Issues With MOSFET bridge, DC motor 24V 200W

It's a bad idea to drive a power mosfet this way. First of all, the Supply Voltage is 24 VDC and the FET's max gate voltage is only 20 V. Therefore, you can expect FET failures via puncturing the gate, though it probably won't happen right away.  Second, as the turn on path is thru a 10K resistor, the maximum gate charge current is only 2.4 mA peak and reduces as the gate charges up. This would translate into a turn on time of about 37 uS, which is much slower than optimal, causing high FET switching losses. It's not as bad for turn off as the T1 NPN can sink 100-300 mA. A proper FET driver should source and sink at least 1.5 amps for a FET this size for fast switching in the 50-200 nSec range. A cascade (totem-pole) NPN-PNP arrangement (or n-CH-p-CH in MOS) using discrete parts can be used, but the drive supply should be limited to 12 to 15 VDC typically. Even if the FET gate is rated at +/-30V, there's no real advantage to driving the gates that high as they're typically fully enhanced at 10-12 V Vgs. One good, low-cost method for gate drive is to use a CMOS hex inverter or buffer and parallel all of them. Note that you cannot use half the inverters to drive the upper FET and half for the lower FET as the H-Bridge FET Sources are at different grounds. Therefore, you would need two inverter or buffer ICs. The 10uF cap used for the bootstrap supply is a bit high. For this size FET, having Qg (total gate charge) of 90 nanoCoulombs max, a 0.1 uF (100 nF) is more appropriately sized. If the switching frequency is relatively low, as it would be for a motor controller, the 10 uF may work fine, but more than 100 nF isn't needed. In many H-Bridge circuits, it's preferable to have the turn-off speed be faster than the turn on. This is usually done by having a separate resistive circuit for the turn-on & turn off using about a two series resistors with a diode across one of them with the cathode pointing toward the driver. For example, a 5 ohm and a 15 ohm resistor with a 1N4148 across the larger resistor. This effectively gives the turn on circuit 20 ohms series and the turn off just 5 ohm series resistance so that the turn off is substantially faster than turn on. This will help to avoid or minimize Cross-Conduction (aka shoot thru current) in the H-Bridge. 

It's a bad idea to drive a power mosfet this way. First of all, the Supply Voltage is 24 VDC and the FET's max gate voltage is only 20 V. Therefore, you can expect FET failures via puncturing the gate, though it probably won't happen right away.  Second, as the turn on path is thru a 10K resistor, the maximum gate charge current is only 2.4 mA peak and reduces as the gate charges up. This would translate into a turn on time of about 37 uS, which is much slower than optimal, causing high FET switching losses. It's not as bad for turn off as the T1 NPN can sink 100-300 mA. A proper FET driver should source and sink at least 1.5 amps for a FET this size for fast switching in the 50-200 nSec range. A cascade (totem-pole) NPN-PNP arrangement (or n-CH-p-CH in MOS) using discrete parts can be used, but the drive supply should be limited to 12 to 15 VDC typically. Even if the FET gate is rated at +/-30V, there's no real advantage to driving the gates that high as they're typically fully enhanced at 10-12 V Vgs. One good, low-cost method for gate drive is to use a CMOS hex inverter or buffer and parallel all of them. Note that you cannot use half the inverters to drive the upper FET and half for the lower FET as the H-Bridge FET Sources are at different grounds. Therefore, you would need two inverter or buffer ICs. The 10uF cap used for the bootstrap supply is a bit high. For this size FET, having Qg (total gate charge) of 90 nanoCoulombs max, a 0.1 uF (100 nF) is more appropriately sized. If the switching frequency is relatively low, as it would be for a motor controller, the 10 uF may work fine, but more than 100 nF isn't needed. In many H-Bridge circuits, it's preferable to have the turn-off speed be faster than the turn on. This is usually done by having a separate resistive circuit for the turn-on & turn off using about a two series resistors with a diode across one of them with the cathode pointing toward the driver. For example, a 5 ohm and a 15 ohm resistor with a 1N4148 across the larger resistor. This effectively gives the turn on circuit 20 ohms series and the turn off just 5 ohm series resistance so that the turn off is substantially faster than turn on. This will help to avoid or minimize Cross-Conduction (aka shoot thru current) in the H-Bridge.