H bridge for PWM

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 (I_D² × R_DS(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 (V_CC/2)×(I_D/2) and whose base is equal to the transition time T_RISE or T_FALL. The area of these two triangles is the total switching energy dissipated during each full PWM cycle: (T_RISE + T_FALL) × V_CC × I_D / 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 R_DS(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.

1. Yes.
2. a) Yes. b) Possibly; see answer #4.
3. R1 limits the current for low-speed forward and reverse operation. T9 provides high-speed forward operation.
4. It raises the switching threshold for P1 by one diode drop. My guess would be that if you drive P1 and P2 high simultaneously (from a dead stop), this allows the low-speed circuit to get the motor started before the high-speed circuit kicks in, providing smoother acceleration.
5. The resistor-capacitor-diode combination provides a soft-start, fast-stop characteristic for the motor. When starting, the resistor and capacitor have a time constant of about 0.224 seconds, which helps prevent the motor from abruptly spinning the wheels. When stopping, the diode dischages the capacitor immediately for faster response.
6. See my comment on your question. When hooked up correctly, T1 feeds base current from the high-side PNP (Q2) directly to the corresponding low-side NPN (T5).
7. Only if you remove the three capacitors. You'll also need to add freewheeling diodes across each of the five power transistors.