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.

Nick Gammon High-Side 2N3904 and FQP47P06 schematic


1 Answer 1


Q2 is a MOSFET which should have an isolated gate, so you should be able to go much higher (something like 100k will not be a problem) and things will still work. Q1 collector is not losing any appreciable DC current into the gate.

However, take into account that the MOSFET has appreciable gate capacitance, and there will be a CR constant between that and R2. So things will still work, but they may work more slowly. Whether or not that is an issue, depends on the application.

  • \$\begingroup\$ I am using this to turn on and off some lights very slowly, maybe once per second, so not trying to do PWM or anything like that. \$\endgroup\$ Feb 23 at 9:19
  • \$\begingroup\$ So with R2=100K, this would be 0.12mA at 12V, which is much nicer. Should I also adjust R1 so the Q1 base current is 1/10th or 0.012mA? Or should I just leave R1 as-is because Ice is so small it doesn't matter? \$\endgroup\$ Feb 23 at 9:23
  • \$\begingroup\$ what does and does not matter depends a lot on your power source. Unless you are running off a coin battery or something, I wouldn't personally try to optimise much, though it can be done. \$\endgroup\$
    – danmcb
    Feb 23 at 9:39
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    \$\begingroup\$ I noticed this because my thermal camera saw a heat bloom on R2 and it was exceeding the 0805 1/8W specs, which was a concern because the rest of the board was cold. I will test out R2=100K, but just wanted to make sure the R1 value didn't need to be adjusted as well due to some dependency I don't understand. Power is not a huge issue but heat is, and I prefer to understand where all the values come from for future use. Thanks. \$\endgroup\$ Feb 23 at 19:02
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    \$\begingroup\$ I replaced the 1K resistor with a 100K resistor and everything seems to be working fine. I did an analysis of this and mouser.com/datasheet/2/149/FQP47P06-1009447.pdf shows the maximum input capacitance as 3600 pF, so calculated the switching time: mustcalculate.com/electronics/… as 863 microseconds. So in theory could support 1159 Hz switching, which is plenty for my application. Thanks for your comments! \$\endgroup\$ Feb 25 at 8:36

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