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.
To turn on any of the MOSFETs properly in your circuit, the gate voltage has to be significantly higher than the source voltage. For the lower position devices Q2 and Q4, the sources are grounded to 0V therefore they can easily turn on into "saturation" by applying a gate voltage of a few volts above ground.
For the devices connected to the top rail, if you want the on-resistance between drain and source to be really low you have to obey the same rule - the drive voltage to the gate has to be several volts above the source. Now for low volt drop you want the source to be switched virtually to 6V - where in the circuit can the gate receive a voltage of maybe 9 or 10 volts?
There isn't one so please consider two options: -
- Using P channel MOSFETs at the top rail, source to 6V. P MOS is slower than N MOS but in your application it won't make a difference (1.7kHz).
- Using a drive circuit derived from a supply voltage that is at least 9V and quite possibly higher to get the best saturation from the MOSFETs.
Now the relays. What are you hoping to achieve here? The MOSFET whose gate is disconnected will float to some almost random voltage level and possibly turn on unexpectedly or just get hot. Get rid of the relays unless you have some cunning plan for their use which eludes me.
This is a standard H bridge using P and N MOSFETs: -
Please ask if you need recommendations for devices.
Best Answer
MOSFET transistors do not need a current to stay "ON" so potentially they can be lower power. Although you could build your own H-bridge using 2 NMOS and 2 PMOS transistors, I would not recommend that since there are ICs that will be easier to use and have everything you need included.
I found an example of such an IC, the L9110 of course there are more but this a cheap one I could find easily. You can buy ready-made modules using this IC on ebay.