# Resistor and current limits for 2N3904 with FQP47P06

I have been using Nick Gammon's excellent tutorial on doing a high-side driver with a 2N3904 and FQP47P06. http://gammon.com.au/motors

However, I've noticed that the R2 resistor value seems to be arbitrarily selected as 1K, and at 12V it will draw about 12 mA of current and dissipate 0.144 watts, which exceeds the limits of many 1/8W 0805 SMD resistors. Furthermore, I would like to operate this circuit at say 14V (maybe up to 20V), which makes the power consumption even higher. The resistor burns way more than the calculated 0.00625 watts used in the transistor, but this is not discussed. I'm not sure why Nick selected 1K as the value for R2, it seems like the other values followed from this arbitrary choice.

I was wondering if it is possible to select a much higher value of R2? Instead of 4mA for Ibe, I was going to use 1mA (3.3V input, R1=2350, Ibe=1mA). Given a gain of 10, this would be a maximum through the transistor of 10 mA. I would aim to use 5 mA, and so this gives an R2 of approx 2.8K. My power consumption in R2 is now 0.07 watts, which is much lower than 1/8W SMD resistors, and less wasteful in general.

Is there a reason using R2=2.8K would not work? How much higher can I go with R2 before something stops working? Could I divide the currents by 10x to save more power? If I provide 20V I would expect the circuit to still work with these higher margins. The FQP47P06 seems to not require much current, so I'd like to understand more what the limits are for R1,R2 here and not waste so much power that exceeds SMD resistors. 