majenko:
oric_dan(333):
Now I don't know what you guys are talking about again. The drive transistor is a MOSFET, not
a bjt. All the 10K pullup does is hold the MOSFET "off" when the NPN is off, and bias the MOSFET
gate to approx -12V when the NPN is on.The only time current flows through the pullup is when the NPN is on. If you lower the value of
the pullup, all you are doing is causing the NPN to work harder, and waste electrons. With
110 ohms, the NPN will have to pull 12V/110 = 109mA, just a waste.10K is a good value for the pullup.
Also, you get economy of scale - the more components of the same value you have, the less they cost. I regularly use 10K in various combinations to make other resistances - 5K, 20K, etc - because I buy 10K in bulk they are considerably cheaper.
Economy is a secondary concern - you need a working circuit first.
With FET's we need to pay equal attention to switching it on as we do to switching it off. In the proposed circuit, switch-on is fast (only limited by the NPN on-state resistance). Switch-off however is slow as the FET is left to discharge its gate through the 10k resistor. For a switching application (such as PWM) this is inadequate even for moderate loads.
A proper design requires a push-pull driver for the gate - not a 10k resistor.