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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,
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.
Solved! Go to Solution.
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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 greater than or equal to Qgtot.
Please let me know if you have further queries.
Regards,
Abhilash P
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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 greater than or equal to Qgtot.
Please let me know if you have further queries.
Regards,
Abhilash P