Driving LEDs with a battery pack

[David Eather]

ehsjr wrote:

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

snip

Show me how he can set the current to 20 ma by controlling only one
variable.

If you can then ohms law is wrong and Kirchoff's law is also wrong and a
Nobel prize is yours for the asking.

Lets get an arbitrary voltage and strap it to the LED and see what
happens! Lets go the whole hog and tell him to plug the LED straight
into the wall socket.

Vexcesssupply = Vbattery - Vf.

The only way he can control the current through the LED is to control
Vexcesssupply AND the effective series resistance of the circuit. He
controls the Vexcesssupply by selecting a specific battery. A passive
resistor works just like any other fixed resistor, a light globe works
as a variable resistor and a JFET constant current source is an active
variant of a variable resistor.

Show me how he can set the current to 20 ma by controlling only one
variable.

 

[fungus]

On Jul 7, 2:00 am, David Eather <eat...@tpg.com.au> wrote:

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.

Yes, but in this context there's only two variables under
my control. I can't change the properties of the LED.

 

[fungus]

On Jul 7, 1:29 am, Peter Bennett <pete...@somewhere.invalid> wrote:

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

As I said, I got them off eBay so there's no manufacturer's datasheet.

The advert said something like this (eg. for green):
Color: Green
Voltage(V): 3.1-3.3
Current(mA): 20
MCD: 13000

I assumed 20mA was 'normal' current, not absolute maximum.

 

[Peter Bennett]

On Tue, 7 Jul 2009 02:53:53 -0700 (PDT), fungus
<openglMYSOCKS@artlum.com> wrote:

On Jul 7, 1:29 am, Peter Bennett <pete...@somewhere.invalid> wrote:

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


As I said, I got them off eBay so there's no manufacturer's datasheet.

The advert said something like this (eg. for green):
Color: Green
Voltage(V): 3.1-3.3
Current(mA): 20
MCD: 13000

I assumed 20mA was 'normal' current, not absolute maximum.

I found a LED datasheet at Digkey (
http://media.digikey.com/pdf/Data%20Sheets/Lite-On%20PDFs/LTL-4232N.pdf
)

The second page of that data sheet has a table headed "Absolute
Maximum Ratings" which indicates that the maximum continuous current
for that LED is 30 mA.

There is no such thing as a "normal" current for any LED. The
manufacturer specifies a maximum, and the user will normally design
the circuit to provide somewhat less than maximum, just to be safe.

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

 

[Sjouke Burry]

fungus 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?

http://cds.linear.com/docs/Datasheet/3598fa.pdf

 

[David Eather]

George Herold wrote:

On Jul 7, 1:20 am, David Eather <eat...@tpg.com.au> wrote:
ehsjr wrote:
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
snip
Show me how he can set the current to 20 ma by controlling only one
variable.

If you can then ohms law is wrong and Kirchoff's law is also wrong and a
Nobel prize is yours for the asking.

Lets get an arbitrary voltage and strap it to the LED and see what
happens! Lets go the whole hog and tell him to plug the LED straight
into the wall socket.

Vexcesssupply = Vbattery - Vf.

The only way he can control the current through the LED is to control
Vexcesssupply AND the effective series resistance of the circuit. He
controls the Vexcesssupply by selecting a specific battery. A passive
resistor works just like any other fixed resistor, a light globe works
as a variable resistor and a JFET constant current source is an active
variant of a variable resistor.

Show me how he can set the current to 20 ma by controlling only one
variable.- Hide quoted text -

- Show quoted text -

Hmm what about a constant current source. You can make one from a
JFET and resistor. They sell these as two terminal devices. A
1N5314 will give you 4.7mA. put four in parallel and that's close to
20mA. (They are not giving these away however. mouser's price each
is $4.70).

George H.

If you read my posts in this thread you will see I suggested using a
JFET in a similar way to solve the OT's, problem. Applying the
description of "constant current source" to this device is however
common, also incorrect. If I hook the JFET + resistor or the more sexy 2
terminal integrated device to the LED it will not provide a constant
current. No, it just sits there and does nothing. Go ahead, get one of
these devices and connect it to an LED - but don't use a voltage source
anywhere - the constant current source description is correct only when
applied to a voltage source and a variable resistor as the JFET is.

Current through the JFET cause a voltage drop (the source- drain channel
is a semiconductor and it has resistance) this voltage drop is applied
to the gate junction which cause a depletion region to partly extend
across the source-drain channel and partly pinch it off the usable
region of the channel. This increases the resistance of the source gain
channel and by using ohms law you can see that this increased resistance
has the effect of reducing current.

Lots of smart guys might be able to make assumptions and leave bits of
information out, but to a newbie that can be disastrous.

 

[David Eather]

fungus wrote:

On Jul 7, 2:00 am, David Eather <eat...@tpg.com.au> wrote:
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.


Yes, but in this context there's only two variables under
my control. I can't change the properties of the LED.

You decide how much current you need. In this case it is *your choice*
of LED that makes you say "I need 20 milliamps of current"

You use a resistor to limit the current to 20 milliamp or less. What
value do you think it needs to be? The answer is meaningless unless you
also specify a voltage and *you have chosen* 3.75 volts or some such figure.

So you have chosen the current and the voltage, which is exactly two of
the three variables in ohms law and the necessary condition for a
meaningful calculation of the unknown resistance.

You are overly fixated on the voltage drop of the LED. This is solved as
I have shown you before by the simple fineness of subtracting that value
from the battery voltage before. You have 4 volts, the LED will "stop"
or "drop" or "consume" 2 of them and therefore you need a resistor to
control the current caused by the remaining unaccounted for 2 volt
potential.

The people who have "abbreviated" this have done so because "every body
knows" this "general knowledge" - except you don't because you are brand
new. They have done you no favours.

 

[David Eather]

Peter Bennett wrote:

On Tue, 7 Jul 2009 02:53:53 -0700 (PDT), fungus
openglMYSOCKS@artlum.com> wrote:

On Jul 7, 1:29 am, Peter Bennett <pete...@somewhere.invalid> wrote:
If the maximum current rating of the LED is 20 mA, you should design
for 20 mA or less at the maximum supply voltage.

As I said, I got them off eBay so there's no manufacturer's datasheet.

The advert said something like this (eg. for green):
Color: Green
Voltage(V): 3.1-3.3
Current(mA): 20
MCD: 13000

I assumed 20mA was 'normal' current, not absolute maximum.

I found a LED datasheet at Digkey (
http://media.digikey.com/pdf/Data%20Sheets/Lite-On%20PDFs/LTL-4232N.pdf
)

The second page of that data sheet has a table headed "Absolute
Maximum Ratings" which indicates that the maximum continuous current
for that LED is 30 mA.

There is no such thing as a "normal" current for any LED.

Bullshit.

The

manufacturer specifies a maximum, and the user will normally design
the circuit to provide somewhat less than maximum, just to be safe.

nominal = normal and almost always specified for an LED, as is a

maximum current which is an intermittent pulsed current.

 

[George Herold]

On Jul 7, 1:20 am, David Eather <eat...@tpg.com.au> wrote:

ehsjr wrote:
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

snip

Show me how he can set the current to 20 ma by controlling only one
variable.

If you can then ohms law is wrong and Kirchoff's law is also wrong and a
Nobel prize is yours for the asking.

Lets get an arbitrary voltage and strap it to the LED and see what
happens! Lets go the whole hog and tell him to plug the LED straight
into the wall socket.

Vexcesssupply = Vbattery - Vf.

The only way he can control the current through the LED is to control
Vexcesssupply AND the effective series resistance of the circuit. He
controls the Vexcesssupply by selecting a specific battery. A passive
resistor works just like any other fixed resistor, a light globe works
as a variable resistor and a JFET constant current source is an active
variant of a variable resistor.

Show me how he can set the current to 20 ma by controlling only one
variable.- Hide quoted text -

- Show quoted text -

Hmm what about a constant current source. You can make one from a
JFET and resistor. They sell these as two terminal devices. A
1N5314 will give you 4.7mA. put four in parallel and that's close to
20mA. (They are not giving these away however. mouser's price each
is $4.70).

George H.

 

[Jon Kirwan]

On Tue, 7 Jul 2009 21:18:03 +0000 (UTC), don@manx.misty.com (Don
Klipstein) wrote:

1 microsecond

Did you mean 1us? Two sheets I just saw today listed 100us pulse
width, 10% duty.... So a total period of 1ms.

Jon

 

[Jon Kirwan]

On Tue, 7 Jul 2009 14:57:42 -0700 (PDT), fungus
<openglMYSOCKS@artlum.com> wrote:

On Jul 7, 6:01 am, Jon Kirwan <j...@infinitefactors.org> wrote:
(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'll put it together again with my variable resistor see what happens.

Make sure you wire up the transformer, correctly, too. The base side
must _aid_ battery voltage when the transistor is on, not oppose it.
If it isn't working at all (or well), try reversing the leads there.

Can a joule thief power multiple LEDs?

Yes, but don't wire them in parallel. You'd wire them in series.
There is a limit, but with no more than 6 (I think you wrote that) the
required voltage ... at worst ... is 6*3.3V or 19.8V, which I think is
doable. I could play with one here and see, to be sure.

The peak current, by the way, is pretty much set elsewhere in the
circuit. What having the LEDs stacked up like that does, is to
dramatically reduce the duration of time they are pulsed. The whole
things works by winding up to a peak current in the collector winding
-- that is set by the circuit itself and the transistor, not by the
LED. However, once the transistor reaches that peak current in the
collector winding, it turns off. And when it does, that current goes
into the LED (or chain of LEDs.) Assuming that the voltage is low
enough that it doesn't "break" the transistor (20V shouldn't, in most
cases), the voltage will self-adjust to whatever is needed to drive
the LED(s).

However, with higher voltages, the transformer will dump its energy
faster. (The time is inversely proportional to the required voltage.)
So the pulse is narrowed by stacking the LEDs. This shouldn't change
the transistor ON time by much of anything, but it changes its OFF
time, and thus, reduces the effective duty cycle. Which may mean you
need a higher peak current and thereby a lower base resistor to get it
(or more windings in the base circuit part of the transformer.)

How important are the number/neatness/type of wire of the windings
on the transformer? (or does it make little difference at 50kHz?)

Within reason, I think you are fine. You cannot use bare wire, for
obvious reasons (shorts itself all over the place.) If you use
insulated wire, your spacing will be pretty wide. Better is 'magnet
wire' which has just enough insulation and not much more than that.

A similar number of windings is probably a good idea for both
windings. I assume that's what you did. But it isn't critical. You
can wind more or less on either side and get by, just fine. Too many
extra windings (like a multiplying factor of 5 or more) on the winding
that attaches to the base and you might get into areas where the
transistor base is so reverse biased when the LED gets pulsed that it
zeners and injures the transistor. But as you can see, there is a lot
of room for error. The main thing is to wire it up so that the base
winding _aids_ the battery voltage. If you get it wrong, just wire
the other way and see how it goes.

I think 50kHz should be fine for most BJTs. I don't recall the
frequency I measured before, but it seems it was in the 10's of kHz,
so your question is in the right ballpark. There are also some very
fast rise/fall times going on, and probably some minor other effects
due to those. But nothing I've needed to get concerned about, that I
recall. I think you'll be fine.

Jon

 

[Don Klipstein]

In <fJydnZ0QQs13Ms7XnZ2dnUVZ_jdi4p2d@supernews.com>, David Eather wrote:

Peter Bennett wrote:
On Tue, 7 Jul 2009 02:53:53 -0700 (PDT), fungus
openglMYSOCKS@artlum.com> wrote:

On Jul 7, 1:29 am, Peter Bennett <pete...@somewhere.invalid> wrote:
If the maximum current rating of the LED is 20 mA, you should design
for 20 mA or less at the maximum supply voltage.

As I said, I got them off eBay so there's no manufacturer's datasheet.

The advert said something like this (eg. for green):
Color: Green
Voltage(V): 3.1-3.3
Current(mA): 20
MCD: 13000

I assumed 20mA was 'normal' current, not absolute maximum.

I found a LED datasheet at Digkey (
http://media.digikey.com/pdf/Data%20Sheets/Lite-On%20PDFs/LTL-4232N.pdf
)

The second page of that data sheet has a table headed "Absolute
Maximum Ratings" which indicates that the maximum continuous current
for that LED is 30 mA.

There is no such thing as a "normal" current for any LED.

Bullshit.

The
manufacturer specifies a maximum, and the user will normally design
the circuit to provide somewhat less than maximum, just to be safe.

nominal = normal and almost always specified for an LED, as is a
maximum current which is an intermittent pulsed current.

Actually, most LED datasheets mention 3 currents. One is
characterization current - the current at which voltage drop and
photometric specifications apply. Another is maximum continuous current,
and a third is maximum peak current.

Most 3 and 5 mm through-hole visible LEDs have characterization current
of 20 mA, maximum continuous current of 30 mA, and maximum peak current of
100 mA (maximum pulse duration 1 microsecond, maximum duty cycle 10%).

- Don Klipstein (don@misty.com)

 

[fungus]

On Jul 7, 6:01 am, Jon Kirwan <j...@infinitefactors.org> wrote:

(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'll put it together again with my variable resistor see what happens.

Can a joule thief power multiple LEDs?

How important are the number/neatness/type of wire of the windings
on the transformer? (or does it make little difference at 50kHz?)

 

[Jon Kirwan]

On Tue, 7 Jul 2009 23:30:11 +0000 (UTC), don@manx.misty.com (Don
Klipstein) wrote:

In article <icj755dl1ko20kue3fpgfuni3v2kavukul@4ax.com>, Jon Kirwan wrote:
On Tue, 7 Jul 2009 21:18:03 +0000 (UTC), don@manx.misty.com (Don
Klipstein) wrote:

1 microsecond

Did you mean 1us? Two sheets I just saw today listed 100us pulse
width, 10% duty.... So a total period of 1ms.

OK, checking to see if I brain farted here...

Nichia NSPW500DS: Pulse width <= 10 ms, duty <= 1/10

Panasonic LN28RPX: Duty 10%, duration 1 msec.

Kingbright WP7113SRD/F: 1/10 duty cycle, .1 ms pulse width.

Lumex SSL-LX5093PGD: <10 uS (for 150 mA peak)

Cree C503B-GAS-CB0F0791: Pulse width <=.1 msec, duty <= 1/10

Looks to me that the maximum pulse width for specified peak current both
is usually longer than the 1 microsecond I was thinking, and varies more
from one manufacturer to another than I thought.

- Don Klipstein (don@misty.com)

Other things being equal, a longer pulse width would seem to be a
harder spec to hit, wouldn't it? (If I can do an hour's run at an
average of 10 MPH, with periodic spurts hitting 14.5 MPH and not going
below 9.5 MPH [with a duty cycle at 14.5 MPH being 10%], it would be
more impressive if I maintained the 14.5 MPH for 6 minutes straight
and 9.5 MPH for the rest of the time than it would if I had to break
it up into 120 separate 3-second bursts.)

Jon

 

[Don Klipstein]

In article <icj755dl1ko20kue3fpgfuni3v2kavukul@4ax.com>, Jon Kirwan wrote:

On Tue, 7 Jul 2009 21:18:03 +0000 (UTC), don@manx.misty.com (Don
Klipstein) wrote:

1 microsecond

Did you mean 1us? Two sheets I just saw today listed 100us pulse
width, 10% duty.... So a total period of 1ms.

OK, checking to see if I brain farted here...

Nichia NSPW500DS: Pulse width <= 10 ms, duty <= 1/10

Panasonic LN28RPX: Duty 10%, duration 1 msec.

Kingbright WP7113SRD/F: 1/10 duty cycle, .1 ms pulse width.

Lumex SSL-LX5093PGD: <10 uS (for 150 mA peak)

Cree C503B-GAS-CB0F0791: Pulse width <=.1 msec, duty <= 1/10

Looks to me that the maximum pulse width for specified peak current both
is usually longer than the 1 microsecond I was thinking, and varies more
from one manufacturer to another than I thought.

- Don Klipstein (don@misty.com)

 

[Jon Kirwan]

On Sun, 5 Jul 2009 11:22:29 -0700 (PDT), fungus wrote:

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)

Let me summarize in my own words. "Your supply is a battery pack,
whose voltage varies from 4.6V down to 3.3V over the 'useful' life
you've defined for it. You want to use various mixtures of LEDs as
'modules' with it and have that work reasonably well. Your question
is about making that work well over the range of your voltage source's
useful life. Efficiency would be a valuable option, but not a
necessary one."

The basic 'model' of an LED, the one talked about so far, is just a
'voltage.' When someone writes, 3.3V, that implies a lot of other
things, too. But you can often get away with just assuming that the
LED will somehow automatically find itself using 3.3V if you get the
right current level going.

A slightly better model of an LED is to use: V(I) = V_start + I *
R_diode. Sometimes R_diode is small. Sometimes it is big. For the
older red LEDs (I'm recollecting something 20 years ago here), V_start
would be about 1.55V and R_diode would be about 21 ohms. Assume 20mA
and that works out to 1.97V. Call it 2V. Which is a number folks
sometimes use for red LED estimates, when 'on.' But the main thing
here is that the model now at least provides a varying voltage
depending on the current and provides a floor below which you can
expect to see zero current (close to it.)

It's by no means a perfect model, but it's decent when the equation is
set up nearby where you expect to use the LED. For example, you might
use a 5V power supply and a 3.3V green LED and pick out two resistors
-- one designed to provide about 30mA and one designed to provide
about 5mA. You don't really know the voltage of the LED, except that
it is supposed to be about 3.3V at 20mA, so use that voltage as a
broad guess (0th order model) and calculate resistor values of
(5-3.3)/30mA or ~56 ohms and (5-3.3)/5mA or ~330 ohms. Precise isn't
important -- just convenient values for the resistors that are about
in the right place. Then you run them and measure the voltage across
the resistor (or LED.) Let's say the voltage across the 56 ohm
resistor reads 1.61V and the voltage across the 330 ohm resistor reads
1.9V. Then you can first estimate the slope as ((5-1.61)-(5-1.9)) /
(1.61/56-1.9/330) or about R_diode=12.6 ohms. Knowing that, you can
now realize that (5-1.9)=V_start+12.6*(1.9/330), or V_start=3.03V. So
your model might be simply V(I)=3+12.6*I. You know you can't go much
below 3 volts, then.

Of course, do this modeling for a few of them and you'll find
different numbers. Let's see how this simple modeling idea applies in
the case you mentioned where you appear to want to drive perhaps 6 at
a time... in parallel. Let's set up two different models for two
supposed green LEDs that are similar, but not exactly the same.

D1: VD1(ID1) = 3.06 + 12.8 * ID1
D2: VD1(ID2) = 2.94 + 12.4 * ID2

Each factor is <= 2% up or down from the model worked out as an
example in an earlier paragraph. Now let's assume we wire them in
parallel and supply an exact voltage of 3.2V to them:

D1: ID1(VD1) = (VD1 - 3.06) / 12.8
= (3.2 - 3.06) / 12.8 or 11mA
D2: ID2(VD2) = (VD2 - 2.94) / 12.4
= (3.2 - 2.94) / 12.4 or 21mA

Note that we've assumed a batch of LEDs with all their parameters
close by, only 2% variation or less. But wiring them up like this
illustrates almost a factor of two in the current they draw when wired
in parallel. Definitely, there will be a noticeable difference in how
they 'look' when you stare at them. Not a whole lot, but enough to
see it.

Now let's assume you wanted 3.2V across those two LEDs but were using
your battery pack, which varies from 3.3V up to 4.6V. Use 3.7V as
your average, okay? You need to then estimate a drop of 3.7-3.2 or
0.5V. But to compute the resistor, you also need to know what current
you need. Since there are two and each takes 20mA, for two you really
want a total of 40mA. So your guessed-at resistor is 0.5V/40mA or
12.5 ohms. Make it 12, because it is available and will pass slightly
more current.

So let's work out the figures using the D1 and D2 models above, for
the situation at both ends of the voltage spectrum... 3.3V and 4.6V.

I cheated and used a matrix calculator (no, I didn't use LTSpice.) At
3.3V, I get ID1=3.3mA and ID2=13mA with a voltage across the two LEDs
of 3.1V. At 4.6V, I get ID1=38mA and ID2=49mA with a voltage across
the two LEDs of 3.55V. Quite a variation, not only on the supply
range but also across the LEDs. Especially when the supply voltage
droops to the lower end. Then there is quite a difference!

This is one reason to wire up LEDs in series. The current in each
will be the same that way. The problem then becomes how to get the
needed voltage?!

Which is one reason the Joule Thief can look good. You can stack LEDs
with it. A downside is that with the higher stacked voltages comes a
lower duty cycle and thereby also higher peak currents which may
stress the LEDs when trying to achieve a desired brightness.

Someone here recently suggested this page for LED drivers.

http://www.onsemi.com/PowerSolutions/parametrics.do?id=197

It may have some options you might want to look at. There are too
many there for me to want to sort through them. But look for ones
accepting a Vin that covers your range of 3.3-4.6V.

Jon

 

[ehsjr]

David Eather wrote:

ehsjr wrote:

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

snip


Show me how he can set the current to 20 ma by controlling only one
variable.

Vin ---LM317---R---+
| |
+---------+
|
LED
|
Gnd ---------------+

Set R to 62.5 ohms. Vin can be anything from Vf + the
headroom the LM317 needs - roughly 2.5 volts IIRC - to Vmax
for the chip, around 37 volts IIRC. No need to regulate Vin,
as long as it is between the min & max values.

If you can then ohms law is wrong and Kirchoff's law is also wrong and a
Nobel prize is yours for the asking.

You need to re-think that.

Lets get an arbitrary voltage and strap it to the LED and see what
happens!

It will work fine if fed from a constant current source,
which was what he stipulated.

Lets go the whole hog and tell him to plug the LED straight
into the wall socket.

That's just bullshit, having nothing whatever to do with
feeding the LED from a constant current source.

Vexcesssupply = Vbattery - Vf.

The only way he can control the current through the LED is to control
Vexcesssupply AND the effective series resistance of the circuit.

That was disproved, above. Any supply voltage feeding the current
regulating circuit input will be fine, provided the supply voltage
falls within the specified input range of the constant current
circuit.

Try it! Hook a variable DC supply up to provide Vin to the circuit
drawn above. Measure the current drawn by the LED as you adjust the
supply voltage from say + 5 to max. Just don't exceed the limits
of the LM317. If you don't have a variable supply, just use
two 9V batteries in series for 18V Vin, and just one of them for
9V Vin. You will be able to verify for yourself that the current
through the LED is the same.

Ed

He
controls the Vexcesssupply by selecting a specific battery. A passive
resistor works just like any other fixed resistor, a light globe works
as a variable resistor and a JFET constant current source is an active
variant of a variable resistor.

Show me how he can set the current to 20 ma by controlling only one
variable.

 

[default]

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.

I have a couple of cheap LED flashlights that just wire the 14 Leds in
parallel and power them from 3AAA cells - apparently depending on the
internal resistance of the battery to keep the 14 Leds alive . . .

Bob Pease published a constant current circuit for leds that only
drops <100 millivolts in his Electronic Design column awhile ago.
http://www.worldtorch.com/LDO-fixed-current.php

the circuit uses a commonly available LM334 three terminal constant
current source to control a PNP pass transistor
--

 

[Jon Kirwan]

On Wed, 08 Jul 2009 08:13:40 -0400, default <default@defaulter.net>
wrote:

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.

I have a couple of cheap LED flashlights that just wire the 14 Leds in
parallel and power them from 3AAA cells - apparently depending on the
internal resistance of the battery to keep the 14 Leds alive . . .

Bob Pease published a constant current circuit for leds that only
drops <100 millivolts in his Electronic Design column awhile ago.
http://www.worldtorch.com/LDO-fixed-current.php

the circuit uses a commonly available LM334 three terminal constant
current source to control a PNP pass transistor

Good selection of a simple arrangement that's been discussed a little,
here. The LM334 is cheap enough. Better still is that there's an
available kit that isn't expensive to buy:

http://www.worldtorch.com/kit-buy.php

What I don't like about it is the idea of wiring up LEDs in parallel.
But that may be just fine for the OP's LEDs. Need to experiment with
the parts in hand, I suppose.

Jon

 

[Jon Kirwan]

On Wed, 08 Jul 2009 19:13:19 GMT, I wrote:

On Wed, 08 Jul 2009 08:13:40 -0400, default <default@defaulter.net
wrote:
snip
Bob Pease published a constant current circuit for leds that only
drops <100 millivolts in his Electronic Design column awhile ago.
http://www.worldtorch.com/LDO-fixed-current.php

the circuit uses a commonly available LM334 three terminal constant
current source to control a PNP pass transistor

Good selection of a simple arrangement that's been discussed a little,
here. The LM334 is cheap enough. Better still is that there's an
available kit that isn't expensive to buy:

http://www.worldtorch.com/kit-buy.php
snip

By the way, I just looked at the general schematic for the LM334 on
National's datasheet and with a quick sweep of my arms came up with a
design Iset/Ibias of 8, not 16 as they show on page 5. I'm off by a
factor of two.

My logic went like this. 1/2 of the I from V+ flows via Q6 to the R
rail. 1/4 via Q4 and 1/4 via Q5. Q5's 1/4*I flows via Q1 to the R
rail, too. So now up to 3/4*I into the R rail. Q4's 1/4*I passes
through two paths. The Ic(Q2)=Ic(Q1)/2... but Ic(Q1)=1/4*I, so that
is 1/8*I, leaving the other 1/8*I for Q3's Vbe conduction, which also
flows to the R rail. So the R rail gets 7/8*I and the V- picks up
1/8*I. Multiplying through by 8 to get rid of the divisor, I see a
factor of 8 for Iset/Ibias... not 16.

Can someone do a quick description about how to arrive at something
more like 16? I'm missing a clue (or two.)

Thanks,
Jon