Al wrote:


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.

That doesn't sound like it would work. Now you have an extra 0.1 V on
the LED. The current could be easily double (or more) what is wanted.
"Close to the right voltage", in an LED or any diode, just doesn't cut
it in terms of limiting the current.

And if the source voltage is NOT a constant, exact 6V, it's even worse.

Resistors are the standard, simple easy way to limit current through an
LED.

Mark