Gate Driver Optocoupler HCPL3120 (or FOD3180) RG ???

Hi
I'm confused when i read HCPL3120 (or FOD3180) mosfet gate driver optocoupler,
the term RG (external gate resistor) calculated based on ohms law and for HCPL3120 it's around 10ohm for typical app. but, based on datasheet, the RG calculation done for 2.5A output current through the RG (p. 17), then the 10ohm RG must deliver 2.5A*2.5A*10ohm = 62.5Watt!!! of power, and such a resistor is so huge!!
however in normal mosfet gate driver RG is around Kohm, but for fast switching the RG must be smaller using HCPL3120, but what is the RG power in this opto-gate-driver?

this seems stupid, am i in deep mistake??!! plz help

that calculation is probably correct, but how long does the opto have to deliver that power? once the mosfet is ON, it does not take any current to maintain the ON state. so, how long does it take for a mosfet to turn ON? maybe 100ns?

so the question is, can the opto deliver big power for an ultra short period of time? the answer is YES.

and in 100ns how hot will the resistor get? not very.. maybe if it is switching 100kHz then it is on/off and during each transition current flows through it.. so now you must use frequency as part of the calculation and also duty cycle. .. then calculate the average power... then double or quadruple it and that is how you size the resistor. gate resistors are often quadruple the calculated value because these resistors are hard to cool.. so you need large thermal mass and big footprint for conduction cooling instead.

Mr.Cool

Thanks for reply Mr.Cool,

datasheet says max switching freq is about 2mhz (500ns); with 80% duty cycle, gate resistor (even big one) will become extremely hot, and even burn. but as i know this opto-gate-driver is used for normal application not for high current switching application, so a big resistor and/or big footprint probably is not the solution.

thanks again.

any idea dudes?

Your gate charge Qgg should be spec'd. Your current is
Qgg*fSW but double that for power dissipation calcs
(in, out, both shed power). How does (2*Qgg*f)^2*Rg
look?

yes you are right, current of 2Mhz (max for HCPL3120) switching for 30nC gate charge is around 60mA (30/500),
and for some IGBTs is about 1.5A. but the problem is not about mosfet gate charge, it's about HCPL3120 datasheet calculation
around RG. (RG=(VCC-VEE+VOL)/2.5A) it's assumed that 2.5A flow through the resistor!!! something is wrong or i am misunderstood??

You have to determine whether this is peak, average or
DC. Or this could be expressing the test conditions for
measuring RG and have nothing to do with application-
realities at all.

You correctly calulated an instantenous power of 62.5 W for a 10 ohm Rg. It corresponds to a maximal pulse energy of a few µW and an average power of some mW up to several 100 mW, depending on the exact current waveform and the switching frequency. In so far, at least the average power is moderate.

But it's also correct to care for the resistor pulse handling capability. If you review respective data sheets and application notes (you won't find plenty of them, e.g. Vishay or Yageo have some data), you'll realize that standard chip resistors have a rather limited pulse handling. But for usual power MOSFET/IGBT gate drivers up to kW switched power, 1206 gate resistors are usually sufficient and MELF or 2512 chips for high power (100 kW range).

yes this is the peak, but i probably made a mistake about power loss in Rg as FvM mentioned.
so i will perform some sort of tests with HCPL3120 and 30nC Qg mosfet.
the resistor package are also important, i have 1206, 805 and 603, and 1206 obviously will be the choice.

thanks all.

The other point is, that you would need a supply voltage of 30V to actually achieve 2.5A peak current with 10 ohm Rg. A supply above 15V would be only reasonable for a bipolar gate driver, which is rarely used for MOSFET. Also the transistor gate charge has to be considered to calculate the actual resistor pulse energy. But I guess, 1206 should be O.K.