Simply, you can get a faster rise-time this way. Whether that's important, depends on the MOSFET, what kind of load is attached, and how fast it should be switched. Conversely, the NPN-only version has fast fall-time and slow rise. Or vice-versa if a P-ch MOSFET is used (given some other considerations: performance is generally ~2.5x poorer than NMOS, and assuming the load can be common-ground instead). We can address these by "boosting" the pull-up/down resistor with an emitter follower: [![NPN gate driver][1]][1] (Resistor values are unimportant.) Note that when Q1 is off, the equivalent circuit is R1 supplying Q2 base. Base current depends on the voltage drop from +V to Vo, and emitter current is about h<sub>FE</sub> times larger (and shifted down a V<sub>BE</sub>). Basically the emitter looks like an h<sub>FE</sub> times smaller pull-up resistor, for no cost to Q1's on-current -- efficient. Or we can boost both ways, replacing the diode with a PNP emitter follower to boost discharge current: [![Complementary emitter follower gate driver][2]][2] This allows less current in Q1 for a given falling speed. Note that the Q1 B-E resistor might not be needed: when the signal source is a CMOS output pin (typical these days), it has a nice low V<sub>OL</sub>, so Q1 will be discharged adequately in the low state (i.e., V<sub>BE(off)</sub> ≪ 0.7 V). The base voltage divider motif does allow faster switching. Optimal switching is with the divider values set for a Thevenin voltage of about 2 V<sub>BE(on)</sub>. So from a 5 V logic source, 10k + 4.7k is pretty reasonable for example. A speed-up cap can also be placed in parallel with the series resistor (your R4), typically 10-100 pF. We can further improve it by replacing the pull-up resistor with a current source, so it pulls up equally as strongly across the whole voltage swing, rather than with a decreasing current. The current itself can also be switched, to save on power dissipation, in which case the circuit starts looking a lot like a CMOS logic circuit (complementary yes, but in bipolar instead!); but then two level shifters are needed, and an inverter, and... it's a lot of components just to drive a gate, and I would definitely reach for a proper gate driver IC before going to those lengths. :) Images from: \ https://electronics.stackexchange.com/questions/414620/driving-mosfet-from-mcu-with-bjt \ https://electronics.stackexchange.com/questions/583716/limiting-the-base-current-to-bjts-in-gate-driver-circuits [1]: https://i.sstatic.net/ZFgYG.png [2]: https://i.sstatic.net/sx7rf.png