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Here is the exact schematic I am using:

enter image description here

Here is the schematic they recommend in the datasheet:

enter image description here

I am trying to drive my dual N-channel MOSFET circuit with a FAN7392N IC. The problem I am having is that I can't drive the upper MOSFET's gate pin by applying a voltage to the "HIN" pin with my Arduino Nano.

If I apply a high signal to the "HIN" pin, the "HO" pin doesn't respond. If I apply a low signal to the "HIN" pin, the "HO" pin doesn't respond.

If I apply a high signal to the "LIN" pin, then the "LO" pin turns on; if I apply a low signal then the "LO" pin turns off. The bottom driver works the way I want it to, but the top driver won't respond to my Arduino signals.

Does anyone know how to properly use a FAN7392 driver IC so I can drive the high-side MOSFET with my Arduino?

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    \$\begingroup\$ What is the Vb-Vs voltage? \$\endgroup\$
    – Tim Williams
    Sep 10, 2022 at 6:55
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    \$\begingroup\$ Are you aware that bootsrapping like this only works by switching Q2 and maintaining some resonable duty-cycle? \$\endgroup\$
    – winny
    Sep 12, 2022 at 15:54

2 Answers 2

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Don't add resistors as R6. What is R4? The load?
Be aware that the upper MOSFET is driven with a charge pump (R1, D1, C1).
So you must drive the upper MOSFET with a PWM waveform, no static voltage.
See this link for why bootstrapping is needed and how.

enter image description here

See also the end of the paragraph :

If the bootstrap capacitor is selected carefully, it will have enough charge stored on it to charge the gate to source capacitor of the MOSFET. However, it will be impossible to provide the continuous gate current required to drive a BJT by using the charge stored on a capacitor, without discharging it completely. This is why the bootstrapping technique cannot be used to drive a BJT.

Only the lowest MOSFET can be driven with static voltage.

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  • \$\begingroup\$ Yes R4 is just the load. You are saying don't add resistors R6 & R5, but PStechPaul's answer says I should keep them? Which one is right haha? \$\endgroup\$
    – Trev347
    Sep 10, 2022 at 18:41
  • \$\begingroup\$ Also could you explain in a little more detail why the top mosfet has to be driven with a PWM signal? Why can't I use static DC voltage? \$\endgroup\$
    – Trev347
    Sep 10, 2022 at 18:43
  • \$\begingroup\$ Ok for R6 & R6, I should not add them because the typical application does not use them. But this is up to you. PWM for the topmost MOSTFET is needed, because of the way that the driving voltage is obtained. If not, there should be no need of D1 and C1. C1 is charged at every cycle of PWM (bootstrap stage) ... and so, voltage driving Q1 is enough for "conduction". Must see into the CI for explaining all. Will search after this. \$\endgroup\$
    – Antonio51
    Sep 12, 2022 at 15:29
  • \$\begingroup\$ Here is some explanation of HOW "bootstrapping" works : a bit complicated, see paragraph at end of web page ... sites.google.com/site/simpletronics/Home/knowledge-centre/… \$\endgroup\$
    – Antonio51
    Sep 12, 2022 at 15:46
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It might be possible to use an isolated DC-DC converter or a small 15V switching supply to provide a gate drive voltage on Vb and Vs (across Cboot).

Otherwise, as suggested, use PWM signals to operate the bootstrap charge pump.

R5 and R6 should be OK to make sure the gates don't float high. R2 and R3 probably should be more like 10-47 ohms, depending on gate capacitance and switching speed.

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  • \$\begingroup\$ I know typically it's better to use a low ohm resistor value for the gate drive, but then I thought about the watts dissipated and I figured a higher value would probably be better. If I use a 10 ohm resistor then the watts dissipated would be V/R = I - 10 V / 10 R = 1A, V*I = P - 10V * 1A = 10W. Isn't 10W quite excessive for a gate drive resistor? Especially because I need 2 of them. Wouldn't it be better to use a higher value resistor so I can save more power? \$\endgroup\$
    – Trev347
    Sep 10, 2022 at 18:39
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    \$\begingroup\$ The high current only lasts long enough to charge the gate capacitance, and that would be a few microseconds at most. Higher value resistors will just slow down the switching of the MOSFET and that can cause high power during the transition. \$\endgroup\$
    – PStechPaul
    Sep 11, 2022 at 3:05
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    \$\begingroup\$ that makes sense thanks for the answer! I'll change it to a lower value resistor. \$\endgroup\$
    – Trev347
    Sep 11, 2022 at 21:08

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