MOSFET operation clarification

[TheWinterSnow]

I am trying to get some clarification on p-channel mosfets.  I am utilizing two mosfets as a switch (series pass transistor), one for a li-on batter disconnect and the soft power switch Dave posted in blog #262.  From my understanding and from what I have seen from a few tutorials and even the Arduino schematics (the circuit where USB power is removed if an external supply is being used) is that the drain should be connected to power and the source connected to the output or load.  It has confused me that in #262 that Dave drew the pfet as the source to power and the drain to the load, maybe that was a mistake


http://arduino.cc/en/uploads/Main/arduino-mega2560_R3-sch.pdf (check the FDN340P circuit towards the bottom right)

My other confusion is about the voltages required to turn on the fet.  So to my understanding a pfet will turn on if Vgs is more negative than the Vgs(th) of the datasheet?  If that is true, with a pfet like the FDN340P, having the gate tied to 0v and Vgs being no more than 0v, is it safe to run a pfet that turns on as low as -0.7v?  If so the pfet mentioned above should work out nicely.

Anyway here is my schematic, I have a battery charger, USB voltage detector, soft power switch, 3.3v Reg and a low battery indicator.  Does this look correct?  I have an odd feeling like I have messed something up.  The 5v node is only connected to the USB 5v input and if there is no USB connected, the node is floating.

[suicidaleggroll]

Source goes to power input, drain to power output.  Pull the gate below the input (source) voltage to turn it on.

You can verify this by looking at your diagram of a pfet.  The body diode should block current from flowing in the intended direction, but allow it in the reverse direction (drain to source).  If you had it the other way around, the fet would be pointless as you'd never be able to shut off the current flow due to the body diode.

[TheWinterSnow]

So I have it backwards then?

This video and the Arduino schematics have it the way I do.



EDIT:  I updated the schematic so it may be a bit easier to read, with arrows pointing electron current flow (conventional flow is for old geezers), maybe that may help a bit.

EDIT EDIT:  If the body diode is supposed to go backwards of cuurent flow, how am I to prevent the 5v from USB from incorrectly charging the li-on battery because if USB was plugged in, the power would then supply 5v minus the body diode drop back to the battery.  I am trying to shut down the battery from VCC (VDD) if there is USB power.  Something tells me a BJT is better off in that position.

[sleemanj]

So to my understanding a pfet will turn on if Vgs is more negative than the Vgs(th)

The fet will START to turn on at the threshold.

The threshold is the point at which the fet just barely starts to conduct, it still exhibits high resistance.

You really need to look at the Rds On figure(s) and the voltages that these were specified at to learn what voltage you should be driving the gate with to get it into a suitably low resistance state.

The fet you specified has a threshold from -0.4 to -1.5v, but have a look at the chart on page 3 of the datasheet...


You can see that it's only really regarded as ON with the gate at least 2v below the source, any higher than that and the the on resistance increases drastically.  This is also borne out by the lowest Vgs that has an Rds(on) specified on page 2 is -2.5v, indicating that the manufacturers think that anything worse than that and you're on your own.

[sleemanj]

If the body diode is supposed to go backwards of cuurent flow, how am I to prevent the 5v from USB from incorrectly charging the li-on battery

Put two of the same fet in series, with one "upside down" so their body diodes oppose each other. 

[TheWinterSnow]

So you mean like this?



I always thought all transistors acted like diodes in that they could only let current pass in one direction, like BJTs, so having current in one fet being source to drain and the other drain to source is a little confusing when with a pfet drain current characteristics are labeled in negative mA.

Sorry for my ignorance, my first deep understanding of semiconductors was through vacuum tubes and later BJT's, all my school textbooks on fets are insulting at best, even for teaching fets at a technician (non-engineering) level and yet I have had a hard time finding simple data about fet operation that is comparable to the easy to use information on the internet about fets and vacuum tubes.

[diyaudio]

So you mean like this?



I always thought all transistors acted like diodes in that they could only let current pass in one direction, like BJTs, so having current in one fet being source to drain and the other drain to source is a little confusing when with a pfet drain current characteristics are labeled in negative mA.

It depends. All Mosfets has parasitic capacitance as a results propagates HF AC from either drain to source OR the gate (or everywhere), this is how gate driver chips are destroyed the large di/dt from an inductive leakage is feed back to the mosfet and thus through the gate channel destroying the gate channel and the driver.. basically it makes a good ac coupler,
that is a tuned resonator with the pcb tracks, mosfet parasitics and external components.

Here is an example of the internal construction.

[TheWinterSnow]

Since I am using the fet as a switch and a diode (in series) I am doing all my analysis in DC.  I need a circuit that will prevent a 5Vdc source from going back to a battery, yet allow the battery to pass if and only if a 5Vdc source is not present.

[sleemanj]

one fet being source to drain and the other drain to source

Mosfets don't care about the direction of current flow (well apart from the body diode of course), they care only about the relationship between their gate and source.

See quick demonstration.  Imagine the top power rail is your battery, the bottom switched one is your power supply (don't worry about the voltages, just don't have a logic-level fet in that software)

http://youtu.be/A2euWvTVjAM

First I show that with the series P-fets with one upside down (the lower one) if the external power is switched on, current can not flow backwards through the fets into the battery.

Then I show if that bottom fet was not in place (by me bypassing it), then current can freely flow back through that body diode of the first fet.

Now, it has to be said, that theory and practice can be two different things in this case, try it for real, with real parts, and things may not be quite so easy.

There are IC's out there that do this function for you.