I'm not sure why you think BJTs are significantly slower than power MOSFETs; that's certainly not an inherent characteristic. But there's nothing wrong with using FETs if that's what you prefer.
And MOSFET gates do indeed need significant amounts of current, especially if you want to switch them quickly, to charge and discharge the gate capacitance — sometimes up to a few amps! Your 10K gate resistors are going to significantly slow down your transitions. Normally, you'd use resistors of just 100Ω or so in series with the gates, for stability.
If you really want fast switching, you should use special-purpose gate-driver ICs between the PWM output of the MCU and the power MOSFETs. For example, International Rectifier has a wide range of driver chips, and there are versions that handle the details of the high-side drive for the P-channel FETs for you.
Additional:
How fast do you want the FETs to switch? Each time one switches on or off, it's going to dissipate a pulse of energy during the transition, and the shorter you can make this, the better. This pulse, multiplied by the PWM cycle frequency, is one component of the average power the FET needs to dissipate — often the dominant component. Other components include the on-state power (ID2 × RDS(ON) multiplied by the PWM duty cycle) and any energy dumped into the body diode in the off state.
One simple way to model the switching losses is to assume that the instantaneous power is roughly a triangular waveform whose peak is (VCC/2)×(ID/2) and whose base is equal to the transition time TRISE or TFALL. The area of these two triangles is the total switching energy dissipated during each full PWM cycle: (TRISE + TFALL) × VCC × ID / 8. Multiply this by the PWM cycle frequency to get the average switching-loss power.
The main thing that dominates the rise and fall times is how fast you can move the gate charge on and off the gate of the MOSFET. A typical medium-size MOSFET might have a total gate charge on the order of 50-100 nC. If you want to move that charge in, say, 1 µs, you need a gate driver capable of at least 50-100 mA. If you want it to switch twice as fast, you need twice the current.
If we plug in all the numbers for your design, we get: 12V × 3A
× 2µs / 8 × 32kHz = 0.288 W (per MOSFET). If we assume RDS(ON) of 20mΩ and a duty cycle of 50%, then the I2R losses will be 3A2 × 0.02Ω × 0.5 = 90 mW (again, per MOSFET). Together, the two active FETs at any given moment are going to be dissipating about 2/3 watt of power because of the switching.
Ultimately, it's a tradeoff between how efficient you want the circuit to be and how much effort you want to put into optimizing it.
I think it's important to mention that as the circuit stands it won't work in other words your basic idea is not OK: -
- When the lower transistor turns on, the base voltage to ground will be about 0.7V and the associated upper transistor will not turn on hardly at all - maybe you might get a couple of mA through it.
- If the above problem is solved the upper transistors might see 3.3V on their bases and this means the emitter voltage will be at a lowly 2.6V for any decent current into the load.
Both of these are reasonable show-stoppers so go the sensible route and use P channel mosfets for the upper devices and N channel mosfets for the lower devices. I'll not labour this because the question has been marked accepted. Alternatively buy a H bridge IC like the DRV8803
Best Answer
There are plenty of circuits for controlling a motor via a MOSFET H bridge - just google "mosfet h bridge". Here's the first one in the list: -
Here's the website I got this from - it'll give you loads of good ideas.
Basically, the motor spins one direction when inputs B and C are taken to the appropriate logic level. To reverse the motor take inputs A and D to the appropriate logic level and make sure you don't try and control it both ways because you might short out and damage your power supply or battery.
Obviously if you are running your logic signals at 3V3 you'll need to choose a MOSFET that reacts to this sort of gate voltage correctly. If using switches then having them controlled from the 12 V supply is fine but, in reality, performance and circuit is determined by your power supply voltage.