I'm working on the open source ignition system design and this exactly how I have it set up. It's simple and pretty cool.
Here is a summary from the design document (with help from Claude)
The OR-diode is one of those circuits that looks almost too simple for what it does. Here's what's actually happening, built up from first principles.
The basic idea
You have two voltage sources — the aircraft main bus and the backup battery — and you want whichever one is higher to supply the load, automatically, with no relay, no logic, no control signal, no moving parts.
You get this for free from a fundamental diode property:
a diode only conducts in one direction, and only when its anode is more positive than its cathode by at least the forward voltage drop (Vf).
View attachment 123428
Each source has its own diode, both cathodes tied together at the output rail. The source with the higher voltage wins — its diode conducts, the other diode is reverse-biased and blocks. No current flows backward through the losing source.
Walking through the three operating states
State 1 — Normal flight (main bus healthy, ~14.2V; backup at ~13.3V):
- Main bus anode: 14.2V. After D1's forward drop (~0.3V Schottky): cathode = 13.9V at the output rail.
- Backup anode: 13.3V. D2's cathode would need to be below 13.3V to conduct. It's sitting at 13.9V. So D2 sees 13.3V on its anode and 13.9V on its cathode — reverse biased, blocks completely.
- Result: main bus supplies everything. Zero current drawn from the backup battery. It just sits there fully charged, waiting.
State 2 — Main bus fails (drops toward 0V):
- D1's anode falls. The moment it drops below (backup voltage − Vf), D2's forward voltage condition is met and D2 starts conducting.
- The transition is not instantaneous but it's very fast — diode switching happens in nanoseconds. There is no relay to energize, no comparator delay, no firmware involved. The physics just works.
- Result: backup battery takes over the load. The output rail sags slightly (by D2's forward drop) but stays live. The ignition system never sees a dropout.
State 3 — Main bus partially degraded (e.g. 12.5V; backup at 13.3V):
- Main bus anode: 12.5V. D1 cathode would be ~12.2V.
- Backup anode: 13.3V. D2 cathode would be ~13.0V.
- D2's cathode (13.0V) is higher than D1's cathode (12.2V), so D1 is now reverse biased. D2 conducts.
- The backup has taken over even though the main bus is still partially live — because its voltage is higher after the diode drop. This is actually the correct behavior: a degraded main bus is not something you want to draw from.
Why Schottky specifically
Standard silicon diodes have Vf ≈ 0.6–0.7V. Schottky diodes have Vf ≈ 0.2–0.5V depending on current. Two reasons this matters:
- Less voltage lost on the output rail. Your backup battery at 12.8V becomes 12.3–12.5V at the load with a Schottky vs 12.1V with silicon. At the low end of battery discharge this headroom matters for the regulators downstream.
- The voltage differential determines the switchover point. With two Schottky diodes of matched Vf, switchover happens when the main bus drops below the backup voltage. With mismatched diodes the math shifts slightly — worth using the same part for both D1 and D2, which is why a dual-common-cathode Schottky package like the MBR20100CT is so clean: both diodes are on the same die, thermally matched, identical Vf, in one TO-220 footprint.
The MBR20100CT specifically
This is a
dual Schottky, common-cathode package. Both anodes are separate pins (Pin 1 and Pin 3), both cathodes are tied internally to Pin 2 (and to the metal tab). This is exactly the OR-diode configuration in one part:
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The 20A rating is massively oversized for OASIS's actual current draw — which is intentional. At low currents a 20A Schottky runs cool and its Vf is at the low end of its curve, minimizing the voltage drop on the output rail.
Oh, and you can get 10 or em for ~8 bux.