LEDs

Discussion in 'Electronic Basics' started by [email protected], Jul 18, 2005.

1. Guest

I have a bunch of LEDs that I want to wire in series of 3, and then
wire each series to my battery in parrellel.

I have 3 LEDs that are rated for a FV drop of 3.0-3.4V, and draw 20mA.
If I wire them in series, and aim for 3.2V each, that would give me
9.6V. So powering them with an 8-pack of NiMH (1.2V) AA's would be
perfect.

Would I still need to use a resistor? I'm thinking yes because of the
current, but I don't know how... one of the formulas I found to
calculate for a resistor would yeild a zero, because zero volts would
need to be resisted:

Resistance = Voltage / Current

0 Ohms = 0 Volts / .020 Amps

So how do I limit the current, or will it be OK to not use resistors?

Thanks,
--Farrell F.

2. John FieldsGuest

---
You have to limit the current because there's no guarantee that the
LEDs will have a Vf of 3.2V at 20mA and if it's lower than that you
could fry the LEDs. If it's higher than that, the LEDs might light up
dimly. The easiest way around that problem is to wire two LEDs in
series, assume a Vf, for the pair,of 6.0V worst case, then bite the
bullet and use a series resistor to drop the remaining 3.6V at 20mA.

The resistance will need to be:

Vbat - Vled 9.6V - 6.0V
R = ------------- = -------------- = 180 ohms
Iled 0.02A

And the power the resistor will dissipate will be :

P = IE = 0.02A * 3.6V = 0.072W

So a standard 180 ohm +/- 5%, 1/4 watt carbon film resistor will be
fine.

3. Guest

I ended up going with 160 Ohm resistors... 3.2V per LED is what I want
to aim for... a little brighter, even if it will be a little more
risky.

i just want to know if its possible to just only current, and not
influence volatage? From what I read about Ohm's law, I'm guessing No.
I'm just curious, since I already order'd the resistors I'm going to
use them, but what if I needed to convserve the most amount of energy.
Is there any other device that could be used to control the flow of
power without using it up (ok, lets be real... using less of it up) ...
more efficient than dissapating it as mostly heat? Something that can
provide a fixed current and voltage?

Sorry if I sound really new, but I sort of am, and I'm slowly
understanding Ohm's law.

Thanks,
--Farrell F.

4. Jonathan WesthuesGuest

I think what you mean to ask is `Is there a device that will put a constant
current through an LED without wasting a lot of power?'. Because the voltage
dropped by the LED is very close to constant, and P = V*I, if you keep the
current through an LED constant then you keep the power dissipated by it
approximately constant as well.

The answer is yes. A resistor will make the current through the LED somewhat
constant, as long as the voltage drop across the resistor is large. Your LED
might be specified as dropping 1.8 V, but it will never be exactly 1.8 V; it
might be 1.7 V or 1.9 V, depending on temperature and process variations.
With the simple resistor circuit, you have to make the voltage across the
resistor large. Otherwise a very small change in the LED's voltage drop will
result in a very large change in the current. You can do some calculations
to convince yourself of that. As you noted, that means that a lot of energy
will be wasted as heat, because you need a lot of voltage `headroom' to
cover the drop across the resistor.

It is possible to build linear constant current sources that will maintain
an almost perfectly constant current through the LED, independent of the
voltage dropped by the LED or the supply voltage. It is not possible to do
this with only resistors; these circuits use transistors and integrated
circuits. They still dissipate the power due to the extra supply voltage as
heat, but they don't need as much headroom; you could run an LED that
dropped something between 1.4 and 1.8 V from a 2 V supply, and the
brightness (current) wouldn't be all over the place. Since you don't need as
much headroom, these circuits can be more efficient.

Still higher efficiency is possible using switchmode power supply
techniques. Depending the supply voltage that you start with and what kind
of LED you are running, you might be able to get efficiencies above ninety
percent. You can buy little prepackaged modules that do this, and there's
lots of special-purpose ICs as well. Nobody would bother with that for a
little front panel indicator, but for the large white LED flashlights enough
power would be wasted that people think it is worth the trouble.

Jonathan

5. Watson A.Name - \Watt Sun, the Dark Remover\Guest

Yes, there is, but that is beyond your understanding at this point (see
the following sentence for the reason why).
Since your LED is dropping 3.2V it is likely that it is a white or blue
LED. If so, I have some important info for you.

I have been running some white LEDs at 20mA and 30mA for a number of
months 24hrs, 7days/wk. I have found that there isn't much "risk" in
running LEDs at or above their maximum current. WHat happens is that
the light output drops dramatically after a thousand or more hours.
Some cheaper LEDs running at only 20mA drop to less than a third, more
like 1/10, their light output after less than a thousand hours, which is
only a month and a third, or 42 days.

If you notice, some white LED specs give an "initial" light output which
might be 10,000 mCd. But after a thousand hours, the output might be
closer to 6,000 or 7,000 mCd, so they use the term 'initial' just to be
safe.

6. John FieldsGuest

---
You misunderstand. Since the LED is a diode, you don't have direct
control of the forward voltage; it just comes out to be whatever it is
when you force a certain amount of current through the LED.

For the LEDs you're using, with 20 mA going through them, the voltage
dropped across any of them them could vary from 3.0V to 3.4V, so for
two LEDs on the low end in series with a 160 ohm resistor, the current
in the circuit would be:

Vs - Vled 9.6V - 6.0V
I = ----------- = ------------ ~ 22.5mA
R 160R

Which should be OK unless your LEDs are spec'ed at 20mA max, as some
white ones are. Running them at a small overcurrent like that will
only degrade their output over time, and shouldn't lead to a
catastrophic failure.
---
---
Good guess, but...

Diodes don't exhibit a fixed resistance in the forward direction, so
while Ohm's Law:

E = IR

still holds, R decreases as I Increases, so E will have to change very
little as it causes very large changes in current through the diode.
---
---
Forget about a fixed voltage. What you need is something which will
provide a fixed _current_ and let the voltage across the load vary as
it wants to.

If you want the most efficient way of doing it, then you need to use a
current-regulating switching supply.
---

7. Guest

Thanks everyone.

Quoting John:

-------

For the LEDs you're using, with 20 mA going through them, the voltage
dropped across any of them them could vary from 3.0V to 3.4V, so for
two LEDs on the low end in series with a 160 ohm resistor, the current
in the circuit would be:

Vs - Vled 9.6V - 6.0V
I = ----------- = ------------ ~ 22.5mA
R 160R

---------

but isn't that assuming the worst... a 3.0V drop from each LED? if it
were 3.2V that would become 20mA and if it were 3.4V it would become
17.5mA.

don't get me wrong, it's best to assume worst, but I just want to make
sure I am understand correctly. since this project of mine is just to
make some headlights, or rather stadium-type lights, for my RC car, I'm
not terribly concerned if they slightly diminish in brightness after
1000 or even 200 hours... as they wont be on for more than perhaps
30-40hrs a month at absolute most, more like 5-6hrs on average.

Thank again,
--Farrell F.

9. Jonathan KirwanGuest

I think the best answer for the OP has been done. But one thought
that crossed my mind, when I allowed it to wander on handling a wide
range of V+ and widely varying Vfwd of the LEDs (different colors,
even) is something like the following. It's very insensitive to
variations in the Vfwd of the LEDs and also relatively insensitive to
the V+ supply, which I kind of guessed might vary over a span of a
volt or more from reasonable use given that it is proposed to be a
stack of 8 NiCds:

V+
|
|
|
\
/ R3
\ 270 V+ V+ V+
/ | | |
| | | |
| | | |
| | | |
| | | |
Q2 e>| |<e Q3 |<e Q4 |<e Q5
|---+--| ,--| ,--|
V+ c/| | |\c | |\c | |\c
| | | | | | | |
| | | | | | | |
| | | | | | | |
| +-----+--- | ----+--- | ----' |
Q1 c\| | | | |
|---+ | | |
e<| | --- --- ---
| | \ / LED1 \ / ... \ / LED N-1
| \ --- --- ---
| / R1 | | |
| \ 1200 | | |
| / | | |
| | --- --- ---
| | \ / LED2 \ / ... \ / LED N
'-----+ --- --- ---
| | | |
\ | | |
/ R2 | | |
\ 10k | | |
/ gnd gnd gnd
|
|
|
gnd

Basically, a cheap multiplying current mirror with Q1 designed to
siphon off most of any current increase in R2 due to V+ changes.
Thus, the current through R2/Ic(Q2) remains fairly fixed over V+
changes, which tends to hold the LED current fixed. I just gave some
gross example values for the resistors (R1 to be 0.6V/500uA, for
example) without thinking much on them. LED strings can be simply
added with just a BJT per string.

Jon

10. Jonathan KirwanGuest

I meant R1/Ic(Q2), as it is the current through R1 that must flow
through Ic(Q2), except for a little to Q2 and Q3's base.

Here's my simplified design steps:

(1) Note the desired LED current: 20mA.
(2) Estimate the base drive current for the BJT sourcing that
LED current. I estimate the beta at 200, so I get 100uA.
(3) If more than one string may be added and all of them must
work about the same as if there was only one string, then
estimate Ic of Q2 at say 50X as much -- 5mA.
(4) Guess at the Vbe of Q1 at 0.66V, for now.
(5) Set R1 = 0.66V / 5mA or 132 Ohms. Round up to 150 Ohms.
(6) Compute new estimated I(R1) at .66V/150 = 4.4mA.
(7) Figure R2 at the minimum voltage to drop at this current.
With an estimated 8.2V minimum (let's say) and a gross
estimate of .7V+.66V for the two Vbe's of Q1 and Q3, this
means (8.2-.7-.66)/4.4mA = about 1550 Ohms. Call it 1500.
(8) Estimate Ic(Q2) as being about R1's current minus two base
drive current loads for modest measure, or 4.4mA - 2*100uA
or 4.2mA.
(9) Since Vb(Q2)=Vb(Q3), Ic(Q2)*R3+Vbe(Q2)+Ve=Vbe(Q3)+Ve. I'm
calling the I*Re=Ve, which is a constant 26mV, or so. They
cancel so, Ic(Q2)*R3=Vbe(Q3)-Vbe(Q2). But this Vbe differ-
ence is 60mV*log10(Ic(Q3)/Ic(Q2)), so we compute R3 as
60mV*log10(Ic(Q3)/Ic(Q2))/Ic(Q2)=60mV*log(20mA/4.2mA)/4.2mA
or about 9.7 ohms. Call it 10 ohms.

All that will probably not give 20mA exactly. But what it does
provide will probably be pretty insensitive to the Vfwd of the LEDs,
up to a limit of course.

V+
|
|
|
\
/ R3
\ 10 V+ V+ V+
/ | | |
| | | |
| | | |
| | | |
| | | |
Q2 e>| |<e Q3 |<e Q4 |<e Q5
|---+--| ,--| ,--|
V+ c/| | |\c | |\c | |\c
| | | | | | | |
| | | | | | | |
| | | | | | | |
| +-----+--- | ----+--- | ----' |
Q1 c\| | | | |
|---+ | | |
e<| | --- --- ---
| | \ / LED1 \ / ... \ / LED N-1
| \ --- --- ---
| / R1 | | |
| \ 150 | | |
| / | | |
| | --- --- ---
| | \ / LED2 \ / ... \ / LED N
'-----+ --- --- ---
| | | |
\ | | |
/ R2 | | |
\ 1500 | | |
/ gnd gnd gnd
|
|
|
gnd

So, how better might one BJT per LED string plus two extra BJTs be
used to improve the insensitivity to Vfwd and V+?? No op-amps please.
That's cheating, big-time! And no JFETs. Just BJTs. I'm interested
in learning.

Jon

11. ehsjrGuest

For a 1 off, if the resistor/2 LED series combo is not
to your liking, I'd do it like this:

----- 250
+Vcc -----in|78L05|out---[R]---+---LED--LED---Gnd
----- |
| |
+---------------+

and repeat as often as needed to run all the LEDs.

I like it better than the mirror, as each LED string
is a "2 wire module". The 78L05's are 4 for a dollar
so the cost isn't great, and it's only 2 parts per
string, so its simple. (The mirror saves on
parts count once the number of strings exceeds 5, if
that's a factor.)

Ed

12. Jonathan KirwanGuest

Actually, I was more interested in playing around with using BJTs, as
I'm just a hobbyist and still at that level of things. I like to
learn to be more facile with the basic tools and I don't learn much
when I use opamps with many dozens of transistors, some of them in
configurations I cannot possibly buy and play with, and yielding
abstract behaviors that are much easier to design with; or with
voltage regulators, themselves of similar complexity and thus both
easier to use but also not nearly as educational to use.

Also, I get 2N3904 and 2N3906 BJTs at one cent per, on sale. This is
\$10 for a thousand of the things. Dirt cheap. So rather than 25
cents per string, it's one cent per string. May not make any real
difference as the labor time is everything, if you are building this
as a one-off for yourself. But still, they are cheaper.

Mostly, though, it's just about learning to think in various ways with
simplified (and somewhat wrong) models. Then learning to build on
that to think about 2nd and 3rd order effects and to deal with those,
etc. I'm curious, right now, about what effects hamper the design I
mentioned and how those effects can be better managed with a different
topology that I haven't considered, using the same number of BJTs to
get there. That is the kind of nuanced thinking I'd like to listen
more to, just now.

Jon

13. Bob MonsenGuest

It wastes a bit of energy. An LM317 would be a better choice, due to its
lower voltage differential.

Also, if you want to vary the brightness on N strings, you need to swap
out N resistors.

If you combine your idea of using the 7805 (or 317) with Jon's circuit,
you get a stable current sink which has relatively good tempco with the
ability to adjust the current on all the LED strings at the same time.
The only thing I'd change in Jon's circuit would be to add some 10 ohm
emitter resistors to even out any differences between the transistors.
Also, the more strings he adds, the worse the matching between source
and string gets, due to base currents. This can be fixed using a
transistor as a follower, in the traditional fashion.

He would have to recalculate the appropriate multiplier resistor (R3) in
this case. He wants 20mA through each string, and 500uA through the
control transistor, so

Ia/Ib = 40 =~ e^((va-vb)/vt)

so

vt*ln(40) = va-vb

but using 10 ohm emitter resistors,

va + 20mA*10 = vb + 500uA * R3

so

vt*ln(40) = 500u*R3-20m*10 ==> R3 = 587 ohms

Or something like that. Then, the resistor between Vout and ground for
the 7805 would be 10k.

--
Regards,
Bob Monsen

If a little knowledge is dangerous, where is the man who has
so much as to be out of danger?
Thomas Henry Huxley, 1877

14. Don KlipsteinGuest

I am just jumping into this thread, and suspect that others may have
already advised against matching voltage. Reason: Current actually drawn
is probably excessively inpredictable and varying excessively with
temperature. Combine with this: LEDs get more conductive as their
temperature increases.

Better off to:

Put these LEDs in series pairs, with each series pair in series with a
resistor, and then put all of these "series strings" in parallel with each
other.

Sounds to me like 3.2V is typically across each resistor, and for 20 mA
the resistance would be 3.2V/.02A which is 160 ohms, and the wattage would
be 3.2V*.02A which is close to 1/10 watt.
Closest popular resistor alue is 150 ohms and for conservative operation
the next one up is 180 ohms. Resistor reliability is best when wattage
rating is well above actual dissipated wattage, so I advise to use
resistors of wattage rating more than 1/8 watt - as in the next popular
wattage rating up from that, which is 1/4 watt. Although I do expect (but
not guarantee) you to get away with resistors rated 1/8 or 1/10 watt in
prototypes or a small quantity production that will not have to survive
adverse conditions (in temperature and/or supply voltage and/or thermal
design of the product).
If the product needs to survive being operated in sunlight or a
sun-baked car when the air temperature is 40 degrees C and you need LED
life in the 10,000's of hours and component packing density is high, then
I recommend using 470 ohm 1/4 watt resistors each in series with two
LEDs [assuming 9.6-10V supply] and hope the LEDs are bright enough at 8-10
mA (which they probably will be).
I advise to redesign to use resistors.

Some guidelines I have in:

http://www.misty.com/~don/ledd.html

- Don Klipstein ()

15. Don KlipsteinGuest

This requires looking into "LED driver" circuits. Some of these
circuits use "LED driver" and "white LED" ICs. This is an extra step of
complication!
Please consider that with diodes including LEDs, it is better to
consider current to be the independent variable and voltage to be the
dependent variable as opposed to the other way around. Voltage as a
dependent variable changes at least sometimes annoying little with current
and at least sometimes annoying excessively with temperature and with
production tolerances of the diodes!
Circuits that regulate (even if only somewhat) current or wattage
generally do well for powering LEDs. Feeding LEDs regulated voltage or
voltage as an independent or regulated variable tend to be less
successful!

- Don Klipstein ()

Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Continue to site