The starter motor is powered by a 12.8V LiFePo4 battery pack (14.2 when fully charged). Sadly, even when the battery is freshly charged, the motor won't turn over unless the charger is still hooked into the circuit. (Somewhat inconvenient at remote locations.) Can I hook a 5 cell 15.0 V (3.2V X 5) "Booster" into the circuit to give the extra push it seems to need?
I have a 12V starter motor that needs extra help at times.
voltage
Related Solutions
Use an inverting buck regulator to create -3V and wire your fan between -3v and +9V. Maybe one of these: -
It's got a -5V output but that can be made -3V by choosing appropriate values for R3 and R4
You will probably find that two 9V wall warts will do the trick - connect the positive output from one to the negative output from another - this connection is the "new" 0V and you'll have +9V and -9V available - regard it as two 9V batteries in series with the middle (common) connection being called 0V.
Because wall warts produce isolated output voltages, they can used like this but always double check with a meter first to ensure everything looks and feels OK.
Looking at your op-amp circuit you should be able to get aways with a single 12V wall wart and a rail splitter made by two resistors across the supply. The junction of those resistors will be 6V up from the negative rail and ALL the points marked with an "earth" symbol on your circuit can connect to this point. This is a much used technique but, as always, it won't suit every target circuit but looking at your circuit it should be OK.
Best Answer
To me it seems like the battery can't deliver enough current to the starter motor. I would not try to raise the voltage since the starter motor is rated at 12V. When the battery is connected to the charger the charger helps the battery to supply enough current. To be sure I would look at the starter motor specification. Look for the wattage or current specification. Also look at how much current the battery is specified for. The battery must be able to deliver as much or more current than what the starter motor needs.
Two solutions to the problem if my suspicion is true: