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ISO5852SDWEVM-017: the isolated power rating for gate driver

Part Number: ISO5852SDWEVM-017

Hi,

In ISO5852SDWEVM-017, the isolated power for gate driver is generated through SN6505, in SN6505 datasheet, its output drive is only 5V 1A.

So I think the power generated can only be less than 5W?

But the source and sink current in the reference design is 20A? Gate drive voltage could be 12V & -5V.

with 12V, 20A, the power rating is 240W!

I know 240W is not constant power, but how do we know whether the 5W SN6505 can handle 240W transient power consumption?

  • Hi Howard,

    I've asked Rais to take a look at this next week, but I think this is due to the fact that the 20A isn't dissipated. You are just moving charge back and forth from the decoupling capacitor to the gate capacitor, so the total current used is much less than the 20A.

    Rais will give you a definitive answer next week.
  • Hi Howard,

    As Don Dapkus has mentioned, driving MOSFET or IGBT is based on charging and discharging of their input capacitance. This capacitance is usually defined in datasheet through plots Qg vs Vgs. Because of capacitive nature of the load, the maximum drive current happens only at the beginning of charge/discharge cycle and the gate current reduces to zero, when gate input capacitance is fully charged/discharged. Thus, SN6505 should be able to provide average drive power only, while peak power during switching time is provided by decoupling capacitors at drive rails, that could be 12V and -5V as in your example.

    To determine average drive power needed, we consider total charge from datasheet plots Qg at drive voltages, drive voltages Vdr and switching frequency Fsw. Assume, Qg = 100 nC, Vdr = (12+5)V or 17V and Fsw = 100 kHz. This gives us 100 nC x 17 V x 100 kHz = 170 mW. Please notice that this average power is distributed between driver IC and gate resistors. The power dissipated by SN6505 is determined by its efficiency at average drive power. Assuming efficiency is 70%, the dissipated power at SN6505 will be 170mW x (1/0.7 - 1) = around 73 mW. Most driver datasheets show how to calculate average drive power dissipated in each component.

    To handle the peak drive power, the decoupling capacitor needs to be chosen at least one order larger than the input capacitance of driven power device. This assuming that the ripple at decoupling capacitor will be 10% or less.

    Regards,

    Rais

  • Rais,
    thanks. Could you please tell how Power= Qg* Vdr* Fsw is derived?
    Charge back and forth the cap should not consume any power at ideal case.
  • Hello Howard,

    Charging the capacitor from voltage source through non-reactive circuit and discharging the capacitor through non-reactive circuit is always the process with energy loss. During charging, the quantum of energy is defined as Qdv. For linear capacitor Q = CV. Thus the quantum of energy is dE = C*V*dv. After integration, the energy of capacitor is E=C*V*V/2, or E=Q*V/2. At the same time, the total energy taken from the voltage source to charge the capacitor is Q*V. Thus in case of linear capacitor, half of total energy is used to store in capacitor and another half is dissipated in non-reactive circuit between voltage source and capacitor. During discharge, the energy stored in capacitor is the loss in non-reactive circuit. Thus, the energy loss during charging and discharging cycle is Q*V, while power loss is going to be Q*V*Fsw.

    The case of ideal lossless charge and discharge of capacitor can be only if additional inductance is used. In this case it becomes resonant charge and discharge of capacitor.

    This is not our case, when simple driver IC is used to drive power device.

    If you have further questions, please contact me directly. There are text books available describing the circuit behavior with capacitors in details.

    Regards,

    Rais