Battery life at 18 volts

[vanhansen]

I'm thinking about running a couple of my builds at 18 volts just to hear the difference and get a little more headroom.  Last night I was searching around and found a post where R.G. explained on how to wire up two 9 volt batteries in series to achieve the 18 volts. 

The question I have is how does this affect the battery life, if at all? 

I know it depends on the current draw of the circuit so I'll lay it out there that these builds (Marsha Valve and AMZ Mosfet Boost) don't drain a lot of current.  One 9-volt lasts months in these things.

[R.G.]

Anything that is resistive in nature pulls twice the current at twice the voltage. So things like LEDs are going to pull (about...) twice the current unless you rework the design to back the current down. Most resistors in such designs will have twice the voltage across them, hence twice the current. The exception is where the resistor is run in a constant current mode by some active device, or clamped by some external parallel device like a zener or diode, etc.

Active devices may or may not let twice the current flow. Opamps tend to be relatively constant current, except for their outputs. An opamp putting out twice the voltage into a load puts out twice the current if it swings twice the voltage.

As a first appoximation, guess at slightly less than twice the current. It may be less than that, depending on the circuit. Sometimes you will break something at twice the voltage and then it pulls a lot more. And dies.

[vanhansen]

Thanks, R.G.  I'll go through the components first and do some math beforehand, pending I don't run in to trouble with the math.    IIRC, I've got a pretty big resistor on the LED of the Mosfet Boost to keep the draw down to begin with, like 47k or something.  It's still plenty bright.  The Marsha Valve has a 10k on the LED (I think, need to double check to be sure).  I'm more interested in how the AMZ Mosfet Boost is at 18 volts more than anything.  If that goes OK, then I'll try it with the Marsha Valve.