Driving LEDs with a battery pack

[fungus]

Hello,

I'm trying to drive some LEDs with standard batteries (eg. from a
battery holder with 3 AAAs in it). Should be easy ... right?

There's various colors of LED which need from 2.2 to 3.3 volts at
around 25mA and I want to drive maybe half a dozen from the battery
pack.

I can calculate the right resistor for any give voltage, no problem,
but how do I deal with the wide range of voltage over the lifetime of
the battery. With a brand new battery the voltage is around 4.6V but
as it discharges it goes down to about 3.3V (with about 10% battery
left). If I pick resistors which work at 3.3V then there's far too
much current when the batteries are new (I measured 60mA on some of
them and they get warm to the touch so I'm guessing that's bad)

So:

a) How delicate are LEDs? Is 60mA going to burn them out?

b) If it is, what's the simplest/smallest circuit which will give me
(eg.) 3.3V @ 150mA from a set of AAA batteries? Size is important as I
want to pack it into a small space.

I did some Googling and tried a 3.3V Zener diode to drop the voltage
but it only dropped the voltage by about 0.2V. I'm guessing the reason
for that is something to do with the the load current being quite high
which makes the Zener resistor very small (two or three ohms).

I also looked at voltage regulators but is seems a 3.3V regulator
needs a higher starting voltage than the batteries can provide.

 

On Sun, 5 Jul 2009 11:22:29 -0700 (PDT), fungus
<openglMYSOCKS@artlum.com> wrote:

Hello,

I'm trying to drive some LEDs with standard batteries (eg. from a
battery holder with 3 AAAs in it). Should be easy ... right?

There's various colors of LED which need from 2.2 to 3.3 volts at
around 25mA and I want to drive maybe half a dozen from the battery
pack.

I can calculate the right resistor for any give voltage, no problem,
but how do I deal with the wide range of voltage over the lifetime of
the battery. With a brand new battery the voltage is around 4.6V but
as it discharges it goes down to about 3.3V (with about 10% battery
left). If I pick resistors which work at 3.3V then there's far too
much current when the batteries are new (I measured 60mA on some of
them and they get warm to the touch so I'm guessing that's bad)

So:

a) How delicate are LEDs? Is 60mA going to burn them out?

b) If it is, what's the simplest/smallest circuit which will give me
(eg.) 3.3V @ 150mA from a set of AAA batteries? Size is important as I
want to pack it into a small space.

I did some Googling and tried a 3.3V Zener diode to drop the voltage
but it only dropped the voltage by about 0.2V. I'm guessing the reason
for that is something to do with the the load current being quite high
which makes the Zener resistor very small (two or three ohms).

I also looked at voltage regulators but is seems a 3.3V regulator
needs a higher starting voltage than the batteries can provide.

1. What are the specs on your LEDs?
If they're generic/Ebay unmarked units, limit the current to a max of
20ma.

2. Use a resistor value that limits the current at maximum battery
voltage. Brightness will drop off some as the batteries age, but
probably not much.

3. Google for
joule thief
This is a circuit that boosts battery voltage with a minium number of
components. The most common place to find a pre-built unit is the
common solar walkway light. Many of these use a single NiCd or NiMH
battery and use a joule thief type circuit to boost that voltage to
the 3+ volts needed for a white LED.

You can buy a kit for $10:
http://www.nifty-stuff.com/LED-boost-circuit-joule-thief.php

John

 

[Jon Kirwan]

On Sun, 05 Jul 2009 15:08:36 -0400, news@jecarter.us wrote:

snip of question about LEDs

1. What are the specs on your LEDs?
If they're generic/Ebay unmarked units, limit the current to a max of
20ma.

2. Use a resistor value that limits the current at maximum battery
voltage. Brightness will drop off some as the batteries age, but
probably not much.

3. Google for
joule thief
This is a circuit that boosts battery voltage with a minium number of
components. The most common place to find a pre-built unit is the
common solar walkway light. Many of these use a single NiCd or NiMH
battery and use a joule thief type circuit to boost that voltage to
the 3+ volts needed for a white LED.

You can buy a kit for $10:
http://www.nifty-stuff.com/LED-boost-circuit-joule-thief.php

I've built these and they work, great. Not hard to make, either.
But....

This link comes up quickly:
http://www.ledsales.com.au/kits/joule_thief.pdf

I read through it. And the description under "How it works" is just a
little bit too simplified for me. Perhaps the shortest possible
explanation that retains important details is: the BJT turns on via
R1 allowing current through the collector winding to rise along a
ramp. During the on-time, rates of flux change in the base winding
actually aids the battery voltage a little in keeping the BJT on still
harder. At some point, though, the fast rise in collector current
demands more base current than can be supplied and the BJT tries to
limit the collector current. In doing so, the collector winding goes
through zero volts across it and reverses its voltage to continue
driving the current through the LED. But the rate of flux change in
the base winding also then reverses, opposes the battery supply, and
acts to turn the BJT completely off. While the BJT is off, the
collector winding's current (and thus, the LED current) declines on a
ramp of its own. As it does, the voltage across it required to drive
the LED also drops. As that drops, so does the rate of flux change in
the base winding and thus it's voltage opposing the battery.
Eventually, the induced voltage isn't enough and the BJT is able to
begin turning back on. As that happens, the voltage across the
collector winding reverses again and the base winding starts aiding
the battery, once again, snapping the BJT into fuller on conditions.
And the cycle repeats.

There are some additional details. As the collector current rises, so
does Vce. As Vce rises, the voltage across the collector winding
declines gradually, which results is lower rate of flux change and
that softens the supporting base current at the very time when more
base current is required due to the rising collector current. So the
BJT doesn't turn off suddenly, but "rolls quickly into" the turn off.

Hobbyist view, anyway.

Jon

 

[Electronworks.co.uk]

"fungus" <openglMYSOCKS@artlum.com> wrote in message
news:8c20623c-6e75-4321-b77c-de3c5a34e681@c36g2000yqn.googlegroups.com...

Hello,

I'm trying to drive some LEDs with standard batteries (eg. from a
battery holder with 3 AAAs in it). Should be easy ... right?

There's various colors of LED which need from 2.2 to 3.3 volts at
around 25mA and I want to drive maybe half a dozen from the battery
pack.

I can calculate the right resistor for any give voltage, no problem,
but how do I deal with the wide range of voltage over the lifetime of
the battery. With a brand new battery the voltage is around 4.6V but
as it discharges it goes down to about 3.3V (with about 10% battery
left). If I pick resistors which work at 3.3V then there's far too
much current when the batteries are new (I measured 60mA on some of
them and they get warm to the touch so I'm guessing that's bad)

So:

a) How delicate are LEDs? Is 60mA going to burn them out?

b) If it is, what's the simplest/smallest circuit which will give me
(eg.) 3.3V @ 150mA from a set of AAA batteries? Size is important as I
want to pack it into a small space.

I did some Googling and tried a 3.3V Zener diode to drop the voltage
but it only dropped the voltage by about 0.2V. I'm guessing the reason
for that is something to do with the the load current being quite high
which makes the Zener resistor very small (two or three ohms).

I also looked at voltage regulators but is seems a 3.3V regulator
needs a higher starting voltage than the batteries can provide.

It sounds like you just need a simple boost converter to boost your battery
voltage up to a voltage that will drive the LED stack with the highest
forward voltage (3v3 in your case). Choose the current limiting resistors
accordingly for each LED stack to give about 20mA flowing through each LED
(25mA seems a bit high for a standard LED, but check the datasheet).

Just Google 'boost converter'. LT and Maxim do them in DIP packages so are
easy to solder. Other parts mentioned in later posts refer to winding
transformers, which can get a bit tricky. A standard boost converter uses an
off the shelf inductor.

You can get clever and use a different boost converter for each LED (LED
brightness is a function of the current going through them, not the voltage
across them), but if you dont mind throwing away some power, powering them
all off one common rail is much simpler.

--
Bill Naylor
www.electronworks.co.uk
Electronic Kits for Education and Fun

 

[fungus]

On Jul 5, 9:08 pm, n...@jecarter.us wrote:

1. What are the specs on your LEDs?
If they're generic/Ebay unmarked units, limit the current to a max of
20ma.

Ebay specials... <g>

2. Use a resistor value that limits the current at maximum battery
voltage.  Brightness will drop off some as the batteries age, but
probably not much.

Quite a lot in my tests. If I select resistors for "brand new battery"
then for most of the life of the battery I see about half brightness.
This would be ok for an on/off indicator but the stuff I'm making is
decorative for a procession so I want them to be as bright as
possible.
I'm not worried if the life of the LED is reduced from 100,000 hours.

3. Google for
joule thief

I found a ferrite bead in an old PC and lashed on up. It works
but isn't very bright (maybe only a quarter of 20mA brightness).

 

[fungus]

On Jul 6, 10:18 am, "Electronworks.co.uk"
<newsgro...@electronworks.co.uk> wrote:

It sounds like you just need a simple boost converter...

I'm not necessarily trying to run off a single battery.
Do you think I should be?

You can get clever and use a different boost converter for each LED (LED
brightness is a function of the current going through them, not the voltage
across them

Isn't it basically the same thing...? Maybe not if I should
be googling for "current limiting circuit" instead of "voltage
limiting circuit".

OTOH, I might give the Zener diodes another try (because
it's easy/small/cheap) with an eye to putting one diode
per LED (or maybe one for every two LEDs if that works).

I'm just waiting for the shops to open because my ham-fisted
experiments made the magic pixie smoke come out of the
one I had...)

 

[David Eather]

fungus wrote:

On Jul 5, 9:08 pm, n...@jecarter.us wrote:
1. What are the specs on your LEDs?
If they're generic/Ebay unmarked units, limit the current to a max of
20ma.


Ebay specials... <g

2. Use a resistor value that limits the current at maximum battery
voltage. Brightness will drop off some as the batteries age, but
probably not much.


Quite a lot in my tests. If I select resistors for "brand new battery"
then for most of the life of the battery I see about half brightness.
This would be ok for an on/off indicator but the stuff I'm making is
decorative for a procession so I want them to be as bright as
possible.
I'm not worried if the life of the LED is reduced from 100,000 hours.

3. Google for
joule thief

I found a ferrite bead in an old PC and lashed on up. It works
but isn't very bright (maybe only a quarter of 20mA brightness).


you may be able to use a miniature light globe as a current regulator.

Look for something like a 15 - 25 ma globe, with a voltage rating of say
3 - 7.5 volts. Wire it in series with the LED. As the battery voltage
reduces the globe dims, as the lamp filament cools it's resistance
lowers and hence somewhat compensates for the sagging voltage. (such odd
light globes are often supplied for illuminated switches and sometimes
for Wein bridge oscillators

Alternatively, connect a JFET (plus a low value resistor)as a constant
current source. Most JFET's will not work perfectly with so few volts to
play with, but it will be better than nothing.

For something sophisticated you could use almost any of the charge-pump
IC's to double the voltage and give you enough volts for a proper
constant current source or as a simple voltage doubler for when the
battery / led combo is just too low.

 

[David Eather]

fungus wrote:

On Jul 6, 10:18 am, "Electronworks.co.uk"
newsgro...@electronworks.co.uk> wrote:
It sounds like you just need a simple boost converter...


I'm not necessarily trying to run off a single battery.
Do you think I should be?

You can get clever and use a different boost converter for each LED (LED
brightness is a function of the current going through them, not the voltage
across them


Isn't it basically the same thing...?

Ouch! you need to start at the very beginning i.e. Ohms law. V=I*R or
I=V/R or R=I/I. (and P = V*I)

"V" is analogous to pressure and too much pressure blows holes in
things. "I" is probably the same as you concepts of a river current. "R"
is a measure of how much a substance resists current flow - think of
water pressure at one end of a clean pipe - water goes though quickly
because there is no resistance in the pipe and lots of current flows,
next stuff the pipe full of tightly packed sand - this has high
resistance - little current flows. If you increase the water pressure
then more current flows and with enough pressure the whole thing blows
to bits and lets out the pixi water.

Something like your case:

you want 20 milliamp (0.02 of an amp) supplied from a 4.5 volt battery
to flow through an LED. The simple application of R=V/I gives the wrong
result

4.5v /.02A = 225 ohms (nothing wrong with the formula - just wrong
application)

because there is no account for the necessary voltage drop accross the
LED. Assume in this case it is 2.2 volts if you used a 220 ohm resistor
you would find the current is too low at something around 10 ma. This
2.2 volt drop is somewhat constant regardless of current thought the LED
(in water terms it is a weir across the river). If you had to use the
exact voltage drop of the LED you would lash up the above values and
measure the voltage across the LED.

In this case:

R = (4.5v - 2.2v) / 0.02A
= 120 ohm.

To check for power requirements of the resistor (so the pixi smoke stays
inside the resistor)

P = (4.5v - 2.2) * .02A
= .046 watts - any small sized resistor will be fine.

As you have noted the battery voltage drops over time. If you want just
a simple resistor solution you cope for it this way. Find the maximum
current the LED can RELIABLY withstand this is probably 20 milliamp and
calculate R using the new battery voltage.

Hope some of that was useful







Maybe not if I should

be googling for "current limiting circuit" instead of "voltage
limiting circuit".

OTOH, I might give the Zener diodes another try (because
it's easy/small/cheap) with an eye to putting one diode
per LED (or maybe one for every two LEDs if that works).

I'm just waiting for the shops to open because my ham-fisted
experiments made the magic pixie smoke come out of the
one I had...)

 

[Jon Kirwan]

On Mon, 6 Jul 2009 03:15:40 -0700 (PDT), fungus
<openglMYSOCKS@artlum.com> wrote:

On Jul 5, 9:08 pm, n...@jecarter.us wrote:
snip

3. Google for joule thief

I found a ferrite bead in an old PC and lashed on up. It works
but isn't very bright (maybe only a quarter of 20mA brightness).

Was the battery voltage you used __below__ the on-voltage for the LED?
(In the joule thief thing, that's the way it is supposed to be.)

Jon

 

[David Eather]

fungus wrote:

On Jul 6, 9:42 pm, David Eather <eat...@tpg.com.au> wrote:
fungus wrote:
Isn't it basically the same thing...?
Ouch! you need to start at the very beginning i.e. Ohms law. V=I*R or
I=V/R or R=I/I. (and P = V*I)


Yes, I heard about that one...

I was under the impression that if you fixed one of
the values in part of a circuit (volts or amps) the other
value would sort of figure out what it was supposed to
be (assuming the power supply can provide it)..

So if I figure out a way to fix the current at 20mA
the voltage will sort itself out and I don't need to
worry about it.

No (no, no, no, no). If you fix two the other takes care of itself.

...
To check for power requirements of the resistor (so the pixi smoke stays
inside the resistor)

P = (4.5v - 2.2) * .02A
= .046 watts - any small sized resistor will be fine.


The specs on the LEDs aren't very exact so the way I
did it was to put the LED in series with a variable resistor
(a ten turn wire-wound pot - quite fine tuning), set up my
multimeter to measure current, dial "20mA" using
the pot ... then take it apart and measure it.

Yep, sounds good.

Not very scientific but I didn't see any other way to get
exactly 20mA using the loose specs of the LED.

As you have noted the battery voltage drops over time. If you want just
a simple resistor solution you cope for it this way. Find the maximum
current the LED can RELIABLY withstand this is probably 20 milliamp and
calculate R using the new battery voltage.


The batteries give between 3.6 and 3.9 volts for most of their
lifetime so
I was designing around an average value of 3.75V.

Not much good if the LED blows while the battery is still at 3.9 volts

See this chart, which seems quite accurate:
http://www.powerstream.com/z/AA-100mA.png

The problem is that brand new batteries give out 4.7V for a while and
this makes the LEDs get warm (not hot, just warm to the touch).
I discharged two sets of batteries through it and got peak currents
up to 45mA but nothing bad happened.

My idea was to use a Zener diode to take away the extra current

Zener diodes don't do that. You could make it work but the cost would be
horrible power consumption and bad efficiency.

at the beginning but I didn't figure out all the resistors before
the pixie smoke escaped. I'll get another one tonight and try
again...

Hope some of that was useful

Yes, of course. I'm a complete newbie at this...all I know
is what I've read on the web over the last few days.

 

[David Eather]

fungus wrote:

On Jul 6, 9:13 pm, David Eather <eat...@tpg.com.au> wrote:
Alternatively, connect a JFET (plus a low value resistor)as a constant
current source. Most JFET's will not work perfectly with so few volts to
play with, but it will be better than nothing.


Like this?

http://en.wikipedia.org/wiki/Current_source#JFET_and_N-FET_current_source

A lot like that. If you put a resistor between the gate and source you
have some control over what the current limit is.

For something sophisticated you could use almost any of the charge-pump
IC's to double the voltage and give you enough volts for a proper
constant current source or as a simple voltage doubler for when the
battery / led combo is just too low.

g

A few days ago I was thinking like this:

"It's just a battery and an LED, how hard can it be...?"

It's not hard. If you just want to hook up an LED then just a resistor
works fine.If you need constant brightness, maximum life, maximum
reliability or maximum efficiency then it gets more complicated.

 

[David Eather]

fungus wrote:

On Jul 6, 9:13 pm, David Eather <eat...@tpg.com.au> wrote:
Alternatively, connect a JFET (plus a low value resistor)as a constant
current source. Most JFET's will not work perfectly with so few volts to
play with, but it will be better than nothing.


Like this?

http://en.wikipedia.org/wiki/Current_source#JFET_and_N-FET_current_source

The circuit diagram just below that heading does not relate to the
heading at all.

For something sophisticated you could use almost any of the charge-pump
IC's to double the voltage and give you enough volts for a proper
constant current source or as a simple voltage doubler for when the
battery / led combo is just too low.

g

A few days ago I was thinking like this:

"It's just a battery and an LED, how hard can it be...?"

 

[fungus]

On Jul 6, 9:56 pm, Jon Kirwan <j...@infinitefactors.org> wrote:

Was the battery voltage you used __below__ the on-voltage for the LED?
(In the joule thief thing, that's the way it is supposed to be.)

Yes, I used a single battery...

I also tried two batteries to see if it would get brighter (it didn't)

 

[Peter Bennett]

On Mon, 6 Jul 2009 14:34:56 -0700 (PDT), fungus
<openglMYSOCKS@artlum.com> wrote:

On Jul 6, 9:42 pm, David Eather <eat...@tpg.com.au> wrote:
fungus wrote:
Isn't it basically the same thing...?

Ouch! you need to start at the very beginning i.e. Ohms law. V=I*R or
I=V/R or R=I/I. (and P = V*I)


Yes, I heard about that one...

I was under the impression that if you fixed one of
the values in part of a circuit (volts or amps) the other
value would sort of figure out what it was supposed to
be (assuming the power supply can provide it)..

This is true if all the components follow Ohm's Law. Unfortunately,
LEDs (and most other semiconductors) don't follow Ohm's Law.

If you apply a low voltage (below 1.9 volts or so) to a green LED,
very little current will flow. As you increase the voltage slowly, at
some point (2.1 volts or so for a green LED) the current will start to
increase rapidly, and the LED will light. If you attempt to increase
the voltage over ~2.5 volts, the current will increase to a point
where the magic smoke comes out of the LED.

So if I figure out a way to fix the current at 20mA
the voltage will sort itself out and I don't need to
worry about it.

If you carefully read the specs for your LED, I expect that the 20 mA
rating is the maximum recommended current. The LED will work fine at
much lower currents, but won't be quite as bright as at 20 mA. I
normally aim for about 10 mA current in common LEDs - they are bright
enough for most applications at that current, and the current is low
enough that I don't have to bother checking the specs on the
particular part I'm using to see if it is OK (and it makes the math
easier!) I also assume that red, yellow and green LEDs all drop about
2 volts, although I know that red and yellow are lower voltage than
green (but blue and white want 3.3 to 3.6 volts, I think.)

The specs on the LEDs aren't very exact so the way I
did it was to put the LED in series with a variable resistor
(a ten turn wire-wound pot - quite fine tuning), set up my
multimeter to measure current, dial "20mA" using
the pot ... then take it apart and measure it.

Not very scientific but I didn't see any other way to get
exactly 20mA using the loose specs of the LED.

As I said above, a LED isn't really very fussy about the actual
current. You're being much too scientific about it.

As you have noted the battery voltage drops over time. If you want just
a simple resistor solution you cope for it this way. Find the maximum
current the LED can RELIABLY withstand this is probably 20 milliamp and
calculate R using the new battery voltage.


The batteries give between 3.6 and 3.9 volts for most of their
lifetime so
I was designing around an average value of 3.75V.

If the maximum current rating of the LED is 20 mA, you should design
for 20 mA or less at the maximum supply voltage.

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[fungus]

On Jul 6, 9:42 pm, David Eather <eat...@tpg.com.au> wrote:

fungus wrote:
Isn't it basically the same thing...?

Ouch! you need to start at the very beginning i.e. Ohms law. V=I*R or
I=V/R or R=I/I. (and P = V*I)

Yes, I heard about that one...

I was under the impression that if you fixed one of
the values in part of a circuit (volts or amps) the other
value would sort of figure out what it was supposed to
be (assuming the power supply can provide it)..

So if I figure out a way to fix the current at 20mA
the voltage will sort itself out and I don't need to
worry about it.

...
To check for power requirements of the resistor (so the pixi smoke stays
inside the resistor)

P = (4.5v - 2.2) * .02A
   = .046 watts - any small sized resistor will be fine.

The specs on the LEDs aren't very exact so the way I
did it was to put the LED in series with a variable resistor
(a ten turn wire-wound pot - quite fine tuning), set up my
multimeter to measure current, dial "20mA" using
the pot ... then take it apart and measure it.

Not very scientific but I didn't see any other way to get
exactly 20mA using the loose specs of the LED.

As you have noted the battery voltage drops over time. If you want just
a simple resistor solution you cope for it this way. Find the maximum
current the LED can RELIABLY withstand this is probably 20 milliamp and
calculate R using the new battery voltage.

The batteries give between 3.6 and 3.9 volts for most of their
lifetime so
I was designing around an average value of 3.75V.

See this chart, which seems quite accurate:
http://www.powerstream.com/z/AA-100mA.png

The problem is that brand new batteries give out 4.7V for a while and
this makes the LEDs get warm (not hot, just warm to the touch).
I discharged two sets of batteries through it and got peak currents
up to 45mA but nothing bad happened.

My idea was to use a Zener diode to take away the extra current
at the beginning but I didn't figure out all the resistors before
the pixie smoke escaped. I'll get another one tonight and try
again...

Hope some of that was useful

Yes, of course. I'm a complete newbie at this...all I know
is what I've read on the web over the last few days.

 

[fungus]

On Jul 6, 9:13 pm, David Eather <eat...@tpg.com.au> wrote:

Alternatively, connect a JFET (plus a low value resistor)as a constant
current source. Most JFET's will not work perfectly with so few volts to
play with, but it will be better than nothing.

Like this?

http://en.wikipedia.org/wiki/Current_source#JFET_and_N-FET_current_source

For something sophisticated you could use almost any of the charge-pump
IC's to double the voltage and give you enough volts for a proper
constant current source or as a simple voltage doubler for when the
battery / led combo is just too low.

<g>

A few days ago I was thinking like this:

"It's just a battery and an LED, how hard can it be...?"

 

[David Eather]

Peter Bennett wrote:

On Mon, 6 Jul 2009 14:34:56 -0700 (PDT), fungus
openglMYSOCKS@artlum.com> wrote:

On Jul 6, 9:42 pm, David Eather <eat...@tpg.com.au> wrote:
fungus wrote:
Isn't it basically the same thing...?
Ouch! you need to start at the very beginning i.e. Ohms law. V=I*R or
I=V/R or R=I/I. (and P = V*I)

Yes, I heard about that one...

I was under the impression that if you fixed one of
the values in part of a circuit (volts or amps) the other
value would sort of figure out what it was supposed to
be (assuming the power supply can provide it)..

This is true if all the components follow Ohm's Law.

No, this is not true. Ohms law has three variables, two of which must be
defined to calculate the third.

(remainder snipped)

 

[fungus]

On Jul 6, 11:53 pm, David Eather <eat...@tpg.com.au> wrote:

The batteries give between 3.6 and 3.9 volts for most of their
lifetime so
I was designing around an average value of 3.75V.

Not much good if the LED blows while the battery is still at 3.9 volts

3.9 is only 4% more than 3.75...

My idea was to use a Zener diode to suck away the extra current

Zener diodes don't do that. You could make it work but the cost would be
horrible power consumption and bad efficiency.

....only until the battery voltage drops down a bit (which should
happen in the first 10% of the battery life).

 

[Jon Kirwan]

On Mon, 6 Jul 2009 14:06:28 -0700 (PDT), fungus
<openglMYSOCKS@artlum.com> wrote:

On Jul 6, 9:56 pm, Jon Kirwan <j...@infinitefactors.org> wrote:

Was the battery voltage you used __below__ the on-voltage for the LED?
(In the joule thief thing, that's the way it is supposed to be.)


Yes, I used a single battery...

I also tried two batteries to see if it would get brighter (it didn't)

The brightness should also be affected by the base resistor value,
too. Did you try changing it to a lower value -- say about half? I
think that should lower the frequency, leave the duty cycle about the
same, and increase the peak current (not quite by double) -- with the
net effect of raising the brightness a little -- assuming the
transistor and LED can handle the peak. If that works as expected,
you may consider trying even lower values.

Jon

 

[ehsjr]

David Eather wrote:

fungus wrote:

On Jul 6, 9:42 pm, David Eather <eat...@tpg.com.au> wrote:

fungus wrote:

Isn't it basically the same thing...?

Ouch! you need to start at the very beginning i.e. Ohms law. V=I*R or
I=V/R or R=I/I. (and P = V*I)


Yes, I heard about that one...

I was under the impression that if you fixed one of
the values in part of a circuit (volts or amps) the other
value would sort of figure out what it was supposed to
be (assuming the power supply can provide it)..

So if I figure out a way to fix the current at 20mA
the voltage will sort itself out and I don't need to
worry about it.


No (no, no, no, no). If you fix two the other takes care of itself.

Yes (yes, yes, yes, yes). He's talking LEDs. If he controls the
current at 20 mA, the LED will be fine. The LED holds the voltage
to Vf.

Ed

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