Electrical – help need in driving high side nmos without bootstrap circuit

bootstrapmosfet-driver

I am currently planning to upgrade an existing circuit which charges a lead acid battery using a non-synchronous buck topology.
The NMOS which i have used is IRFP4768PbF.Currently my driver is the IR2110 which uses a bootstrap capacitor.

I would like to remove the IR2110 and replace it with an opto isolated gate drive(TLP5701) with a dedicated dc/dc converter.
I have drawn up a shematic and i have a few quiries in it ,regarding he dc/dc converter.

enter image description here
1.I am going to use the MEU1S1212ZC(12v/83mA).The input to this module will be referenced to the system groung(solar-) and the output will be referenced to the source of the MOSFET.
The the current sufficient enough to grive the MOSFET?

2.What should be the values of R1 and R2.Is 10 ohm(R1) and 3k(R2) sufficient enough?

Edit:
I have sketched the DC/DC converter section also to avoid any confusion.
enter image description here

Best Answer

Energy stored in a cap = QV/2.
Power needed to repeatedly charge a cap from a certain voltage:
P = freq * QV/2 * 2.
Average current needed: I = P/V = freq*Q.

Using the total gate charge Qg = 180nC from the datasheet (even though Qg depends on the actual gate and drain voltage), and 10KHz:
I = 180nC * 10kHz = 1.8mA
which is not much.

With the amount of current being switched at the load, it is easy to guess that minimizing the switching loss of the MOSFET is important. Just guessing a number to get some ideas, say 90ns for the gate charge/discharge time.

The gate drive charge/discharge current needed would be Qg/t (at 10V Vgs, same as the condition given in the datasheet): Igs = Qg/t = 180nC/90ns = 2A.

There is a big difference between the average and the gate-charge current because the duty cycle of gate-charge to total time is 90ns/100us = 0.0009. 100us comes from the inverse of 10kHz.

Some conclusions:

  • A DC/DC converter of 83mA average current should be more than enough.
  • Enough bypass capacitance would be needed to supply the gate-charge switching current of thousand times that of the average.
  • Driving the MOSFET directly with a optocoupler may provide only a fraction of the drive current wanted. The drive current probably should be in the amps range. But the actual required depends on the optimization of the MOSFET switching which takes much more to work out.

Edit to include a way to estimate bypass capacitance on floating supply:

Take the Input Capacitance from the datasheet, e.g. Ciss = 10880pF. If you want the voltage ripple caused by charging the base to be less than 1:100 on the supply, then use a bypass cap that is 100*Ciss*2 (the times 2 is for extra margin for output capacitance of MOSFET), that is 2.2uF in this example. It is a lot less than my impression when I wrote the above. Use ceramic capacitor for low ESR and high current capability.

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