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LED wiring

  • Thread starter Ming the Murphyless
  • Start date
M

Ming the Murphyless

Jan 1, 1970
0
I need to wire four white LEDs to run off a 12v dc supply (regulated).
The Vf is 3.2-3.6 The calculation gives a resistor value of 220 ohms for
2 LEDs in series. Can I wire all four in series without using a
resistor or are there good reasons for having it? The LEDS will be lit 24/7

TIA

Bob
 
M

Ming the Murphyless

Jan 1, 1970
0
John said:
I need to wire four white LEDs to run off a 12v dc supply (regulated).
The Vf is 3.2-3.6 The calculation gives a resistor value of 220 ohms for
2 LEDs in series. Can I wire all four in series without using a
resistor or are there good reasons for having it? The LEDS will be lit 24/7

---
There are good reasons for having it and you should run two parallel
sets of LEDs in series with 220 ohm resistors: (View in Courier)

+12V>--+------+
| |
[220] [220]
|A |A
[LED] [LED]
| |
[LED] [LED]
|K |K
GND>---+------+

Do you want to know what the good reasons are?

JF
JF
Yes please and thank you for a sensible, useful answer.
 
M

Ming the Murphyless

Jan 1, 1970
0
John said:
John said:
On Sun, 09 Nov 2008 18:51:04 GMT, Ming the Murphyless

I need to wire four white LEDs to run off a 12v dc supply (regulated).
The Vf is 3.2-3.6 The calculation gives a resistor value of 220 ohms for
2 LEDs in series. Can I wire all four in series without using a
resistor or are there good reasons for having it? The LEDS will be lit 24/7
---
There are good reasons for having it and you should run two parallel
sets of LEDs in series with 220 ohm resistors: (View in Courier)

+12V>--+------+
| |
[220] [220]
|A |A
[LED] [LED]
| |
[LED] [LED]
|K |K
GND>---+------+

Do you want to know what the good reasons are?

JF
JF
Yes please and thank you for a sensible, useful answer.

---
LEDs are diodes, first and foremost, and are designed to be driven with
a constant current, instead of a voltage, because of the sharp rise in
their forward current (If) with a very small change in their forward
voltage. (Vf)

Such being the case, the Vf of LEDs is specified with a particular
current through the LED and Vf can fall anywhere in the range specified.

In your case, if we look at the nominal Vf of 3.4V (and double it
because of the 2 LEDs being in series) we can determine the forward
current (If) by applying Ohm's law:

Vs - 2Vf 12V - 6.8V
If = --------- = ------------ ~ 0.024 amperes = 24mA.
R 220R

Doing the same thing with a Vf of 3.6V will give us an If of about 22
mA, and with Vf = 3.2V, If will be about 25.5mA

That means that the most current the resistor will allow through the
LEDs is 25.5mA, and that'll happen when the Vf of the LEDs is the lowest
it can be so, no matter what, as long as your supply stays stable, the
LEDs will be safe.

Now, if you took 4 LEDs and connected them in series, the drop across
the string (even if Vf for all the LEDs was only 3.2V) would be 12.8V,
so your 12V supply wouldn't be able to drive them properly.

JF
Thank you for that full answer. I amd going to wire the LEDs in pairs
with a 220 resistor per pair. I take it this means I can add one or two
more pairs if the light output from 4 LEDs is too low?

Thanks again

Bob
 
J

John

Jan 1, 1970
0
John said:
On Sun, 09 Nov 2008 18:51:04 GMT, Ming the Murphyless

I need to wire four white LEDs to run off a 12v dc supply (regulated).
The Vf is 3.2-3.6 The calculation gives a resistor value of 220 ohms for
2 LEDs in series. Can I wire all four in series without using a
resistor or are there good reasons for having it? The LEDS will be lit 24/7

---
There are good reasons for having it and you should run two parallel
sets of LEDs in series with 220 ohm resistors: (View in Courier)

+12V>--+------+
| |
[220] [220]
|A |A
[LED] [LED]
| |
[LED] [LED]
|K |K
GND>---+------+

Do you want to know what the good reasons are?

JF
JF
Yes please and thank you for a sensible, useful answer.

---
LEDs are diodes, first and foremost, and are designed to be driven with
a constant current, instead of a voltage, because of the sharp rise in
their forward current (If) with a very small change in their forward
voltage. (Vf)

Such being the case, the Vf of LEDs is specified with a particular
current through the LED and Vf can fall anywhere in the range specified.

In your case, if we look at the nominal Vf of 3.4V (and double it
because of the 2 LEDs being in series) we can determine the forward
current (If) by applying Ohm's law:

Vs - 2Vf 12V - 6.8V
If = --------- = ------------ ~ 0.024 amperes = 24mA.
R 220R

Doing the same thing with a Vf of 3.6V will give us an If of about 22
mA, and with Vf = 3.2V, If will be about 25.5mA

That means that the most current the resistor will allow through the
LEDs is 25.5mA, and that'll happen when the Vf of the LEDs is the lowest
it can be so, no matter what, as long as your supply stays stable, the
LEDs will be safe.

Now, if you took 4 LEDs and connected them in series, the drop across
the string (even if Vf for all the LEDs was only 3.2V) would be 12.8V,
so your 12V supply wouldn't be able to drive them properly.

JF

Also, note the current is predominantly being set by the voltage drop
across the resistor. The knee in the LED voltage-current curve is so
sharp that even if you had four matched LED devices with 3V drops each
at 25mA you still couldn't safely connect them across your 12V
regulated supply. Why? If the temperature went up their operating
point will shift. Say a couple of degrees hotter and 3V on the LED may
correspond to 100mA (for argument's sake). By dropping a sizeable
proportion of the 12V across a series resistor you degrade efficiency
(resistor dissipates heat) but you gain by making things less
sensitive to parameter changes like temperature or device-to-device
variations. And your 12V regulated source will not be 12.00000V over
all loads and temperatures. You need things to be insensitive to
changes.

John
 
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