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Gate Driver ICs Forum Discussions

User20899
Level 1
Level 1
I am using the IPP016N08NF2S MOSFET in a synchornous buck converter. I am willing to use the 2ED2183S06F gate driver IC and i am not able to figure out what should be the bootstrap capacitor.

As per the datasheet(https://www.infineon.com/dgdl/Infineon-2ED2183-4-S06F-J-DataSheet-v02_21-EN.pdf?fileId=5546d4626cb27..., page 14 ) of the gate driver,
5189.attach

I have confusion regarding the ∆VBS is the maximum allowable voltage drop at the bootstrap capacitor within a switching period, typically 1 V.
But if we use the formula of the ∆VBS ≤ (VCC – VF– VGSmin– VDSon), It is impossible to get the 1V for me.

How to deal with that.
I am supplying around 15V to the gate driver. It must have a maximum duty cycle of 99.99% and switching frequency of 100KHz.
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1 Solution
Abhilash_P
Moderator
Moderator 25 likes received 250 replies posted 250 sign-ins
Moderator
Hi,

After considering the formula from the datasheet and the variables that you have mentioned, I could arrive at Vbs<1V

Qg = 170nC
Qls = 1nC
Iqbs = 170uA
Ilkgs = 10nA
Ilk = 1uA
Thon = T * D = 9.9uS

Qgtot = 170n + 1n + (170u + 10n + 1u)*9.9u
19.692uC

Hence to get Vbs <1v, the bootstrap capacitance must be less than or equal to Qgtot.

Please let me know if you have further queries.

Regards,
Abhilash P

View solution in original post

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Abhilash_P
Moderator
Moderator 25 likes received 250 replies posted 250 sign-ins
Moderator
Hi,

After considering the formula from the datasheet and the variables that you have mentioned, I could arrive at Vbs<1V

Qg = 170nC
Qls = 1nC
Iqbs = 170uA
Ilkgs = 10nA
Ilk = 1uA
Thon = T * D = 9.9uS

Qgtot = 170n + 1n + (170u + 10n + 1u)*9.9u
19.692uC

Hence to get Vbs <1v, the bootstrap capacitance must be less than or equal to Qgtot.

Please let me know if you have further queries.

Regards,
Abhilash P
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