2026-07-14
When you need to combine multiple power sources — a battery and a wall adapter, redundant supplies, or backup capacitor banks — the classic answer is OR-ing diodes: two Schottkys, cathodes tied together, each anode to a source. Whichever source is higher wins, and neither can back-drive the other. Simple, robust, and terrible for efficiency at high current.
The problem is the forward drop. A Schottky drops 0.3–0.5 V. At 5 A, that's 1.5–2.5 W wasted as heat, plus a real voltage sag your downstream regulator has to swallow. On a 3.3 V rail, you've just lost 10–15% of your headroom.
An ideal diode controller replaces the Schottky with a MOSFET plus a tiny IC that monitors drain-to-source voltage. When current flows in the "forward" direction (source-to-drain), the controller turns the FET fully on — Rds(on) is typically 5–20 mΩ, so your drop becomes I × Rds(on) instead of 0.4 V. When the drain voltage tries to exceed the source (reverse current), it shuts the FET off in microseconds.
Concrete example: A dual-input 12 V industrial supply with a 3 A load. Using two SS34 Schottkys: 0.5 V × 3 A = 1.5 W dissipation per diode, and you'll need heatsinking. Swap in an LTC4415 controller driving two SiR426DP MOSFETs (Rds(on) ≈ 8 mΩ): drop is 8 mΩ × 3 A = 24 mV, dissipation is 72 mW. You've reduced power loss by 20×, kept the voltage rail cleaner, and eliminated the heatsink.
Rule of thumb for FET selection: Pick Rds(on) such that I_max × Rds(on) < 50 mV in normal operation. That keeps you well below the controller's typical 30 mV regulation threshold (where it starts pinching the gate to prevent reverse conduction). Also verify the FET's body diode can carry your load current for a few microseconds during startup, before the controller enhances the channel.
Watch out for two gotchas:
Common controllers: LTC4352 (single channel, 0–18 V), LTC4415 (dual), LM74700 (automotive reverse-battery protection), TPS2419 (12 V servers). Most cost $1–3 and pay for themselves in eliminated heatsinks alone.