directly driving the IRFP240 n 9240 using the LME49810 or 811

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but how much in general to be used.. I have checked the following website

Using HEXFETs in High Fidelity Audio

but it states that according to the Gate capacitance the calculated gate current is 350ma? isnt it too high? I never saw any IRF240 based amplifier to have such a high gate current for the output stages...

so if that the case one irf610 can drive may be at the max only 2 or 3 IRF240s...

what im trying to do is i would like to use a single lme49830 without a driver to drive the IRFP240 for low power amplifiers like hardly 50w-100w class AB or the reason behind it is i want to reduce the number of components in the path.
350mA? I think that is the bias of the output transistor. Unlike BJT's, Mosfets in general have a widely varying transconductance (Gm). Gm decreases significantly as the device conducts less Id, say under a few hundred mA. Rod is saying it is important to bias these transistors a 350mA in order to lessen the distortion effects of the drop in Gm around the current crossover. (current and voltage may not crossover at the same time because of speaker reactance)
The gate does not require current like the base of a BJT but rather charge as in like charging a capacitor, and this is dependent on frequency. If you look at a complementary source follower output you see that Cgs is bootstrapped by the output. This means the only voltage change that Cgs sees is the change in Vgs that is related to the change in Id. The real internal capacitance that is the problem is Cgd, but this only comes into play at clipping because otherwise it is much smaller. 15-20mA of current in the driver stage per pair is plenty.
I would say it probably will work fine but if there is slew limiting or crossconduction due to slow turnoff, a driver stage could be used. You would want to drive both devices in a pair like in the app note with the Vbe (Vgs) multiplier. Vertical fets need thermal compensation as to BJTs.
240 or 9240


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