C M Baker wrote:
I asked the same thing a couple of months back and most people
responded with the formula (Supply Voltage minus forward voltage) =
voltage drop.
Voltage drop divided by max forward current = value of limiting
resistor.
This works fine for 1 LED, but as soon as you put two or more in
series, that formula does not work.
It does. You just add the voltage drops of all the LEDs together.
Say, you want to operate a red (1.8 V), green (2.4 V) and blue (3.5 V)
LED in series. The total voltage drop across them would be (1.8 + 2.4 +
3.5) V = 7.7 V (the LED act as resistors in series, the total voltage
dropped is the sum of the individual drop voltages, the current flowing
is identical through all of them).
If you use a 9 V supply, you have to drop 1.3 V across the current
limiting resistor. If you want 20 mA to flow through the diodes, you'd
use 1.3 V / 0.02 A = 65 Ohm, or rather the next higher standard value.
The power dropped over the resistor would be 1.3 V * 0.02 A = 26 mW,
thus a standard 1/8 watt resistor would do just fine.
As always, there is a little trick to save oneself the calculation: If
you use a J-FET instead of a resistor to limit the current, the current
depends only on the amplification of the FET (up to the maximum voltage
of the FET, that is), not on the supply voltage. A BF245A will supply
about 10 mA, a BF245B 15 mA and a BF245C about 18 mA:
o +5..30 V
|
|
| |S
|----
G | BF245
------->|----
| | |D
| |
-------------
|
__|__
<-- \ | /
<-- \|/ LED
-----
|
|
|
--- Gnd
If you have a AC supply, you can use a capacitor as a blind resistor to
limit the current through the LED. Since current and voltage are out of
phase, little power is dropped across the capacitor, limiting heat
developed. This is usefull as "power on" indicator in mains circuits:
o 230 V, 50 Hz
|
|
----- 100 nF, 1000 V =, 630 V ~
-----
|
|
---
| |
| | 10 Ohm, 1/4 W
| |
---
|
|
-----------
__|__ | 1N4148
<-- \ | / -----
<-- \|/ /|\
----- / | \
LED | --|--
-----------
|
|
o neutral
Note that in this circuit, all components have contact to mains, thus
appropriate care is required. The capacitor needs to be a type certified
for use across mains. For countries using other mains voltages or
frequencies the value of the capacitor needs recalculation.
Shouldn't we put this in a FAQ somewhere, I recall answering this
question a couple of times.