What do the LED spec mean?

[karotto]

Apologies for asking such basic questions but

Vf means forward voltage and that is the maximum voltage to be
applied? This is confusing to me. I thought it is the current that
matters. Why would the voltage matter? If you use a 100V power supply
you would still have to have a large enough resistor to reduce the
current to 20mA (for most LEDs). So why do they specify the voltage?

I-f means forward current. If an LED only specified the maximum I-f
how do I know what the current is that will make it last for years?

Any other LED specs that I should be aware of?

Thanks so much for your help

Karotto

 

[DJ Delorie]

karotto <biz@theplayspace.com.au> writes:

Vf means forward voltage and that is the maximum voltage to be
applied?

Forward voltage is the voltage drop across the LED when you apply the
drive current. You can use this to determine the load resistor, for
example, or to determine how many LEDs in a string you can get before
there isn't enough voltage.

This is confusing to me. I thought it is the current that
matters.

Current is what you *provide*, voltage is the *result*.

Why would the voltage matter? If you use a 100V power supply you
would still have to have a large enough resistor to reduce the
current to 20mA (for most LEDs). So why do they specify the voltage?

Case #1. Let's say you had a 5v supply and a green LED. If is 20mA,
Vf is 2.2v. What size resistor do you need? How do you compute it?
Well, you know the current you want... can you figure out the voltage
across the resistor? It's 5v - Vf, or 5-2.2 = 2.8v. So the resistor
has 2.8v across it, and needs to have 0.02 amps through it, R = V/I =
140 ohms.

Case #2. My exposure box provides 39 VDC. Each LED has a Vf of 3.8v.
How many can I string together? 39 / 3.8 = 10.2, or 10 LEDs. In
practice, I strung 9 together with an LM317, which needs some volts to
work - 9*3.8 = 34.2v, leaving about 4.8 volts for the LM317.

100V on a red LED (Vf = 1.8v); 98.2v / 0.02a = 4910 ohms. 98.2v *
0.02a = 2 watts.

I-f means forward current. If an LED only specified the maximum I-f
how do I know what the current is that will make it last for years?

Well, zero current will let it last for years. The spec should have
expected lifetime, but it probably depends somewhat on heat - if you
have a lot of LEDs in an enclosed space, they won't last as long.

 

[default]

On Thu, 27 Aug 2009 11:20:20 -0700 (PDT), karotto
<biz@theplayspace.com.au> wrote:

Apologies for asking such basic questions but

Vf means forward voltage and that is the maximum voltage to be
applied? This is confusing to me. I thought it is the current that
matters. Why would the voltage matter? If you use a 100V power supply
you would still have to have a large enough resistor to reduce the
current to 20mA (for most LEDs). So why do they specify the voltage?

I-f means forward current. If an LED only specified the maximum I-f
how do I know what the current is that will make it last for years?

Any other LED specs that I should be aware of?

Thanks so much for your help

Karotto
The forward voltage is usually a range - that is the voltage you

measure across the led with a specified amount of current - and varies
from led to led and from batch to batch. It is really only of use
when you want to know how many you can put in series and get them to
light with a specific voltage across the string.

There's also usually a de rating curve (chart) to show life expectancy
or maximum current or power dissipation at a range of ambient
temperatures the led might encounter. Hotter ambient, less allowed
dissipation just like other semiconductors.

The safest way to drive leds is with a constant current source. A
resistor is cheap and easy. If the led is expensive or needs to last,
it pays to take the time and effort to use constant current. Use an
IC driver intended for leds or roll your own (takes two transistors
and a couple of resistors is all).

Life expectancy is not often specified and there seems to be no
standardization when life is specified. Leds "never" burn out but
some cheaper ones do fail. "Life" is usually specified as the point
where the LED is down to 50% or 70% of the original light output in
years.

To test the life they run them at higher currents - and try to predict
how long they'd last at normal current.

One this groups regular respondents built a led tester to wean out the
good suppliers on Ebay from the bad ones. He runs 20 ma leds with a
constant current source set at 30 milliamps then measures the light
output days or months later to see how fast it is dropping - but the
technique, will useful to separate good ones from bad ones, doesn't
predict the absolute life but the relative life between different
suppliers.

Buy a good brand from a reputable source, if you need longevity and
pay for quality.

If you run them at lower current they last longer and dim more slowly.
So a high efficiency led rated at 20 milliamps, can often be run at
much less current with the same light output as a less efficient one.

Also look at the beam spread - you can get Leds with impressive mcd
numbers that only have very narrow beams (usually measured to the 1/2
power points in degrees). Buy with the beam spread and application in
mind. If you want to see a pilot light from anywhere in the room - it
needs to be a wide beam, if you need to light an area some feet from
the source a narrow beam may do a better job.
--

 

[Rich Webb]

On Thu, 27 Aug 2009 11:20:20 -0700 (PDT), karotto
<biz@theplayspace.com.au> wrote:

Apologies for asking such basic questions but

Vf means forward voltage

Usually given as typical and maximum, to be read as "at the specified
forward current (If), we expect that this part will typically show a
voltage drop of 1.7 V. We further expect that the +3*sigma voltage drop
will be no more than 2.2 V at that current." (Using Vf-typ 1.7 and
Vf-max 2.2 and you hope that they're using 3 sigma and not one).

Why would the voltage matter?

If your voltage source minimum spec is, say, 4.5 V and you wanted to
guarantee at least 10 mA through the LED then you'd need to drop 2.3 V
across your limiting resistor (since the LED might drop as much as 2.2
V). Thus, if you're using common 5% resistors, use a 220 ohm.

If your supply is typically 5 V and you have a typical LED, then you'll
be pushing about 16 mA (if the resistor is at the other end of its own
tolerance band).

You'll generally want the typical If to run at half the rated max or
less. Check for a derating curve of If vs temperature, also.

I-f means forward current. If an LED only specified the maximum I-f
how do I know what the current is that will make it last for years?

One can not know, Grasshopper, one can only estimate the probability.
;-)

--
Rich Webb Norfolk, VA

 

[Rich Grise]

On Thu, 27 Aug 2009 11:51:49 -0700, karotto wrote:

Thanks Delorie,
So this means that the Voltage rating only serves one single purpose:
To calculate. It has no other function. There is no other reason for
needing to know it.

Isn't one enough? ;-)

Cheers!
Rich

 

[Ken]

On Thu, 27 Aug 2009 12:33:41 -0700 (PDT), karotto
<biz@theplayspace.com.au> wrote:

Apologies if I appear a bit dense. I almost got it...

So if an LED says forward currents 350mA (as my Luxeon LED says) and
the forward voltage is 3.4V that means that if I hook up a battery
with an exact voltage of 3.4V directly to the LED then there will be
a current of 350mA flowing through it (no resistor needed), correct?

Don't try.

 

[DJ Delorie]

karotto <biz@theplayspace.com.au> writes:

So if an LED says forward currents 350mA (as my Luxeon LED says) and
the forward voltage is 3.4V that means that if I hook up a battery
with an exact voltage of 3.4V directly to the LED then there will be
a current of 350mA flowing through it (no resistor needed), correct?

The problem here is that the "resistance" of an LED is exponential, so
if your battery offered 3.5V instead of 3.4V, the current could go up
a *lot*. Likewise, if you give it 3.3V you might not see any light at
all.

 

[Jon Kirwan]

On Thu, 27 Aug 2009 11:51:49 -0700 (PDT), karotto
<biz@theplayspace.com.au> wrote:

Thanks Delorie,
So this means that the Voltage rating only serves one single purpose:
To calculate. It has no other function. There is no other reason for
needing to know it.

If you operate the LED at the current for which the voltage is stated
as a range, then you will get that range of voltages most of the time.
Not always, even then. It's a guidance, that's all.

More fully, the voltage is at least a curve over reasonable ranges of
driven current (unreasonable currents make the curve still more
interesting), if not a multidimensional surface relating even more
parameters, such as ambient temperature, aging, heat sinking, the
actual distribution of manufactured LED behavior, and maybe even
incident light from other sources, to name some.

What saves you from too much detail is that the voltage near to the
specification range often goes by a factor that is a simple, linear
function of current (equivalent to a resistor for short excursions
beyond the range given), so that you can estimate the voltage by
something like this:

V_fwd = V_cal + I * R_cal

With only one point specified in the data sheet, you really can't
figure out R_cal. But if you make a two point measurement on some
actual devices, you can average the results and compute a reasonable
value for Rcal that will "work mostly" for a wider-ranging design.

To go much beyond the local area near the specification, you need to
incorporate an LED voltage that varies with the emission coefficient
(N) times 60mV per 10-fold current change plus a little more also for
the change in current times its ohmic resistance. (In LEDs, N is
often higher than 2.)

In other words, you need to go from a linear (resistor) interpretation
to a more 'curvy' view that follows an exponential/logarithmic (same
thing, just depends on your parameter plotting) curve:

V(I) = N*(k*T/q)*ln(I/I_s)+I*R_s

Where k*T/q at room temps is about 25.9mV and I_s and R_s and N are
model parameters for your LED. This is the more sophisticated model
(though it develops from even deeper understandings of the physical
layout of the diode, dopant concentrations and so on.)

Small signal variation wise, rates of change in voltage relative to
rates of change in current will be:

d[V(I)]/dI = n*k*T/q*(1/I) + R_s

That becomes the R_cal mentioned above, if you already know the
approximate (I) for which you want to estimate its value. The V_cal
can be had from the V(I) equation, using the same value of (I).

Jon

 

[karotto]

Thanks Delorie,
So this means that the Voltage rating only serves one single purpose:
To calculate. It has no other function. There is no other reason for
needing to know it.

 

[karotto]

Apologies if I appear a bit dense. I almost got it...

So if an LED says forward currents 350mA (as my Luxeon LED says) and
the forward voltage is 3.4V that means that if I hook up a battery
with an exact voltage of 3.4V directly to the LED then there will be
a current of 350mA flowing through it (no resistor needed), correct?

Thanks again

 

[Nobody]

On Thu, 27 Aug 2009 12:33:41 -0700, karotto wrote:

Apologies if I appear a bit dense. I almost got it...

So if an LED says forward currents 350mA (as my Luxeon LED says) and
the forward voltage is 3.4V that means that if I hook up a battery
with an exact voltage of 3.4V

There's no such thing as "exact" in the real world.

directly to the LED then there will be
a current of 350mA flowing through it (no resistor needed), correct?

No.

Vf=3.4V means that if you apply 3.3V you'll get 1uA flowing through it,
while if you apply 3.5V you'll get 10A. Well; that's not strictly
accurate, but it's a reasonable model.

Vf is the point at which the V-I curve transitions from near-horizontal to
near-vertical.

This is why you need a resistor or a regulator.

Suppose that you have a 3.4V LED which you want to run at around 15-20mA.

If you have a 5V supply, a 3.4V LED, and a 100-Ohm resistor, then the
current will be (5V-3.4V)/100R = 1.6V/100R = 16mA.

If the supply voltage is actually 5.2V, you get 18mA, if it's 4.8V you get
14mA. Not too much difference.

OTOH, if you have a 4V supply, a 3.4V LED and a 33-Ohm resistor, then the
current will be (4V-3.4V)/33R = 0.6V/33R = 18mA.

If the supply voltage is actually 4.2V, you get 24mA, if it's 3.8V you get
12mA. Which is a 2:1 change in current for a +/-5% change in voltage.

 

[Paul E. Schoen]

"karotto" <biz@theplayspace.com.au> wrote in message
news:f7f5ffb2-3031-449b-a588-62e09d727b78@z4g2000prh.googlegroups.com...

Apologies if I appear a bit dense. I almost got it...

So if an LED says forward currents 350mA (as my Luxeon LED says) and
the forward voltage is 3.4V that means that if I hook up a battery
with an exact voltage of 3.4V directly to the LED then there will be
a current of 350mA flowing through it (no resistor needed), correct?

For a string of seven Cree LEDs that I'm driving with a boost regulator, I
get the following voltages and currents (and power):

I(string) V(string) V(LED) P(LED)
0.28A 23.5V 3.36 0.94W
0.32A 24.1V 3.44
0.34A 24.3V 3.47
0.35A 24.4V 3.48 1.22W
0.53A 26.5V 3.78
0.56A 26.8V 3.83 2.14W
0.58A 27.1V 3.87
0.60A 27.2V 3.88 2.33W
0.63A 27.7V 3.96
0.73A 29.0V 4.14 3.02W
0.88A 31.2V 4.46 3.92W

Thus the effective series resistance of each LED will be R =
(4.46-3.36)/(0.88-0.28) = 1.83 ohms, with a zener voltage of V = 4.46 -
0.88*1.83 = 2.85V. The high-power LEDs are rated based on their junction
temperature, which depends on the thermal resistance of the package and
heat sink. I have these mounted on a large heat sink so I can drive them
pretty hard, but they are rated at 350 mA maximum forward current, and
their light output does not increase linearly with current. I think these
are the Xlamp7090 and they produce 100% rated intensity at 350 mA and 170%
at 700 mA, but probably at greatly reduced life. These lamps have 17 deg C
/ W for the package and a maximum junction temperature of about 85C, so at
25C ambient you can have 60C rise or a maximum of 3.5 watts, which
corresponds to about 800 mA.

So, you can drive one of these LEDs with three 1.2 VDC NiCd cells for 3.6
volts, and you could use three 1.5 VDC cells for 4.5 volts and limit with a
small resistor or even a silicon diode. But best results are with a current
regulator which can be linear or PWM and can approach 98% efficiency with
the addition of an inductor.

The forward voltage of the LED is fairly constant with temperature but I
don't have any specs, so again a current regulator is your best bet. I have
made a variation of the "Joule Thief" which provides current regulation and
runs on as little as 2.5 VDC. I posted an LTSpice schematic previously but
I can provide it again. I also made one that uses 6-15 VDC and a power
MOSFET which is what I used for the tests shown above with 7 white LEDs at
up to 27 watts at 77% efficiency. Not bad for a simple circuit with no ICs.

Paul