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I have been looking into the cheap IBT-4 MOSFET H bridge to use it as a base for a costume PCB design.

But there is one thing I don't get; the PCB has a DC-DC step-up converter (MC34063AB by STMicrolectronics) that, across the recommended 5-12V PCB supply voltage, outputs more or less 14V to all other ICs as supply voltage.

I don't get why this DC-DC converter is there. The system has 2 half bridge MOSFET drivers with bootstrap capacitors, shouldn't it be this alone enough to actuate the gates of the high side of the H bridge? Why use a step-up converter as well?

For all else the design seems more or less straight forward it uses a 2 comparator IC(LM393M by Texas Instruments), which in case of a signal from the MCU connects the 14V to the half bridge gate driver (L6384 by STMicrolectronics). This effectively decouples the MCU from the PCB. The gate drivers pin out is as expected, nothing special.

The only thing I'm missing is why is there this DC-DC step up, and under what circumstances should I carry this DC-DC converter over to me PCB design. Thanks enter image description here

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    \$\begingroup\$ Bootstrap capacitors only work when the transistors are switching frequently. \$\endgroup\$
    – user16324
    Commented Aug 7, 2021 at 14:43

2 Answers 2

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Most MOSFETs require at least 10V of gate-source voltage to achieve their rated on resistance. So what is supposed to provide this if you connect the board to a 5V supply if the DC-DC converter isn't there? Check out the MOSFET datasheet.

The L6384 has an undervoltage lockout between 10-12V for this purpose to prevent incomplete proper MOSFET turn which would make it overheat. So the IC itself actively prevents you from trying to drive the MOSFETs with a lower voltage.

Having the DC-DC converter there lets you use the board for lower voltage motors.

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The wide input voltage range allows a wide range of impedance matching with voltage boost function.

While a buck regulator requires a higher input voltage to down regulate. Very low switch resistance permits the 50A peak input currents for the inversion.

You must decide on your conversion ratio and power transfer needed for your load , i.e. <1 or >1 . This design is for >1 voltage conversion ratio.

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