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_{D}^{2} × 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 I^{2}R losses will be 3A^{2} × 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.

So, the friendly folks at IR got back to me, and suggested I swap out the IRS2001 with a IRS2301. The IRS2301 has a VCC requirement of just under 5V, and will thus work for my application.

Now I just have to wait for the priority mail from Digi-Key...

## Best Answer

Your first concern in selecting a gate driver is to find one that can drive enough current to switch your selected MOSFETs fast enough for your application. As a rough estimation, you can divide the total gate charge of your MOSFET by the current the driver can sink/supply.

$$ t_{on}=\frac{Q_g}{I_g}$$

Using the worst-case values for IRF1405 and the slower of your two gate drivers, IRS4253:

$$ \require{cancel} \begin{align} t_{on} &= \frac{260 \cdot 10^{-9} C}{180 \cdot 10^{-3}A} \\ &= \frac{260 \cdot 10^{-6} \cancel{C}}{180 \cdot \cancel{C}/s} \\ &= 1.44 \mu s \end{align} $$

Off is faster, because this driver (which is typical) can sink more current than it can source:

$$ \require{cancel} \begin{align} t_{on} &= \frac{260 \cdot 10^{-9} C}{260 \cdot 10^{-3}A} \\ &= 1 \mu s \end{align} $$

If your switching frequency is 10kHz, each switch period is \$1/10000 = 100\mu s\$ and you will spend \$ (1.44\mu s + 1\mu s) / 100\mu s = 2.44\%\$ of that time switching. Probably acceptable, but you should calculate your switching losses and check.

Also, keep in mind this calculation is an

approximation. The current specified in the gate driver datasheet is current into a short circuit, but a MOSFET gate isn't that. Unlike a short circuit, the gate voltage rises as it is charged, which will reduce the current the driver can provide. Also, your layout may introduce more inductance and resistance than there was in the test circuit the manufacturer used, further reducing current. Consequently, your actual switching losses may be higher than this calculation suggests.In selecting the bootstrap capacitor, you want to make sure it's significantly bigger than the gate capacitance it will be charging, so that the bootstrap voltage doesn't sag appreciably when you switch. It also needs to supply whatever leakage current there is as long as you keep the high-side switched on. You can calculate these leakage currents, or just make the bootstrap capacitor way bigger to be safe. 100 times bigger than the gate capacitance should be good, so at least \$26\mu F\$. Bigger doesn't hurt much, so round up to a standard value or whatever you already have in the BOM or stock.

Since this capacitor is the power supply for the high-side gate current, you also want it to be very low impedance. It wouldn't hurt to parallel your big capacitor with some smaller \$100nF\$ decoupling capacitors very near the gate driver(s).

Selecting a bootstrap diode isn't terribly difficult. It needs to be able to withstand the reverse voltage when the H-bridge is switched high. Also keep in mind that you will lose the diode's voltage drop from the gate voltage. A Schottky diode might be nice for this reason, but depending on your circuit, you may not find one that can take the reverse voltage. A simple 1N4148 can take reverse voltage up to \$100V\$.

The reverse recovery time of the diode can also be relevant if your are switching very fast; 1N4148 has a reverse recovery time of \$4ns\$, so you will have to have the H-bridge switched low for significantly longer than that for the bootstrap capacitor to have time to recharge between cycles.