fet driver supply current

I need some help understanding how much current a MOSFET driver would need.
I saw that for example a half bridge driver can source on (external ) mosfet gate 3-4 Amps (for short duration , in order to charge the gate capacitance), although the input current through the supply pins is far less than that;

What would be the power needed by a driver which have to drive external transistors?
I now that this power is proportional to the switching frequency , the gate capacitance size, and the peak output gate current capability of the driver, but i don't have a formula for that( even an approximate one)

Let s say for example a half bridge driver powered from +15V , namely FAN7390 :4.5A source/sink
If I would design the power supply which generates the +15V needed , how much current will be needed? It will be in range of 100mA? 200mA? 500mA? (Regardless of Mosfet driven )

How about IR21363 : ~0.4A source/sink; +15V supply. What will be power requirement aproximately?

You can estimate the gate drive power from the energy stored in the gate (in Joules) multiplied by the switching frequency in Hz.
The gate discharge energy does not come from the gate supply, only the gate charging energy .

You will sometimes find that the actual dc power to the gate driver chip can be far higher than that.
The reason being that to get very fast switching, the upper and lower gate driver devices often slightly overlap in conduction, creating a short cross conduction current spike.

You can test for that by running the gate driver chip without any external load, and measuring its dc supply current. First not switching, then at the anticipated switching frequency.

Your best bet though is to actually measure the dc supply current under normal expected operating conditions. Then you will know for sure.

Nowadays drivers have built-in dead time generators so cross-conduction it is not a problem

Testing I can't do right now because I am at the design stage...

But for all reasons something tells me that 0.3A dc supply current under normal conditions should be enough. Am I wrong?

I suggest you take 15 minutes to breadboard your driver hooked up to a function generator and do the test, otherwise you are just guessing.

And you can do it by the Qg value , and get the current from Qg * fsw.....make sure Qg is for the gate drive voltage you are using and for the same VDS swing.........you can also use the Qgs/2+Qgd if you dont have Qg....but same applies to the voltages etc.

or...have an engineering guess at Cgd.....abd do a q=cv calcualtion to get that, minding your voltage swing etc...and Cgs can be approx'd as Ciss.....so do another q=cv calc there.....remember that the cgd is only getting the miller charge between vfh and vmiller
where
vmiller = vth + I/transconductance.

if this isnt clear i suspect it isnt, i will send some links soon.

you could always change to a fet which gives you the info you need to easily calculate gate drive power.

ultimately the equations you use are just

q=cv
q=it
i = cdv/dt
energy = 0.5*c*v^2
power = energy * time

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the actual average current to drive the fet on doesnt depend on the peak current of the driver....and remember that the discharge of the fet gate heats up the driver , but doesnt need current to be drawn from the supply as its discharge of what the supply has already supplied.

No it is not totally clear;
My IGBTs have 84nC (i use 3 phase bridge- 6 igbts)
I remember a basic formula for an aproximation of the current ;
i*t=c*u so i=c*u/t
for 16kHz, 84nC, 15V Uge -> i=(84nC*15V)/ ( 0.1*(1/16000) )=0.2A
(assuming rise time is 0.1 * switching period)
And multiplying by two , beacause there are two igbts conducting allways in a period , i came up to 0.4A
Seems reasonable?

While driver designs try to minimize crossover / shoot-through
current, these are never zero and there's a component from
the predriver chain that just won't go away. So expect some
internal CVf current (which ought to be rated, you'd think).

Your MOSFET / IGBT Qgg should contain some Miller term
which varies with working voltage. This may differ from the
component datasheet conditions. Maybe you can find a
family of Qgg vs Vds(off) curves to use in finer figuring.