I built this circuit, works excellent, substituted the red LED's for
some 6500 mcd orange ones, the switch is a push button 250 volt
"lampswitch" this can only be installed on the positive battery lead
not the negative (has only two terminals), this won't drain the
battery when the unit is off will it? If so, I'll just install
another switch on the negative side.
9V battery, 3x 2V (min), 3V to be dropped across a resistor X at 20mA -
V=IR, R=V/I R=3/0.020, R=150 ohms.
9V battery, 3x 2.3V (max) 9-6.9=2.1V headroom - you could use an LM317
set up as a current source. That would allow driving reliably with 20mA
(actual voltage not that big of an issue for LEDs, and not a very
reliable number). The LM317 acts as whatever size resistor is needed to
run 20mA (when it's set up to act as a current source). There are also
newer devices that do that as a unit, with no external parts.