Two weeks ago, user @tmk commented on the weblog post about diodes with a pointer to a Zero Volt Diode.

Here’s the circuit, by Wilf Rigter:

I couldn’t quite wrap my head around it, so I re-drew it in a different way – it’s still the same circuit:

The key is that a MOSFET can switch a voltage with nearly no voltage drop, i.e. a signal on its gate can turn it from near-infinite to near-zero resistance. There’s no “bipolar junction” involved, therefore no 0.6V .. 0.8V threshold.

I’m not going to explain the circuit, but I’ve built it up and did some measurements to show its behavior:

Since a supercap has all sorts of odd behavior w.r.t. deep discharge and such, I replaced it with a 6800 µF electrolytic cap for this experiment, in parallel with a 1 kΩ resistor to simulate a load of a few mA.

Instead of the solar panel, I’m using a 2.9V power supply, limited to supply at most 10 mA. In other words, when connected to the capacitor, it will reduce its voltage a bit while the charging current is high, and then end up as 2.9V once the capacitor has been fully charged up. This makes it similar to a solar cell with limited capacity.

Enough talking. Let’s see how this thing behaves, while tracking a number of voltage levels at the same time:

There’s a lot to describe here:

• the RED line is the voltage from the power supply, i.e. Vsolar
• the YELLOW line is the voltage over the capacitor, i.e. Vcc
• the GREEN line is the voltage between drain and source of the MOSFET
• the BLUE line is the voltage on the gate of the MOSFET
• all signals have their zero origin at 2 divisions from the bottom
• power was turned on after 2 seconds from the left edge of the screen
• power was turned off again about 5 seconds later

The RED line is actually the YELLOW line minus the GREEN line.

The first thing to note is really the whole point of this circuit: the voltage on the capacitor (YELLOW) rises up to 2.84V, while the input voltage (RED) reaches 2.90V, so there’s only a 0.06V voltage drop over that FET while it conducts. That’s a ten-fold improvement over a silicon diode, and three-fold over a Schottky diode.

I’m using a BC557 for the PNP transistor, a BC549 for the NPN transistor, and a VN2222 as MOSFET, just because I happened to have those lying around. That MOSFET in particular is not a great fit here, really.

The other peculiar thing about this circuit, is that the MOSFET is used for current flowing through it in the wrong direction, from drain to source! But most MOSFETs won’t mind, they really act a bit like (controllable) resistors.

The most interesting bit is the GREEN signal, i.e. the voltage over the MOSFET. At input levels under about 1.5V it does conduct, but with a substantial voltage drop – first from the built-in diode conducting, and then gradually the MOSFET turns on and its low resistance takes over, with a total voltage drop of some 30..60 mV.

When the input voltage is completely switched off, the MOSFET goes into high impedance mode within a fraction of a second, which can be deduced from the fact that the GREEN and YELLOW lines meet up and overlap.

Lastly, once the voltages drop below about 0.4V, the gate voltage on the MOSFET rises a bit, but this is not enough to turn it on, and also not very important since the whole circuit is now essentially “dead”.

Here’s the second event in greater detail, i.e. when the input voltage drops – or in this case, gets switched off:

The MOSFET is switched off within milliseconds, with the cap now holding a higher voltage than the input.

The result of it all is that the capacitor soaks up almost all the voltage it can get, with no diode forward voltage drop involved. When the input voltage drops, the circuit disconnects it from the cap so it’ll retain its charge. Brilliant!

Update – Wanted to get a bit more info on-screen, so here’s another scope capture (oops, green is now yellow):

It better displays the elegant “swirly” charge ramp, with an odd little 64 mV bump on the MOSFET when the cap is full (maybe that’s the power supply switching to constant voltage mode). Given that the cap voltage reaches 2.82V and is loaded down by a 1 kΩ resistor, we can deduce that the current through the MOSFET must be 2.82 mA at that point, and therefore that its resistance is 23 Ω in this circuit (64.57 mV / 2.82 mA, Ohm’s law again!).