3
\$\begingroup\$

I have a FAN7392N driver circuit which is used for driving high-side, N-channel MOSFETs. Here is the pin example circuit / block diagram in the datasheet:

enter image description here

enter image description here

When I apply a Vcc voltage of 13 V and a LOW signal to the HIN pin, then the HO pin will output 2.5 V. If I apply a Vcc voltage of 13 V and a HIGH signal to the HIN pin, then the HO pin will output 2.5 V. If I PWM the HIN pin, then the voltage will increase anywhere from 2.5-11 V depending on the PWM duty cycle that I apply to it.

Is the HO pin supposed to be outputting a voltage even when the HIN pin is LOW/HIGH? I have hooked up my circuit exactly like the diagram listed in the datasheet above and tried multiple ICs to make sure the one I am using isn't broken.

\$\endgroup\$
8
  • 1
    \$\begingroup\$ Do you have a load on the output of the half-bridge to GND? If not, and if LIN is low, both IGBTs will be off, and a voltage may appear there (on VS). This could cause current to flow through the body diode of the upper driver to the gate drive. It is a good idea to add resistors from gate to emitter of the IGBTs. Bootstrap circuits do not work if duty cycle is insufficient to maintain gate drive voltage on VB. \$\endgroup\$
    – PStechPaul
    Oct 18, 2022 at 4:35
  • \$\begingroup\$ You shouldn't need to change anything if you have wiring parts and have wired it up like the diagram. I have used several of the fan793x parts and so far they've all behaved as expected. \$\endgroup\$
    – David Molony
    Oct 19, 2022 at 7:23
  • \$\begingroup\$ @DavidMolony does yours also output a voltage level when you the HIN pin is applied a LOW voltage? I think my problem might just be I need higher rated Vgs MOSFETS instead of mine which turn on at 4 volts because there is always atleast 2.5 volts being applied to my MOSFETs. \$\endgroup\$
    – Trev347
    Oct 22, 2022 at 14:48
  • \$\begingroup\$ When Hin is low, the Vgs (differential measurement between the gate and source of the high side MOSFET/IGBT) should be 0V. If it's not, you're either measuring wrong, have blown parts or wrong circuit. \$\endgroup\$
    – David Molony
    Oct 24, 2022 at 1:27
  • \$\begingroup\$ @DavidMolony What about your capacitor values? What do you use for the Cboot and bypass capacitor? \$\endgroup\$
    – Trev347
    Oct 24, 2022 at 2:29

1 Answer 1

Reset to default
1
\$\begingroup\$

This is because of the current consumption by the gate driver high side. It (the switch node and consequently the gate voltage) rises until the bootstrap supply voltage minus the switch node voltage causes it to reach the under voltage lockout, at which point they consume far less current.

It's normal on these half bridges, different gate drivers result in different floating voltages.

I've got 4 or 5 different designs like this and the switch node voltage ranges from 2V to about 9 or 10V.

Your description of the behaviour is a bit odd, can you clarify what the switch node does when you pwm it?

\$\endgroup\$
5
  • \$\begingroup\$ Is there a way to stop it from floating this high? Or do I need to buy a different IC? The problem I am having is it is turning on all the high side mosfets when they shouldn't be on in my circuit. I am combining 3 of these IC's in a BLDC motor circuit. I am considering buying a IR2101 instead. \$\endgroup\$
    – Trev347
    Oct 18, 2022 at 6:15
  • \$\begingroup\$ Try putting a 10k resistor from gate to emitter of the IGBTs. You can use a lower value resistor if needed, but that will require more current for the bootstrap circuit. Also, maybe a resistor from Vs to GND. \$\endgroup\$
    – PStechPaul
    Oct 18, 2022 at 6:30
  • \$\begingroup\$ Trev, to be clear, are you also posting the Lin pin? You have to toggle that one in antiphase (with dead time) to the Hin pin to allow the bootstrap capacitor to charge \$\endgroup\$
    – David Molony
    Nov 7, 2022 at 13:30
  • \$\begingroup\$ @DavidMolony I actually just figured this out yesterday. I am planning on PWMing the HIN and LIN pin on the first FAN7392 IC with 2 different pins that will be in sync together, then use an inverter on one of them and somehow implement a 1 us dead time when switching on / off. Then for the 2nd FAN7392N model I will apply a LOW signal to HIN and a HIGH signal to LIN. This is the correct way to use this right? \$\endgroup\$
    – Trev347
    Nov 10, 2022 at 22:36
  • \$\begingroup\$ You should find a microcontroller that implements complimentary pwm with dead time. They're very common. Stm32 for example, most of them have a timer1 peripheral where you can program all this and more with minimal effort. \$\endgroup\$
    – David Molony
    Nov 11, 2022 at 5:46

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.