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مثال The driver must source 1 A at the start of each edge, and tr + tf = 60 ns (estimated from Qg and Rg) — gate drive power 30 mW, which the driver and Rg burn, not Q1.

مثال Q1 itself dissipates 995 mW = 720 mW switching + 275 mW conduction (72.36% of it is the edges); 1.025 W once the gate drive is counted too. Rg is the knob between the two failure modes: more of it damps the gate loop but stretches the edges — at 20 Ω the switching loss doubles to 1.44 W — and less of it doubles the drain's dV/dt and lets the gate ring against the loop inductance — at 5 Ω it halves to 360 mW. Conduction is the temperature-sensitive half: Rds(on) is roughly 1.5× higher at 100 °C, so budget 412.5 mW for it.

A MOSFET draws no steady gate current, so holding it on is free — but every edge has to shovel the whole gate charge in and back out again, and $P_{\mathrm{gate}}=Q_g\times V_{\mathrm{قيادة}}\times f_{\mathrm{sw}}$ is real power out of the driver's supply. The mistake is then treating the gate resistor as free too: it is the only damping in the gate loop, and it is also what stretches the edges that the switching loss is directly proportional to — $P_{\mathrm{sw}}=\frac{1}{2}{V}_{\mathrm{DS}}{I}_D(t_r+t_f)f_{\mathrm{sw}}$ — while the part only earns its keep conducting, at $P_{\mathrm{cond}}={I_D}^2{R}_{\mathrm{DSon}}D$.

With no measured edge times, the estimate comes from the gate loop itself: the driver starts at Ig = Vdrive/Rg and sags towards zero as the gate fills, so it averages about half of that, and moving Qg at that average takes $t_r+t_f\approx \frac{2Q_g{R}_g}{V_{\mathrm{قيادة}}}$. It is a first-order figure — the real Miller plateau is what sets the drain's transition — so put your datasheet's or your scope's numbers into the optional field whenever you have them.