Ideal-diode OR-ing: USB plus battery
Two supplies feed one load and must never feed each other. Schottky diodes would waste 0.3 V. Here each supply goes through a P-MOSFET that a matched PNP pair turns fully on while its side is higher, and off the moment the other side is higher: a diode with almost no drop.
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How it works
Both PNPs carry the same current when their emitters are equal. Q_A’s side is higher, so its mirror transistor carries less, R2 pulls the gate to about 1 V and the MOSFET is fully on: the drop is I × R_DS(on) ≈ 2.3 mV at 50 mA.
On the battery side the output is above the battery, so that mirror transistor conducts hard, lifts the gate to the output and holds Q_B off: no current flows back into the battery.
Try it on the bench
- Swap in Schottky diodes (variant) and compare the output: 0.3 V lower.
Parts on the bench
- Bench power supply
- 18650 Li-ion cell
- 2 × P-channel MOSFET
- 2 × Matched transistor pair
- 5 × Resistor
- Electrolytic capacitor
In the course
Electronics Engineering, module “Transistor amplifiers”.