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Over the past week I have been facing this problem where my high-side driver isn't going all the way down to 0 as seen by the simulation below.

enter image description here

My low-side driver puts out a good a PWM signal, however, when it comes on it causes shoot through due to the high side not being fully off as seen below (low side is blue and high side is black).

enter image description here

My full circuit can be seen below:

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I am unsure as to what steps to take next as I have tried many things without much success. Any advice would be appreciated.

Vbus = 120 V

The MOSFETs: https://www.mouser.com/datasheet/2/240/Littelfuse_Discrete_MOSFETs_N-Channel_HiPerFETs_IX-1856276.pdf

High-side gate drivers: https://www.onsemi.com/pdf/datasheet/fan73711-d.pdf

Low-side gate drivers: https://www.infineon.com/dgdl/Infineon-2ED24427N01F-DataSheet-v02_00-EN.pdf?fileId=5546d46275b79adb0175c0b0b3833ee7

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  • \$\begingroup\$ Can you get a plot of V_HO vs V_S, please? \$\endgroup\$
    – horta
    Jul 10 at 3:32
  • \$\begingroup\$ What specifically have you tried? The high side drivers do have low voltage cutoff, but it might be that the bootstrap capacitor (400 nF) is too small, or the surge limiting resistor (150R) is too large. You might also start up your PWM with low duty cycle to stabilize the boost voltage before using a higher duty cycle. And make sure you have a good size bypass capacitor on the driver power supplies. The schematic does not show any. \$\endgroup\$
    – PStechPaul
    Jul 10 at 3:38
  • \$\begingroup\$ Yeah, I would agree with PStechPaul. First thing I would try is to increase the bootstrap capacitor. Your symptoms show that the internal driver mosfet between HO and VS isn't getting activated so your HO voltage is falling through the internal shunt first and after that, it's draining through R6(10k) which is why it's taking so long. \$\endgroup\$
    – horta
    Jul 10 at 3:55
  • \$\begingroup\$ Thank you for the advice, I increased the value of the bootstrap cap but it didn't change the outcome. \$\endgroup\$ Jul 10 at 5:29
  • \$\begingroup\$ You need to make sure that both bootstrap capacitors are charged before first switching on one of a high side MOSFETs. I would measure the bootstrap cap voltage and high side gate-source directly to see whats going on. \$\endgroup\$ Jul 10 at 7:11

2 Answers 2

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The issue with this circuit is possibly that a FAN73711 employs an under-voltage lockout feature that disables the output when the bootstrap voltage is too low. The bootstrap voltage measured across C1 or C3 in your circuit must achieve the Under Voltage Lock Out (UVLO) voltage of around 9 volts, and remain above this voltage to prevent the output driver from switching off. C1 is charged through the 150-ohm resistor and diode during the time when your low side driver brings U1-VS to ground, and the driver can't operate until the voltage on this capacitor is above the UVLO level. This appears to require more than one PWM cycle. You will have to make sure of two things: that your low side driver initially stays low long enough to fully charge this capacitor, and that after being initially charged, it comes on often enough and stays on long enough to keep the voltage above this level. Increasing the capacitance increases the charge time, so this doesn't solve your start-up problem. From your first picture, you can see that the gate high voltage is a little higher with each pulse, indicating that the capacitor is still being charged with each pulse. A fully charged bootstrap would show almost the full 20 volts above Vbus.

If UVLO is active, the driver will shut off, and you will drive the FET "off" with a 10K impedance. These FETs have very high gate capacitance, so they won't shut off quickly. You can see the RC exponential decay in both of your curves.

You can solve your problem by allowing an initial start-up charging period during which the capacitors may charge, and then always running with PWM pulses sufficient to keep the charge refreshed. You can never run this type of bootstrap circuit without PWM. To do so, you would need a seperate, floating DC power source for each bootstrap capacitor.

Good Luck!

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    \$\begingroup\$ Well answered. The present values of 150 ohms and 400 nF for bootstrap comprise a time constant of 60 uS and thus about 300 uS for full charge. The waveform seems to show a frequency of 5 kHz with 50% duty cycle and on time of 100 uS. So the low side driver should stay on for at least 300 uSec before starting PWM on the high side. Larger bootstrap capacitor would only extend the TC. Changing the resistor from 150 ohms to perhaps 15 ohms will still limit charge current sufficiently while decreasing the TC. \$\endgroup\$
    – PStechPaul
    Jul 11 at 21:38
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The reference cicuits in datasheet for the 73711 fail to show how the bootstrap circuit should be configured when Vbus is much greater than Vdd. Basically, the bootstrap capacitor only gets fully charged when the MOSFET is switching, but the MOSFET won't switch properly until the bootstrap capacitor is charged.

App Note (AN-6076) shows a circuit that circumvents this by adding a diode and resistor from Vbus to the Vb pin on the driver chip, and a Zener diode in parallel with the bootstrap capacitor.

enter image description here

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  • \$\begingroup\$ That is a good reference about bootstrap design, but the selection of Rboot is the most important reason for the issues experienced by the OP. Using 10-15 ohms in place of 150 ohms should solve the problem. The start-up circuit seems to be necessary only when there is initial voltage on the output, as for a battery charger. \$\endgroup\$
    – PStechPaul
    Jul 11 at 22:57

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