A power MOSFET looks like the perfect switch for a microcontroller: the gate draws no steady current, so a 5 V pin should be able to turn it on and off. What the pin cannot do is turn it on and off fast. The gate is a capacitor, and charging a capacitor takes current — far more current than a microcontroller pin can supply. This post works out the numbers for an IRFZ44N switching a 12 V, 5 A load at 20 kHz, and shows why a gate driver IC is not optional at that speed.

Figure 1: The gate is a capacitor (a), and charging it slowly costs watts (b). The same MOSFET and the same load lose 3.78 W from a microcontroller pin and only 0.52 W with a driver IC. (Schematic, not to scale.)

The gate is a capacitor

The IRFZ44N's datasheet gives a total gate charge of Q_g = 63 nC (at V_GS = 10 V). To turn the MOSFET on, that charge has to flow into the gate; to turn it off, it has to flow back out. The time each edge takes is roughly:

t ≈ Q_g × R_g / V_drive

A microcontroller pin can supply at most about 40 mA, so the gate resistor cannot be smaller than 5 V / 40 mA = 125 Ω. With a safe choice of R_g = 220 Ω the peak gate current is 5 / 220 = 22.7 mA, and each edge takes 63 nC × 220 Ω / 5 V = 2.77 µs. During those microseconds the gate voltage pauses at the Miller plateau while the drain voltage falls — the MOSFET is half on, carrying full current with full voltage across it, and that is where the heat comes from.

Worked example — switching loss at 20 kHz

Take a 12 V load drawing 5 A (2.4 Ω), switched at 20 kHz by the IRFZ44N (R_DS(on) = 17.5 mΩ). Three losses add up:

Conduction: P = I² × R_DS(on) = 5² × 0.0175 = 0.44 W

Switching: P = ½ × V × I × (t_r + t_f) × f = ½ × 12 × 5 × 5.54 µs × 20,000 = 3.33 W

Gate drive: P = Q_g × V_gs × f = 63 nC × 5 V × 20,000 = 6.3 mW

Total = 0.44 + 3.33 + 0.01 = 3.78 W — and 88% of it is switching loss. In a TO-220 package without a heatsink (about 62 °C/W to the air) the junction reaches 25 + 3.78 × 62 = 259 °C, far above the IRFZ44N's 175 °C maximum. The MOSFET would destroy itself — not because it is too small for 5 A, but because the gate was charged too slowly.

The fix — a gate driver IC

A driver IC such as the TC4420 sits between the pin and the gate and can deliver amps, not milliamps. With R_g = 10 Ω and a 12 V drive:

Peak gate current = 12 / 10 = 1.2 A  ·  edge time = 63 nC × 10 Ω / 12 V = 52.5 ns

Switching loss = ½ × 12 × 5 × 105 ns × 20,000 = 0.063 W

Gate drive power = 63 nC × 12 V × 20,000 = 15.1 mW

Total = 0.44 + 0.063 + 0.015 = 0.52 W, and the junction sits at 25 + 0.52 × 62 = 57 °C — safe with no heatsink. The edges are 2.77 µs / 52.5 ns = 53 times faster, and the total loss is about 7 times smaller, for the price of one small IC.

When direct drive is fine

Direct drive is not always wrong — it depends on frequency. At 1 kHz the same pin-driven circuit loses only ½ × 12 × 5 × 5.54 µs × 1,000 = 0.17 W in switching, for a total of 0.61 W and a junction temperature of about 63 °C: perfectly fine. The rule of thumb is: below about 1 kHz, or for a simple on/off switch that changes state a few times a second (where switching loss is near zero), a pin and a 220 Ω resistor are enough. Above that, or whenever the load current is large, use a driver IC — or at least a small push-pull transistor pair — and keep the gate resistor small.

Key points to remember

• A MOSFET gate is a capacitor: the IRFZ44N needs Q_g = 63 nC of charge per switching edge, and the edge time is roughly Q_g × R_g / V_drive.
• Worked example: a 5 V pin through 220 Ω gives 22.7 mA peak and 2.77 µs edges; at 20 kHz and 5 A the switching loss is 3.33 W, total 3.78 W, junction ≈ 259 °C — the device fails.
• With a driver IC (R_g = 10 Ω, 12 V drive): 1.2 A peak, 52.5 ns edges, switching loss 0.063 W, total 0.52 W, junction ≈ 57 °C — safe with no heatsink.
• Conduction loss (0.44 W) is the same either way; the difference is entirely in how fast the gate is charged.
• At 1 kHz the pin-driven circuit loses only 0.17 W switching — direct drive is fine for slow switching, but not for PWM at tens of kilohertz.
• A microcontroller pin must never drive a gate without a resistor: 5 V / 40 mA = 125 Ω minimum, and 220 Ω is a safe practical choice.