High-Side N-MOSFET Actuator Driver Overheating
A production batch of 24 V industrial actuator boards uses an N-channel MOSFET as a high-side switch. The 24 V rail feeds a 2 A resistive-actuator load; the MOSFET source connects directly to the load, and the drain to +24 V. A comparator pulls the gate to 24 V through a 10 kΩ resistor when the command is asserted; a 100 kΩ resistor holds the gate low when unasserted. The MOSFET's datasheet specifies R_DS(on) = 8 mΩ at V_GS = 10 V, 12 mΩ at 4.5 V, V_GS(th) = 1.0–2.5 V, and a maximum gate-source rating of ±20 V. On the bench, the board delivers only 20.8 V to the load at 1.9 A, and the MOSFET tab reaches 95 °C within two minutes. The comparator output is a clean 0–24 V square wave. The layout is compact, and the gate trace is short. The customer requires full 24 V at the load and a MOSFET case rise below 30 °C.
1. Explain why the load receives only 20.8 V and why the MOSFET heats up. Include an operating-point calculation and state why the 8 mΩ R_DS(on) specification does not apply.
Read the worked answer
The MOSFET is wired as a source follower, not as a low-side or fully enhanced high-side switch. The gate is fixed at 24 V, while the source is the load node. For 1.9 A, the gate-source voltage needed to support that current is roughly 3.2 V, so the source settles at 24 V − 3.2 V ≈ 20.8 V. The drain-source voltage is therefore about 3.2 V, and the dissipation is P = 3.2 V × 1.9 A ≈ 6.1 W. The 8 mΩ value is specified at V_GS = 10 V in the ohmic region; here V_GS is only about 3.2 V, so the device operates in saturation. The 10 kΩ gate resistor causes no DC voltage drop because gate current is negligible; it only slows switching. The 100 kΩ pull-down is also irrelevant when on. Raising the gate above 24 V is impossible from the existing rail, so the source can never reach 24 V with this topology.
2. Propose two workable fixes and compare them quantitatively. State any important constraint for each.
Read the worked answer
Fix 1: add a high-side gate driver with a charge pump so the gate can sit about 10 V above the source. Then R_DS(on) ≈ 8 mΩ, so P ≈ (1.9 A)² × 8 mΩ ≈ 29 mW and the load sees essentially 24 V. A bootstrap driver alone cannot hold the switch on continuously at 100% duty; a charge pump or an always-on supply is needed. Fix 2: move the MOSFET to the low side, with the load connected between +24 V and the drain and the source grounded. The gate can then be driven from 0–10 V relative to ground, giving full enhancement. This requires the load to tolerate a floating high side when off, or a freewheel path. A P-channel high-side switch with a level-shifted gate is a third option; its R_DS(on) is usually higher, so check dissipation. In all cases, verify gate-source ratings, startup behavior, and the gate resistor value if the switch is PWM-driven rather than static.
3. The actuator is replaced by a 24 V relay coil with 120 mA hold current and 1.2 H inductance. What additional design checks become mandatory, and why?
Read the worked answer
An inductive load stores energy when on, so turning off the MOSFET forces the coil current to find a path. Without a flyback diode or an equivalent clamp across the coil, the drain voltage rises until something breaks down, which can destroy the MOSFET or cause erratic relay release. Add a diode across the coil, cathode to +24 V, or use an active clamp; check that the resulting release time is acceptable. The off-state gate pull-down must hold the gate low while the source node moves, or the MOSFET can turn on momentarily through Miller coupling. Check the safe operating area for the turn-off transient, and check that the 10 kΩ gate resistor gives acceptable switching loss if the relay is pulsed. The relay coil's 120 mA is well below the actuator current, so steady-state heating is not the issue; the inductive transient and off-state integrity are.