1
\$\begingroup\$

schematic

The above circuit utilizes a bootstrap design similar to this next illustration,

high-side drive circuit

Image source: Toshiba - MOSFET Gate Driver Circuit - Application Note

I understand quite well how the capacitor C2 is charged when the low-side MOSFET provides a path to ground. D1 clips this voltage to a safe Vgs for the logic level MOSFETS. Once M2 is switched out it provides a threshold voltage across R2.

Sure enough if I probe Vboot-Vout I can see the clipped bootstrap voltage as it charges to 4.7V.

clipped bootstrap voltage

Now, if I remove the charge path R13 and D4 as well as the zener diode which is no longer necessary.

changed schematic

Now if I probe the same Vboot-Vout I see that the capacitor is charged in reverse to a much lower voltage, averaging about -450mV. Yet the circuit continues to work, almost better than before as the shoot-through is self mitigated since unlike the original circuit this one charges much quicker, albeit in reverse.

Now if I'm analyzing this correctly, I would suggest that the same current that charged C2 forward can be used in reverse when placed across R2, but I can't reason why.

enter image description here

\$\endgroup\$
4
  • \$\begingroup\$ Compare saturation voltage Vout-VCC in the high state, before and after. \$\endgroup\$ Commented Sep 13, 2023 at 22:27
  • \$\begingroup\$ That would be the PWM voltage sent to the motor, are you asking to see that they both accomblish the task? They looks almost identical, in my case a 24V square wave 50% duty cycle. \$\endgroup\$
    – user349610
    Commented Sep 13, 2023 at 22:33
  • \$\begingroup\$ Not the full waveform, the difference (Vout - VCC) in the high state specifically. In other words, conduction loss. \$\endgroup\$ Commented Sep 13, 2023 at 23:56
  • \$\begingroup\$ @DanielCanaday - Hi, Please note the site rule which requires that when a post includes content (e.g. text, image, photo etc.) copied or adapted from elsewhere, that content must be correctly referenced. As a minimum, for online material the source webpage or PDF etc. should be linked (see that rule regarding references for books / articles etc.). In order to help you, I found what I believe to be the source webpage link for the copied image & added it. For the future, please remember it's your responsibility to do that :) Thanks. \$\endgroup\$
    – SamGibson
    Commented Sep 29, 2023 at 0:23

1 Answer 1

0
\$\begingroup\$

For your circuit, you only need the low-side switch. You have some strange open-loop "boost convertor" topology. M1 and its drive circuitry isn't doing anything useful. Try connecting your load to ground - the standard buck configuration. And update your findings. You will begin to see why the circuit malfunctions without the bootstrap components where they ought to be - The top transistor (m1) isn't sourcing any power to the load so your circuit modifications can't truly be realized.

\$\endgroup\$
6
  • \$\begingroup\$ Not sure what you mean by "M1 not doing anything useful" You can use a MOSFET like this (short its gate to its source) or you can use a power diode. The MOSFET drops less voltage than a forward biased diode so the MOSFET needs less heatsinking and wastes less battery power. \$\endgroup\$
    – user349610
    Commented Sep 13, 2023 at 22:28
  • \$\begingroup\$ @DanielCanaday Well let me ask you, what is your load on M1? What is it switching? \$\endgroup\$
    – MOSFET
    Commented Sep 13, 2023 at 22:31
  • \$\begingroup\$ Its acting as a flyback MOSFET giving the current stored in the motors inductance a place to circulate when the low-side is switched off. \$\endgroup\$
    – user349610
    Commented Sep 13, 2023 at 22:36
  • \$\begingroup\$ There is a body diode in M1 that is conducting. M1 is not actually turning on. After removing the bootstrap circuitry, you have no way to put a potential across Gate to Source. To check, remove M1 from the circuit and put in a reverse-biased diode (where drain and source used to be). Leave the Gate connection disconnected. Run the simulation again. I bet you get the same exact results. \$\endgroup\$
    – MOSFET
    Commented Sep 13, 2023 at 22:44
  • \$\begingroup\$ @DanielCanaday Or even better, connect the gate directly to the transistor's Source and only it's Source. This will prove your bootstrap circuitry isn't doing anything and that the transistor isn't turning on. \$\endgroup\$
    – MOSFET
    Commented Sep 13, 2023 at 22:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.