5.3 Example Using MOSFET

The follow example uses a power MOSFET as the switching device with the following desired parameters:

• Power device = DMNH6021SK3Q
• HVIC gate driver = Microchip’s MCP14H2304
• V_CC = 12V
• QG = 20 nC
• I_GSS = 100 nA
• t_HON = 10 μs
• R_DSON = 25 mΩ max, 125°C
• I_OUT = 5A
• I_QBS = 150 μA
• I_{LK_IC} = 50 μA
• QLS = 10 nC
• V_F = 1.0V
• I_{LK_DB} = 100 μA
• V_GSmin = 10V

From equations above:

$∆V_{BS}=12V-1V-10V-0.625V=0.375V$

$QT=QG+QLS+\left[I_{LKN}\right]\times t_{HON}$; where $I_{LKN}\times t_{HON}=3nC$

Thus $QT=20\mathrm{ n}C+10\mathrm{ n}C+3\mathrm{ n}C=33\mathrm{ n}C$

Therefore $C_{Bmin}=\frac{33\mathrm{ n}C}{0.375V}=88\mathrm{ n}F$

The bootstrap capacitor calculated in the above problem is the minimal value required to supply the needed charge. It is recommended that a minimal margin of two to three times the calculated value is used. Utilizing values lower than this could result in over charging of the bootstrap capacitor especially during –VS transients. Typically for motor driver applications C_BS = 1 μF to 10 μF are used, and for power supplies typically C_BS = 0.1 μF to 2.2 μF. Also it is recommended to use low ESR ceramic capacitors as close to the VB and VS pin as possible (see PCB layout suggestions section).