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I am designing a class D audio amplifier using an HIP4080A integrated circuit to drive an H-bridge, simulating in Proteus. My problem is that the output pins of the integrated circuit, although rated for 2A or so, only supply 300mA (peak) into the gates of my power MOSFETs (this happens for both high and low side devices). Consequently, the transistors turn on very slowly and I get some really bad shoot through (500A for 100ns).

Things I tried:

  • Changing the gate resistors - current peaks at 300mA below 12 ohms
  • Lowering the frequency of my PWM signal
  • Ditching my PWM signal and using separate test sources at 600, 100 and 50 kHz, 50% duty
  • Modifying the bootstrap capacitors
  • A whole bunch of different MOSFETs
  • Supplying the IC with a higher voltage
  • Increasing the dead time via the HDEL and LDEL resistors
  • Lowering the high voltage supply from 75 to 50 V

Maybe this is just a glitch in the software?

enter image description here

VBAT is 12 and HV is 75 volts. The shortest pulses from the PWM signal are ~80 ns (the IC needs at least 50).

Any help would be much appreciated!

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  • \$\begingroup\$ Maybe the model is weak in this area? \$\endgroup\$ – Andy aka Jun 9 '18 at 15:08
  • \$\begingroup\$ That is indeed a possibility. If no other causes can be found I will try using a different software. \$\endgroup\$ – Andrei Jun 9 '18 at 17:12
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You are trying to go really fast! The specified turn-off delay time is 175 ns and the turn-on delay time is 40 ns. 80 ns pulses are going to be tricky if not impossible.

You will note that the HIP4080A gives has the following spec:

enter image description here

From this you can see that the output voltage drop is about 1 volt when the output current is 100 mA, implying an output impedance of 10 ohms. Since the gate threshold is between 2 and 4 volts, you will only be able to pull down at between 200 and 400 mA once you get to the Miller area (where the turn-off current is needed to avoid shoot-through). You just can't push these muscular FETs around with the driver in this device. To get to the rated sink current, you would need to have the operating point at (i.e. shorted to) the 12 volt rail. I give the spec writer some credit for pointing this out in the spec, but I can see that it is still a little misleading since there is no practical application I can think of that would operate here.

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  • \$\begingroup\$ So basically this driver can sink 2A in the same way that a power MOSFET is rated for 500W when perfectly mounted on an ideal, infinite area heatsink to keep the case at 25C, I guess? How disappointing... I would try connecting it to a totem-pole to increase the current but if it's too slow anyway there is no point. Thanks a lot! \$\endgroup\$ – Andrei Jun 16 '18 at 23:36
  • \$\begingroup\$ You are right. Good luck! But no matter how fast or slow you go, you will have to have a dead time adjustment in your logic to avoid shoot-through, because these type of FETs take longer to turn off than to turn on. \$\endgroup\$ – John Birckhead Jun 18 '18 at 13:11

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