Ok, I understand what you are saying... but how do i calculate the farad value for the capacitor?You want the cap to be large enough so that the required gate charge (from the data sheet) does not lower the cap voltage below 10V. You place C1 in parallel with the gate capacitance to determine the drop in voltage as the charge on C1 is now shared between C1 and the gate capacitance.
25V rating is fine for the cap.
VLS is the voltage drop across the transistor when it is on (don't know why they also say "or load"). Just multiply the maximum load current times the maximum ON resistance of the MOSFET (from the data sheet). That should be no more than a few tenths of a volt.
Yes, that is the Vgs voltage to fully turn on a typical (non-logic-level) MOSFET.Cool thanks man!
and the VMin = 10V, that is the 10V diff required to switch mosfet correct?
Quick question, u say tenths of a volt, where my answer is way higher..VLS is the voltage drop across the transistor when it is on (don't know why they also say "or load"). Just multiply the maximum load current times the maximum ON resistance of the MOSFET (from the data sheet). That should be no more than a few tenths of a volt.
Your calculations are correct. I said "should be no more than a few tenths of a volt" because you normally pick a MOSFET with a low enough ON resistance to keep the ON voltage low at the operating current to minimize transistor dissipation and maximize converter efficiency. Your transistor will be dissipating 38W when conducting 10A.Quick question, u say tenths of a volt, where my answer is way higher..
Final product will be a buck converter, which should be able to supply 10A. Thus using the Rds(on) = 0.38 ohm, value from the STP14NK50Z datasheet alongside the maximum amount of current the transistor can handle, Id (@25 degrees) = 14A. This results:
Vls = Rds(on) = Id = 0.38 * 14A = 5.32V.
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Ok thanks!Your calculations are correct. I said "should be no more than a few tenths of a volt" because you normally pick a MOSFET with a low enough ON resistance to keep the ON voltage low at the operating current to minimize transistor dissipation and maximize converter efficiency. Your transistor will be dissipating 38W when conducting 10A.
I suggest you find a MOSFET with a lower ON resistance to keep the ON voltage below a volt and the dissipation below 10W.
Edit: I would use a larger Cboot capacitor, since the value you calculated is the minimum.
If you peak current is 10A then Vls = 0.044 * 10A = 0.44V.I found a alternative: IRF540NSPbF
Thus my Vls = 0.044 * 33A = 1.452V
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YIP i totally agree, I just went with the absolute worst case, to see if everything is still fine. From what I could see in the MOSFET gate driver datasheet is that the high side and low side are connected to one another through the PULSE Generator block (Vcc passing through) on the high side... Now I hope I can just use the high side output without connecting anything to the low side output.If you peak current is 10A then Vls = 0.044 * 10A = 0.44V.
by Duane Benson
by Jake Hertz
by Jake Hertz
by Duane Benson