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Right now I try to reverse engineer a piece of electronic which consists of 3-phase (BLDC) motor driver and a microcontroller with FOC running on it. My goal is to replace that piece of firmware with my own. Until now I have a good part of the schematic drawn but have some quiestionmarks still.

One of those is: where the heck is the bootstrapping diode connected to? Usually it has to be connected to the drain. Are there any other possibilities? I know this is a broad question but still. Maybe there is a bootstrapping technique I haven't seen yet.

enter image description here

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  • \$\begingroup\$ Probably +5V if the low side runs off it. Seems suspiciously low though. \$\endgroup\$
    – winny
    Commented Aug 1, 2023 at 11:54
  • \$\begingroup\$ Yeah, one would like to squeeze the most (or less) of the Rds_on usually. \$\endgroup\$
    – stowoda
    Commented Aug 1, 2023 at 11:56
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    \$\begingroup\$ @winny Considering you lose 0.7V in the emitter follower, then another 0.4V or whatever in the diode, also minus whatever Vds of the low side driver is when driving the motor, also minus the drop across R124.. seems low especially since that MOSFET is not specified below 4.5V. \$\endgroup\$ Commented Aug 1, 2023 at 12:11
  • \$\begingroup\$ @winny I'm agreeing with you and enumerating why. \$\endgroup\$ Commented Aug 1, 2023 at 12:41
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    \$\begingroup\$ If you have access to +12 or +15, I don't see any reason to not tie both drivers to that rail instead. \$\endgroup\$
    – winny
    Commented Aug 1, 2023 at 12:44

2 Answers 2

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It's typically connected to a ground-referenced power supply of 10V or whatever so that when the low-side driver turns on the source of the high-side transistor is grounded and the bootstrap capacitor charges.

That supply is missing from your schematic.

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    \$\begingroup\$ Or considering the low side is run at 5V, it could be that as well. (Which would be suitable for a FDMC8032L logic-level dual FET.) \$\endgroup\$ Commented Aug 1, 2023 at 11:34
  • \$\begingroup\$ @TimWilliams, I have considered that as well but can not measure "continuity" with a DMM between the diode's cathode and that 5V supply. \$\endgroup\$
    – stowoda
    Commented Aug 1, 2023 at 11:49
  • \$\begingroup\$ @Spehro Pefhany, yeah, a Voltage source which adds on top to the source voltage so it is well above the mosfet's gate-source threshold voltage. I have thought about that as well but was not able to figure out that connection on the PCB yet. Maybe, considering that it is a battery powered device, it gets turned on at some point by firmware.. But that seems strange to me (power saving reasons.. ?!). \$\endgroup\$
    – stowoda
    Commented Aug 1, 2023 at 11:53
  • \$\begingroup\$ @stowoda Hmm, could be? That 2.2k pullup won't exactly be sipping microamperes if it were hard-wired. I'd rather a CMOS driver myself; but that might well do the job, if a bit unorthodox. \$\endgroup\$ Commented Aug 1, 2023 at 13:35
  • \$\begingroup\$ @TimWilliams They can drive the outputs low though, which only leaves the 47k resistors. They're emitter followers so the base current will be negligible in comparison. \$\endgroup\$ Commented Aug 1, 2023 at 13:39
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Allright, I have found it. Thanks for all your suggestions.

The MCU turns T33 on through T32. So indeed +5V is the Voltage we were looking for. Regarding the comments about the relatively low V_gs_th: the power transistor is an FDMC8030 and not xx8232 as stated before, which has a lower threshold voltage:

enter image description here

enter image description here

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