UCC27211 : ucc27211

Hello Do Seoung,

I will be happy to answer your question, but the picture of the schematic did not show in the last message. You can try inserting a file, or picture in the next message.

Regards,

Hi Do Seoung,

Thanks for reaching out and thanks for the schematic!

You will want to place the resistor in location A, so that the bootstrap capacitor is still directly referenced to the switching node.  This is to make sure you have sufficient V_GS on the high-side FET.

A resistor value you can start with can be similar to your gate resistors, around 10 Ohms.  Also, note that when you add a resistor on the HS pin, you will have increased resistance on the turn-off path of the high-side FET (therefore increasing the switch-off time), but having the turn-off diode on the HO gate helps.

One more thing to consider (if you are having large negative voltage spikes on your switching node), is to put a clamping diode from the HS pin to GND, along with the resistor (as shown below).

I hope this helps and answers your question!  If so, please press the green button, otherwise feel free to follow up!

Thanks,

Aaron Grgurich

Hi Do Seoung,

Sorry for the image problem, I fixed it above.

Regardless of the of the additional components, you should consider output loads, switching times, etc. to see how your system will react to the load and how efficient it will be.

I hope this answered your question; if so, please press the green button, otherwise feel free to follow up again!

Thanks,

Aaron Grgurich

Hi Do Seoung,

You are welcome!

Regarding your questions:

1)  I only suggested the resistor or resistor-diode combination, assuming you had large negative voltages spikes on your switching node that was damaging the part.  Otherwise, the additional circuitry is not needed assuming your circuit stays within the recommended operating conditions in Section 7.3 of the UCC27211 datasheet.  So, if your switching node voltages stay within the specifications below, the additional circuit is not suggested.

The technical reason is that excessively large negative voltage spikes on the switching node (HS pin) can damage the part.  The resistor on the HS pin allows for some voltage to be dropped across it, therefore less voltage is seen at the pin; but this comes at a cost of limiting the current in the turn-off loop.  The diode works by shorting the HS pin to ground when the switching node falls below ~ -0.5V (a diode drop).  The resistor is still included with the diode to limit the in-rush current into the diode, so it does not burn up.

2)  Sorry, I do not understand what you are asking here...  If you are asking about what gate resistor value to use, please keep in mind that the gate resistor value affects the FET turn-on speed, power losses, and etc. If the gate resistor is too big, the FET cannot turn-on fast enough which will then affect the switching frequency range.  I think a good starting point is somewhere between 0-10 ohm, depending on the gate charge of your FET's (because this will dictate how much charge is needed to turn on and off the FET.  If able, bench-top testing of your system will be a good start.

3)  Yes, and for your current schematic, I overlooked the external bootstrap diode you have.  The external bootstrap diode is not needed for both parts (UCC27211 and UCC27282), because they both have internal bootstrap diode (Please refer to their respective datasheets).

Also, if you were to use the UCC27282 instead, you would have to consider the extra feature of the ENABLE pin, which must be pulled high in order to enable the driver (again, is specified in the data sheet).

My final comment is to again say that the HS pin does not need the complication of adding the protection circuit if you system will not have abnormal levels of negative voltage spikes (these specifications of how long and how large the spikes can be, are again, specified by the datasheets.)

I hope this answered your questions!  If so, please press the green button.

Thanks,

Aaron Grgurich