On Sun, 01 Oct 2006 15:31:55 GMT, Al wrote:

In article ,
John Fields wrote:


big snip

---

And, this method is not for manufacturing millions of items. It's for
solving a special home problem. Like I said, I do it and it works for
me. No, I don't bake the circuit nor do I freeze it. And as I mentioned
above, diodes with soft knees should be used.

---
Ridiculous. Take a diode and straighten out its knee and what have
you got? a resistor. The point is that the softest knee you can
find won't help you regulate (or even limit) the current into the
LEDs if the source voltage varies.


The poster said his source voltage was 6 volts. Since he was using it to
drive LEDs, I presumed, perhaps falsely, that it was a steady DC. He
should have specified a range, such as 6Vdc +/- 0.5V.

Is it a true DC as derived from a dry cell? Is it pulsating DC derived
from either a half-wave or a full wave diode bridge? Does he have an LC
filter on the output of the bridge? Or is it just a big rectifier across
the bridge? Is there a linear regulator or a switcher involved? As
someone else in this thread had suggested, it might be the output of a
6.3V filament transformer that is rectified.

All of these factors, and probably many others, would have to be
considered.

The brightness of an LED is a function of the current through it.
Typically it specified to have a certain light output level at a
specified current. You may increase or decrease the current as you will.
The lifetime of the LED will depend on the current as will its light
output. Even the specified current is just a normalization of the
readings from a large sample. Your specfic LED may need more or less
current for the specified light output.

So, if the LED is specified to give a certain light output at 10mA at
4.5V and if the source voltage is a constant 6V, I would use two diodes
whose forward voltages are specified as 0.75V at 10mA to give me a 1.5V
drop.

The forward voltage drops of a typical 1N914 diode are shown as:

Vf If
volt ma

0.6 3.0
0.7 10.0
0.8 30.0

Two diodes in series would give me the approximate 1.4V drop close to
what I would need.


---
The problem with thinking that it's the voltage which is what must
be controlled is that it isn't. What must be controlled/limited is
the current.

LEDs are specified to operate at a certain current, and when that
current is pumped through them then the voltage dropped across the
LED will vary according to the range given in the data sheet.

For example, take an LED rated at 20 mA with a minimum voltage of 2V
and a maximum voltage of 3V across it.

That means that with 20mA through the diode the voltage dropped
across it can vary from 2 to 3V. It _doesn't_ mean that if you put
a voltage source across the LED and crank it up to 3V everything
will be OK. More than likely you'll toast the LED.

With that in mind, the proper way to limit the current through an
LED is to subtract the _maximum_ LED Vf from the supply and then to
divide that by the LED current.

For example, let's say you have a white LED rated for 20mA with a Vf
somewhere between 3.5V and 4.5V and you have a 12VDC supply.

The current limiting resistor would be:


Vcc - Vf (max) 12V - 4.5V
Rs = ---------------- = ------------ = 375 ohms
If 0.02A

The closest standard 5% resistors are 360 and 390 ohms, but to err
on the side of caution the 390 ohm resistor should be chosen.


--
John Fields
Professional Circuit Designer