A Schottky diode is an excellent way to connect multiple batteries or dc power supplies to a load. With a forward-biased “or” diode between each supply and the load, faults are isolated. Weak, low, or shorted supplies can’t sink current from the other supplies. Unfortunately, the heat produced by a diode can be difficult to remove in some applications. This circuit was created to reduce the heat dissipated by a power-supply “or” diode in a redundant 50-V supply.
We can replace an “or” diode with an n-channel power FET. To match the behavior of an “or” diode using a FET, the FET must be off when the diode is reverse-biased. Therefore, we connect the FET’s gate to the output of an amplifier, driving the gate low when the drain is higher than the source (Fig. 1).
To prevent oscillation as the bias switches from reverse to forward, a 60-mV input offset is created by unequal voltage dividers on the op amp’s inputs. This small offset voltage becomes the forward voltage drop of the ideal diode. The basic circuit has only a few parts, yet the FET-based ideal diode must be carefully controlled. Unlike an actual diode, we must consider stability, transient response, and supply voltage range.
The feedback loop must be stable. The feedback path includes a power FET. Its gate capacitance adds a pole to the open-loop response, which can cause instability with a weak op amp or a high gate capacitance. As long as the op amp can drive the capacitive load of the FET gate with plenty of current, this pole will be above the bandwidth of the op amp and won’t affect the loop’s stability. Spice modeling can quantify the gain and phase margins of a particular op-amp and FET combination.
In addition, transients must be handled gracefully. The ideal diode response is slower than that of an actual diode, so it might be damaged if it absorbs too much energy during a transition. If the forward current increases quickly, the FET is protected. Its body diode responds practically instantly and clamps the voltage across the FET to less than 1 V. A few microseconds later, the op amp drives the gate high to reduce voltage drop even further.
However, if the current switches from forward to reverse very quickly, reverse current will flow until the op amp can turn off the FET. The power dissipated during this transition can be unsafe for the FET. The op amp must be fast enough to respond quickly to this transition.
The circuit’s response to a fast reversal of current can be tested with two power supplies and the 10-O load (Fig. 1, again). While power-supply 1 delivers current to a load through the ideal diode, it’s suddenly short-circuited by a mechanical switch (SW1). After the switch is closed, the amplifier takes 30 µs to fully turn off the FET. Reverse current peaks at 300 A, but the impulse energy is only 0.15 joules, which is within the FET’s specified maximum single-pulse avalanche energy. Figure 1 shows the test points, while Figure 2 shows the oscilloscope traces immediately following the short circuit of SW1.
Finally, operation must be safe across the entire range of possible input voltage. If the op-amp supply voltage falls close to or below the FET’s threshold, the ideal diode will never turn on and will dissipate as much power as an actual diode. Its heatsink must therefore handle the full power dissipated by the body diode alone. If that’s not feasible, then brownout conditions must be detected and avoided with an additional supply voltage comparator and power switch (not shown here).
This diode was designed for low (6 mW) quiescent power, supply voltage between 40 and 60 V, and current up to 10 A. Used in place of a fan-cooled diode, the ideal diode reduces maximum power dissipation from 7 W to less than 1 W, enabling passive cooling without a fan.