snip

I just made up random numbers because it depends on too many unknown
factors.
snip

Well, I used two different bits of knowledge to come at it
from two different directions, both of which yielded about
the same result -- which I suspect strengthens what I guessed
at. It's slightly better than 'random numbers' in this case,
I think.

Jon

 

[Jon Kirwan]

On Mon, 16 Aug 2010 12:31:34 -0500, "Jeff Johnson"
<Jeff_Johnson@Hotmail.com> wrote:

snip
IIRC White LED's have a very gradual knee voltage and hence more less likely
to have issues with current sharing. They design the LED's this way to
reduce the cost of using extra resistors. Of course current sharing is still
an issue but depending on the design it may be relatively safe to use.

There are others here who know better than I do, but so far
as I'm aware single-LED whites (as opposed to RGB style) use
a blue LED and some phosphor to get the job done. I'm not
aware that there is an outright design goal in making the
blue LEDs as you suggest above. Can you quantify your
meaning? For example, provide a specific white LED example
and explain what specific part of an accurate model of it is
as you say, above? A gradual knee voltage might imply a
modeled series R value that is higher than would be expected,
for example. I am curious.

Jon

 

[Jeff Johnson]

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news:io6j66d061tvs71ufopg5ce4p1r5s3g8t9@4ax.com...

On Mon, 16 Aug 2010 12:26:45 -0500, "Jeff Johnson"
Jeff_Johnson@Hotmail.com> wrote:

"Jon Kirwan" <jonk@infinitefactors.org> wrote in message
news1rg66tgp2iimod476eild7vmscajh253a@4ax.com...
snip
A change of .5V across 10 ohms would be a change of 50mA, not
500mA.

It wasn't a change but a maximum current assuming 0 led voltage drop. I
don't know the LED voltage drop nor the configuration(parallel or series)
so
I didn't take it into consideration.

It has to be included to make any kind of reasoned
suggestion, though. And white LEDs are based upon blue ones,
which operate in the roughly-3.5V-4V range, depending
somewhat on operating current and design.

snip

I just made up random numbers because it depends on too many unknown
factors.
snip

Well, I used two different bits of knowledge to come at it
from two different directions, both of which yielded about
the same result -- which I suspect strengthens what I guessed
at. It's slightly better than 'random numbers' in this case,
I think.

You made an estimation while mine was not an estimation but a demonstration.
Your estimation may be "correct" but if it happens to be too low then you
can ruin his LED's. I'm not going to take that chance since it is not mine
to ruin. My numbers wern't to give him any values to use which is why I made
them higher than nominal just incase he did use them. The idea is that if he
uses them and they produce an intensity that is to low he could say "I used
10Ohms and 30Ohms so it must be somewhere inbetween". He then might try 20
ohms or so and realize it probably would work. If the resistor was cool he
might decide to go 15ohms or stick with 10 and "risk it". In any case I
would bear no responsibility for his actions since I didn't tell him what to
use.

After all, if the heat of the resistor was the only concern then there are
simple ways to fix that. If it was something I was doing I probably would
have doubled the resistance simply to reduce power draw since the light
intensity would probably be sufficient.

 

[Robbob]

and I that taught that it was a simple matter of changing one resistor
for an other!

so if i want to keep the current configuration (it is on a nice little
pcb with a switch) i need to change the one resistor with a single
15ohms 1/4 watt?

 

[Jon Kirwan]

On Mon, 16 Aug 2010 14:15:28 -0700 (PDT), Robbob
<robbob1er@gmail.com> wrote:

and I that taught that it was a simple matter of changing one resistor
for an other!

so if i want to keep the current configuration (it is on a nice little
pcb with a switch) i need to change the one resistor with a single
15ohms 1/4 watt?

As someone else wrote, I might go for a 1/2watt, not 1/4.
That will give you a little margin. You could instead
parallel two 33 ohm, 1/4 watt resistors; or perhaps 3 of the
47 ohm or 56 ohm variety at 1/4 or 1/8 watt and get by.

Chances are, you have a full 1/4 watt to dissipate there. So
you don't want to have only 1/4 watt capability in the part,
as that would not give you any margin and the part will be
pretty hot when it runs. Better to keep the temp down a bit.

Assuming you really do only want to replace that one
resistor, of course.

Jon

 

[Robbob]

I'll try just that thanks you all