battery draining

**broken link removed**
Hello. I have a couple of questions. I am using the above circuit to create an adjustable delayed output pulse using a NO switch that will remain closed after the output pulse is generated. This circuit meets all of those requirements with no problems at all. However, my meter indicates that the circuit is drawing about 20 mA after the pulse is generated. I wonder how long a 9V battery will last with the switch closed and if there is a better way to achieve what I want without draining the battery so much. I drew the circuit with the positive to ground (I hope that is ok).

Thank you,
Mike

There is a CMOS version of the 555 chip that is very stingy about using current. Two of those could replace a 556 chip. My book on batteries says a 9V alkalne battery has 440 milliamp hours of current. 440/20 = 22 hours.