battery-chargingmosfet
I would like to use a PFET as a diode to only let current from the 3.3V to 4.2V area (when 4.2V is not present).
When 4.2V is present, it should not charge the battery.
Is it possible to do this with mosfets. I would like minimal voltage drop and up to 600mA can be drawn.
Circuit 2 is what I am trying to achieve. Circuit 1 is my attempt, but it isn't working.
Btw, this is a mod on a PCB, so I am limited in the part count.
It is not difficult to integrate current. If you're willing to roll your own you'll have complete control over the specs.
As you probably know, a capacitor has the relationships Q = CV and \$Q = \int i \cdot dt\$ .
One way I see off the top of my head to do this is to create a current mirror to charge a cap. Reading off the voltage of the cap is all that is needed. You can get caps as accurate as you need and there are many accurate current mirror configurations.
With such a method you can really get any amount of complexity that you need. You could have multiple resolutions(multiple mirrors and caps of different sizes). You can use op amps to improve the resolution and create a simple reset.
Of course it's not as simple as using a chip but as you have already stated, you can't find any chips that suit your needs.
It may be possible use current sensing(even proximity) but I'm not sure the accuracy you will get. For example, if your load is fairly low you could stick a 1ohm resistor in series. The voltage across the resistor is then equal to the current. Integrate this(say, using an op amp) and you have the charge. The efficiency here would be much larger, almost near unity while the current mirror method will be slightly less than 50%.
For the voltmeter protection, use a series resistor to limit current and a shorting diode to prevent reverse voltage. It will let a reverse voltage equal to the forward voltage drop of the diode through. Assuming the series resistance of the voltmeter is very high, the series resistor will not require any correction in the voltage measurement.
simulate this circuit – Schematic created using CircuitLab
When the diode is reverse biased (up to the breakdown voltage) it effectively be open.
Best Answer
As the user The Photon already pointed out in the comments, there are some ics which can do what you want. For example, this evaluation board from Texas instrument uses two ideal diodes LM66100 to connect a load to two different supply voltages. If the second supply voltage is larger than the first one, the ideal diode of the latter is disabled and the load is supplied by the former.
If for some reason, you still want to implement something similar using discrete components, you can try the following. I must say upfront what this circuit was not optimized and may have some flaws, which I might have overseen.
How it works:
The upper mosfets act as ideal diodes operating in a complement fashion. The lower long tailed pair compares both supply voltages and turns on, the one which is higher. Some positive feedback might be needed somewhere in order to add some hysteresis during the transition.