A voltage of about 3 times the supply (minus about 5v) can be created by the following
circuit. The output will deliver about 50mA. Pin 3 goes LOW to about 0.5v and HIGH to
about 11v.
When pin 3 goes LOW, capacitor “a” charges via diode “a” to about 11v.
When pin 3 goes HIGH, capacitor “a” has about 11v across it, plus the voltage on pin 3.
The 22v on the positive lead of capacitor “a” passes through diode “b” to charge capacitor
“b” with 21v – 12v = 9v. This produces a voltage of 21v on the anode of diode “c.” When pin
3 goes LOW, both capacitors “a” and “c” will charge via diodes “a” and “c” with capacitor
“a” being charged via diode “a” and capacitor “c” charged via diode “c.” When pin 3 goes
HIGH, the 9v across capacitor “c” will be added to the 22v to charge capacitor “d” to 31v.