Selection

RDS(on) vs Gate Charge: Which Loss Dominates and When

Conduction loss scales with RDS(on), switching loss with gate charge and frequency. A 48 V, 20 A buck stage crosses from one to the other near 100 kHz.

OCENME Engineering · Published November 15, 2026 · 6 min read

Short answer

Conduction loss (I²·RDS(on)·D) dominates below roughly 80–100 kHz, switching loss (½·Vin·Iout·(tr+tf)·fsw) dominates above it. In a 48 V to 12 V, 20 A synchronous buck, conduction is about 0.8 W and switching about 1.9 W at 200 kHz — the crossover sits near 100 kHz.

Key points

  • Conduction loss is set by on-resistance and duty cycle; switching loss is set by voltage, current, edge times and frequency
  • The two losses cross over near 100 kHz in a typical 48 V to 12 V, 20 A stage — below that, RDS(on) decides; above it, gate charge decides
  • RDS(on) × Qg (the figure of merit) only ranks parts correctly at one operating point; check it at your own current and frequency
  • RDS(on) rises 40–80% from 25 °C to 125 °C, so a quote-sheet comparison at 25 °C can pick the wrong part

Two MOSFETs sit on the same quote sheet. One has 8 mΩ on-resistance and 60 nC gate charge, the other 12 mΩ and 22 nC. Which one runs cooler in your converter depends entirely on switching frequency, and the datasheet will not tell you which to pick. This article gives the two loss equations, the frequency where they cross, and the cases where the simple comparison misleads.

The two loss terms

Conduction loss is the easy one. While the device is on, it behaves as a resistor:

P_cond = I² × RDS(on) × D

where D is the duty cycle. In a synchronous buck the high-side duty is Vout/Vin, so a 48 V to 12 V stage has D = 0.25, and the low-side device carries the complement, 0.75. The low-side FET usually dominates total conduction loss even though it never sees the full input voltage, because it is on three times as long.

Switching loss is the overlap between voltage and current during each transition:

P_sw ≈ ½ × Vin × Iout × (tr + tf) × fsw

The factor of ½ comes from treating the voltage-current crossover as a triangle. Multiply by two transitions per cycle and you have the loss in one device; in a synchronous stage the low-side body diode and its reverse recovery add a further term that becomes significant above 200 kHz.

Add gate drive loss, Qg × Vgs × fsw, and you have the whole picture. At 200 kHz with a 30 nC part driven at 10 V, that last term is only 0.06 W — small, but it is why gate charge still matters when you push frequency.

Where the crossover sits

Take a real example: 48 V input, 12 V output, 20 A, high-side device with 8 mΩ at 125 °C, rise plus fall time of 20 ns, 30 nC of gate charge.

Switching frequencyConduction lossSwitching lossDominant term
65 kHz0.80 W0.62 WConduction
100 kHz0.80 W0.96 WBalanced
200 kHz0.80 W1.92 WSwitching
500 kHz0.80 W4.80 WSwitching, 6× conduction

Conduction loss does not move with frequency. Switching loss is linear in frequency, so it doubles at 200 kHz and multiplies by five at 500 kHz. The crossover in this stage is just above 100 kHz. Below it, every milliohm you remove from RDS(on) is worth real watts. Above it, you want the smallest gate charge and the fastest edge you can tolerate, even at the cost of on-resistance.

The same arithmetic at 20 V to 5 V and 3 A — a point-of-load converter — puts the crossover much higher, because current is small and V×I×t is small. Conversely, a 400 V stage at 10 A crosses over well below 100 kHz: high voltage makes every nanosecond of overlap expensive.

The figure-of-merit trap

Suppliers like to cite RDS(on) × Qg as a single number that ranks parts. It is a useful screen, but it assumes the two loss terms weigh equally, which only happens at one specific current and frequency. A part with half the RDS(on) and double the Qg has an identical figure of merit and will be better at 50 kHz, worse at 300 kHz. Ask for the figure of merit at your gate drive voltage, because Qg is quoted at 10 V and at 4.5 V and the ranking can flip between them.

Gate charge also has structure. The Miller plateau section — Qgd — is what actually sets the crossover time, while Qgs and the tail charge mostly set drive loss. Two parts with the same total Qg can have very different Qgd, and the one with lower Qgd switches faster at the same drive current. This is why the gate-charge curve in the datasheet is more informative than the single Qg number in the summary table; see how to read a MOSFET datasheet for the sections worth checking first.

Temperature changes the answer

RDS(on) is quoted at 25 °C and rises with junction temperature. For a 100 V trench device the multiplier is typically 1.5 to 1.8 at 125 °C; for a super-junction part it is closer to 1.3 to 1.5, but its RDS(on) also rises steeply at low current. Switching loss barely moves with temperature, so the balance shifts toward conduction as the die heats. A design that is loss-balanced at 25 °C will be conduction-dominated at 100 °C — which usually means the thermal design should be sized on the hot number, not the headline.

The families differ in more than temperature coefficient. Planar and super-junction parts trade on-resistance against charge storage in different ways; super-junction vs planar MOSFET covers which structure suits which voltage class. If you are still deciding between device types rather than between two parts of the same type, start with MOSFET, IGBT or SiC.

When to use this comparison — and when not to

Use the conduction-versus-switching split when:

  • You are choosing between two parts with different die sizes at the same voltage class
  • Your switching frequency is within a factor of two of the crossover point
  • The converter runs continuous current at a duty cycle between 0.2 and 0.8

Do not rely on it when:

  • The converter is in discontinuous or burst mode at light load — then fixed losses, output capacitance and controller behaviour dominate
  • The device is used as a load switch or in a hot-swap path, where switching loss is negligible and only RDS(on) and SOA matter
  • You are comparing across voltage classes, where the die structure itself changes

Limits of this estimate

These equations screen candidates; they do not replace measurement. They ignore reverse recovery of the body diode, which can add 20–40% to switching loss in a hard-switched synchronous stage at high current. They ignore package and layout inductance, which lengthens effective edge times well beyond the datasheet value — a 20 ns datasheet fall time commonly measures 35 ns on a real board. They also ignore the current derating that comes from the thermal path.

Treat the numbers as a way to eliminate the obviously wrong part, then build the two candidates and measure. If your application is a 65 kHz flyback, choose on RDS(on) and thermal resistance and stop reading gate charge curves. If it is a 300 kHz LLC or a 500 kHz POL, choose on Qgd and layout, and accept the higher on-resistance that comes with a smaller die. Our high-voltage MOSFET and low-voltage MOSFET families are specified with both numbers at 25 °C and 125 °C so the comparison can be made at the temperature the part actually runs at.

This article describes general manufacturing practice for machined and fabricated components. It is not a substitute for a drawing-specific engineering review — tolerances, material selection and process capability depend on the individual part.

Questions

Frequently asked

Is lower RDS(on) always better for a switching MOSFET?

No. Lower RDS(on) usually means a larger die, and a larger die carries more gate charge, so drive and switching losses rise. In a converter running above roughly 200 kHz, a part with 30% higher RDS(on) but half the gate charge often runs cooler overall. The right comparison is total loss at your switching frequency and your actual junction temperature, not the on-resistance in the headline.

How do I estimate switching loss before I have a prototype?

Use the rise and fall times from the datasheet gate-charge curves, not the switching-time table alone, and assume the current commutation and voltage commutation overlap. A first-order estimate of ½·Vin·Iout·(tr+tf)·fsw, plus gate drive loss Qg·Vgs·fsw, typically lands within 30% of bench measurement. Treat it as a screen to choose between two candidates, then confirm on a real board with a current probe and a thermal camera.

Does gate charge matter if I use a strong gate driver?

It matters less for switching loss and more for drive loss and EMI. A 4 A driver shortens the Miller plateau and cuts crossover time, but the charge still has to be moved every cycle, so Qg·Vgs·fsw is a hard floor on drive power. It also sets how much dV/dt and dI/dt you generate, which determines whether you need a snubber or a slower turn-on resistor.

Send the part number and the operating conditions

Tell us the voltage class, current, switching frequency and what the device is replacing. You will get a datasheet, a parameter comparison at matched conditions, and a quotation with MOQ and lead time.