I have ALWAYS heard .7V for Si. Bridge rectifiers drop 1.4V (2 diodes
in series). I have even measured this in real circuits.
That applies to resistors (including the series resistor used with a
zener diode or LED). A diode is a voltage regulator, keeping a
constant voltage across it.
It you have a wall-wart DC supply, with a diode in series with the
output (correct polarity). The output voltage will always be .7V lower
than the output of the wall wart. The diode does have a current limit
(1A with the common 1N4001 diodes) that needs to be observed.
BTW, I've also used diodes on AC to make a bulb light with half
brightness. Maybe you can figure out why 2 diodes DON'T give you
quarter brightness. The diode is a 1N4004 (1A, 200PIV).
LEDs are diodes (although with higher forward voltages). A LED would
destroy itself trying to maintain that voltage, without a series
Yes. A shunt regulator would normally be used only for a small load,
because of the inefficiency of it.
One time I needed to supply 12V @ 1A and 6V @ 50mA to a circuit. I
used a series regulator (7812) for the 12V and a small shunt regulator
(zener diode) for the 6V.
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