# LED Questions (just two: )

Discussion in 'Electronic Basics' started by [email protected], Apr 15, 2007.

1. ### Guest

Ok - since one person did offer here as a good place to ask some of my
ignorance to knowledge questions concering my LED Brake Light fantasy:

Ok, so I have learned that you vary voltage - preferable with pulse
width? - to vary the lunines output of an LED
AND that voltage drops from one LED to the next if wired in series
1. Then would not the LEDs in a series be dimmer the further on the
series they are? (ie The 1st one brighter than the next with the last
being the dimest)
I know this must be wrong so
2. What am I missing?

Thanks
would be enough. Typing is a pain!! : )

2. ### Jan NielsenGuest

skrev:
The leds will have the same brightness and share the voltage (about)
evenly untill the voltage eventually drops too low to keep the leds
turned on.

/jan

3. ### Anthony FremontGuest

Well you can vary the pulse width to vary the brightness, but it is the
average current (not voltage) that is being varied. Generally a resistor
(or some other current limiting device) is placed in series with the LEDs to
limit the current flow to some predetermined maximum. If you hook LEDs
straight to a voltage source without something to limit current, the LED
will flow massive amounts of current and then they will likely burn out
immediately.
More of less yes depending upon the color.
The part mandating that the current flowing thru devices arranged in series
is the same in each device. Therefore each LED should be the same
brightness if they are the same type.

4. ### Guest

So - it is the Current not the voltage that determines the
brightsness.
Voltage can change and not affect the brightness just so long as it
remaines with an operating range.
Am I closer?
Thanks for the replies

5. ### Peter BennettGuest

LEDs should be driven from a current-limited supply,, not directly
from a fixed voltage. The most common way to do this is to connect a
resistor in series with the LED (or several LEDs), and connect the
combination across a fixed voltage supply.

The LEDs can be dimmed by pulsing the voltage.
The voltage drop across an operating LED (red, yellow or green) is
about 2 volts (red a bit less, green a bit more)
The brightness of a LED depends on the current flowing through it, and
the current is the same at all points in a series circuit, so if you
have several identical LEDs in series, they will all get the same
current, and will all be the same brightness.
--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter

6. ### Guest

Well - look Iike I sent my responce to the Author - Sorry!

So, brightness would be determined by the Current.
Voltage just need to remaine within the operating range of the LEDs.

Am I closer?

Thank you all for your help

7. ### Anthony FremontGuest

The voltage (all of it, whatever it is) will be "dropped" across your
circuit. Some will be dropped by the LED (about 2V for each LED in series,
just as you said), and the rest needs to be dropped by a resistor. Using
Ohms Law, you can determine how much resistance you will need given the
voltage being dropped and the desired amount of current flow for the LEDs
(resistor and LEDs will flow same amount of current since they are in
series).

As an example, lets say we have a 5V voltage source and the LED we wish to
illuminate has a forward voltage drop (Vf) of 2V. That means the resistor
we will use will need to drop (5-2) 3V total. If we wish to pass 20mA thru
the LEDs (this would be determined by looking at the datasheet for the LEDs
you have, but 20mA is usually a safe amount of current), we would use Ohms
Law to find the value of resistance that will drop 3V when 20mA is flowing
thru it. Arranging Ohms Law such that R is on one side of the equal sign,
we come up with R=E/I. So R (the resistance in Ohms) will be equal to E
(voltage being dropped in Volts) divided by I (current in Amps).
Substituting our numbers, we come up with R = 3 / .020 or R= 150Ohms.

So using a 5V source, we can use a 150Ohm resistor to drop the 3V required
to allow 20mA to flow thru both the resistor and the LED. The LED will drop
the other 2V.

8. ### Peter BennettGuest

LEDs work quite differently than incandescent lamps - they do not obey
Ohm's Law (and I'm sure some pedant will jump on me for that
statement.)

The voltage across an operating LED is determined primarily by the
chemistry (and therefore colour) of the LED, and not by the current
through the LED, so an LED _must_ _not_ be operated from a fixed
voltage, but rather from a higher-than-necessary voltage, with some
means (usually a resistor) to limit the current to an acceptable
value.

If you want to operate a bunch of red LEDs (typical voltage 1.8 volts)
from a 12 volt power supply, you might connect 5 LEDs in series, to
give a 9 volt drop across the LEDs, then put a 150 ohm resistor in
series to limit the current to 20 mA (9 volts across the LEDs leaves 3
volts for the resistor - 20 mA in 150 ohms gives a 3 volt drop.)

When reading LED datasheets, note that the voltage stated is typical,
for some stated current, and the datasheet will usually state a
maximum current, beyond which the LED may be damaged. LEDs will work
fine well below that specified maximum current. Unless I _really_
need maximum light from the LED, I'll usually plan on operating it at
10 mA or less - that's well below the maximum rating for any LED I've
looked at, and generally gives enough light, so I don't have to study
the datasheet for the particular LED I have to calculate the
appropriate resistor value.

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter

9. ### ehsjrGuest

No. The supply voltage can be much higher than the operating
range of the LED, provided you keep the *current* within
the operating range of the LED. Look at this circuit:

+12V ---[470R]---+
|
[LED]
|
Gnd ------------+

If the LED Vf (forward voltage) spec is 2 volts, that means
that there will be ~2 volts across the LED when it is operated
in its current range. The 470 ohm resistor will limit the
current to ~ 21 mA. If the supply was 9 volts, you would
change the resistor to 330 ohms. If the supply was 100
volts, you would change the resistor to 4700 ohms. In each
case, the resistor limits the curret through the LED to about
21 mA.

Focus on controlling the current through the LED - the voltage
across it will automatically be correct, as long as you limit
the current properly.

Ed

10. ### Wim LewisGuest

Yes. For a LED, the brightness is directly proportional to
the current. (Other kinds of lamps behave differently.)
Closer, but not quite there. The current depends on the voltage across
the LED.

The other thing I think you're missing is that "absolute" voltage
isn't important, only voltage differences between one place and
another are important. It's kind of like altitude --- climbing
a flight of stairs at street level feels exactly the same as climbing
a flight of stairs at the top of a tall building, as long as you
don't look out the window. What matters is how tall or steep those
stairs are. In other words, the difference in altitude between
the top and the bottom, not the absolute altitude.

People also talk about "voltage drop", which is just a way of talking
about the difference in voltage on each side of a component. If one
terminal is at 5v and the other is at 4v, then you can say that the
voltage (or potential) has dropped by 1v as you move from one terminal
to another.

LEDs have a funny current/voltage relationship. If you put a small
amount of voltage across them, (almost) no current will flow. If you put a
bit more voltage, still no current will flow. If you keep increasing
the voltage, then at some point --- around 2 volts, for a green LED ---
the current will suddenly start to flow, and the LED will light up.
Increase the voltage any more, and a huge amount of current will flow,
and the LED will burn up, pfft!

So you can't just attach the LED directly to a voltage source, because
you're unlikely to get exactly the voltage that will light up the LED
without destroying it. Plus, the voltage (the "forward voltage drop"
of the diode) is different at different temperatures, for different
LEDs, and so on. You need some way to keep the right amount of current
flowing even if you don't know exactly what voltage to apply.

The simplest way to do this is with a resistor. Unlike a diode, a resistor
has a very simple relationship between current and voltage; they're
directly proportional. In other words they follow Ohm's law.

So, imagine this circuit:

[ Battery + ]-----/\/\/\/\/\----->|------[ Battery - ]

(that's a resistor and a diode in the middle there, with a battery
that it's operating correctly?

- The voltage between [battery +] and [battery -] will be the
battery's voltage. Let's assume we have a 12-volt battery.

- The current through the LED will be 10 mA, because that's how
much current we want. (The right amount depends on the LED ---
bigger LEDs can handle more current.)

- The voltage across the diode is unknown, but it's roughly 2 volts,
depending on the kind of LED.

- The voltage across the resistor is (current times resistance).

So, we can subtract out the "roughly 2 volts" of voltage across the
diode, and that tells us that the remaining voltage across the resistor
will have to be "roughly 10 volts". It's a series circuit, so the
current through all parts is the same: we've decided we want it
to be 10 mA. Ohm's Law tells us how big the resistor must be: 10V/10mA
equals 1 kOhm. Now we know enough to build the circuit and get
10 mA flowing through the LED.

Why does this work? Because now, if the diode's voltage varies by 0.1
volt, or if the battery's voltage varies by 0.1 volt, the current through
the diode doesn't change much. It stays around 10mA (maybe 9.9 mA, maybe
10.1 mA, but that's close enough). Most of the variation is being taken
up by the resistor, which has a very smooth variation of current
with voltage, unlike the LED's very sudden reaction.

Unfortunately, most of the power is being taken up by the
resistor, too. Only about 1/6 of the energy is going to the LED,
and the rest is going to the resistor (and turning into heat). For a
brake light, this isn't much of a problem, but in other cases, it
can be. In those cases, you can use more complicated techniques,
like pulse-width modulation (where you keep the *average* current
correct by adjusting the duration of pulses of current).