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Shoot-through 3 phase bridge? gate drive waveform

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ghoetic:
i did tests at both 125 and 230VAC. should a load really be neccesary?

jmelson:
I build both brush and brushless motor drives, using IR gate drivers.  One of the things about them is that they cannot stand the midpoint (U, V, W)
or VS1, 2, 3 going much below zero.  I had to put ultrafast diodes across the low-side FET to prevent that.  Now, in my case, it only was a problem with an inductive load.  So, assume you have the high-side transistor on, sourcing 10 + A current out to the load, and then shut off the high-side transistor.
The load inductance will try to continue accepting current from the drive, and the midpoint of the two transistors drops below ground.

It turns out the body diodes in these fets will allow forward voltages of 10-12 V for well over a microsecond before they start to conduct.  That's why an unltrafast diode is needed.  I use an ESD3 for this.

What happens is that as the midpoint drops to around -7 V currents start leaking all over the gate driver chip, and eventually, it turns on BOTH transistors at the same time.  Using a digital scope and creeping up on the current where this started to happen, I was able to capture the event without catastrophic failure.

Jon

ghoetic:

--- Quote from: jmelson on June 09, 2020, 04:41:59 pm ---I build both brush and brushless motor drives, using IR gate drivers.  One of the things about them is that they cannot stand the midpoint (U, V, W)
or VS1, 2, 3 going much below zero.  I had to put ultrafast diodes across the low-side FET to prevent that.  Now, in my case, it only was a problem with an inductive load.  So, assume you have the high-side transistor on, sourcing 10 + A current out to the load, and then shut off the high-side transistor.
The load inductance will try to continue accepting current from the drive, and the midpoint of the two transistors drops below ground.

It turns out the body diodes in these fets will allow forward voltages of 10-12 V for well over a microsecond before they start to conduct.  That's why an unltrafast diode is needed.  I use an ESD3 for this.

What happens is that as the midpoint drops to around -7 V currents start leaking all over the gate driver chip, and eventually, it turns on BOTH transistors at the same time.  Using a digital scope and creeping up on the current where this started to happen, I was able to capture the event without catastrophic failure.

Jon

--- End quote ---

Thanks for your input. This i will def keep in mind.
But i aint even there yet with load :(

duak:
Ghoetic,

Further to what jmelson points out, page 23 of the datasheet has another solution to Vs going negative by adding a series resistor Rvs and clamp diode Dvs.  link: https://www.infineon.com/dgdl/Infineon-IRS2334-DataSheet-v01_00-EN.pdf?fileId=5546d462533600a40153567aa9fe280b

Questions for JMelson:
1.) any thoughts on why the IGBT's diode is so slow to turn on?  I note that the data sheet does not show the forward recovery time and the diode's Vf graph shows a higher than expected voltage.
2.) have you observed this effect in other IGBTs?

Jay_Diddy_B:
Hi,

You may benefit form the resistor highlighted here:



There will be one resistor for each of the diodes D3, D4 and D5.

If U,V or W goes negative this voltage is added to the Vcc voltage in the boostrap circuit. There are internal clamps at 25V.

The series resistor will help prevent the bootstrap circuit peak detecting the negative spike.

They may simply damp the resonance between the bootstrap capacitor, and the wiring inductance. This would also cause over-voltage of the high side drivers.


Regards,
Jay_Diddy_B

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