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JA.1
Associate III
Jan 5, 2021 1:32 PM UTC
Solved

Why is the B-G431B-ESC1 board using L6387E as gate driver? (incompatible logic level)

I have spend some time to thoroughly understand the B-G431B-ESC1 board, which is a BLDC motor controller. According to the L6387E datasheet the high level logic threshold voltage is minimum at 3.6V. Looking at the schematic of the discovery board, the high and low level logic pins of the gate driver are directly connected to the microcontroller (STM32G431). These pins are 3.3V when high.

How come the L6387E is chosen as gate driver instead of L6388E which has a high level logic threshold voltage minimum of 1.8V?

Also the L6388E is branded as "3.3 V, 5 V, 15 V CMOS/TTL input comparator swith hysteresis and pull-down".

0693W000006HZ1CQAW.png0693W000006HZ17QAG.png

This topic has been closed for replies.
Giuseppe Scuderi
Giuseppe ScuderiBest answer
Associate II
Mar 1, 2021 8:44 AM UTC

Dear @Community member​,

the L6387E gate driver is CMOS/TTL compatible as shown in the datasheet (description). The High level logic threshold is compatible with 3V3 MCU, it is possible that the datasheet will be updated.

Regards,

Giuseppe

2 replies

Laurent Ca...
Senior III
Mar 1, 2021 3:47 AM UTC

Dear @Community member​ 

I forward your question internally.

Best regards

==

Laurent Ca...

Giuseppe Scuderi
Giuseppe ScuderiBest answer
Associate II
Mar 1, 2021 8:44 AM UTC

Dear @Community member​,

the L6387E gate driver is CMOS/TTL compatible as shown in the datasheet (description). The High level logic threshold is compatible with 3V3 MCU, it is possible that the datasheet will be updated.

Regards,

Giuseppe