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LMG1210: External Bootstrap circuit - higher voltage and MOSFET instead of Gan FET

Part Number: LMG1210

Hello, 

I want to use LMG1210 as a half-bridge driver in our application. Our switching frequency is 1Mhz and external MOSFETs are required to carry load current of 2A. 

If I use recommended bootstrap circuit from the datasheet then, high side Vgs will be around ~4.5V (5V internal LDO output – 0.4V bootstrap diode drop – 0.1V Mosfet drop). 4.5V will not be sufficient to fully turn ON MOSFET and it will be hanging around Miller's plateau. I don't wish to GaN FET because of high cost. 

I was looking at the external synchronous bootstrap circuit shown in the attachment. I have 2 questions as below: 

1. Can I give a voltage higher than 5V (10V for ex.) to fully turn on the external high side MOSFET? 

2. Is it necessary to use GaN FET as the synchronous switch?

Thanks in advance. 

  • Hello Rahul,

    Thanks for reaching out.

    1. In the schematic you shared, the bootstrap FET Q44 is driven through the high-side output channel GLH1 which is generated from the 5V driver rail VDD. You may be able to use a Si FET instead driven at 10V which will require a stepping up the 5V voltage which will require additional circuitry and cost which would equate to the GaN cost essentially.

    2. We use GaN FETs on Q1 and Q2 because of the higher switching frequency capabilities (>1MHz) to accomplish higher power density.

    Regards,

    -Mamadou    

  • Hello Mamadou, 

    Thanks for your reply. It certainly cleared my queries. 

    Please see the attachment. I have 2 queries as below: 

    1. What advantage does a synchronous switch in circuit 2 gives over circuit 3? Can I go with circuit 2 using a low reverse recovery charge diode? 

    2. Since I'll give an external supply of 10V, is there any limit of voltage across HB-HS / HB-VSS which could exceed? 

  • Hello Rahul,

    When using configuration 2, depending on boot diode selection, the high-side channel may have losses induced by the reverse recovery losses of the bootstrap diode (internal diode or external). However your current configuration as shown will damage the HB-HS supply as this is not meant meant for >5V (check abs max). Configuration 3 eliminates the need for a bootstrap diode using the N-ch to provide charging and discharging path without the diode losses. Configuration 3 as shown on your image is also different from the initial picture where source of the N-ch FET goes to HB-HS supply (without the need for the boot diode) and gate is driven by the low-side channel.

    2. Please review recommended operating conditions and absolute max rating section of the IC pins.

    Regards,

    -Mamadou

  • Hello Mamadou, 

    Thanks for your support. 

    Regards,

    Rahul.