This is a basic lithium battery protection circuit, but looking at the dual mos-fet part of the circuit, It doesn't make sense to me. It's a 8205A dual mos-fet, with its drain connected together and each of its source connected to the negative of the input and output. Normally, the drain of a mosfet is connected to the positive. in this case both are connected to the negative if they are turned-on. I don't understand! what voltage will be the drain with respect to the ground?
Best Answer
Here is some reference reading about MOSFETs.
When an "enhancement-mode" MOSFET gate charge exceeds \$VGS_{TH}\$, the channel from D to S goes into a low impedance or "closed" state. From the device's perspective, current can go through this channel either way - it doesn't matter - as long as \$VGS_{TH}\$ is met, current can flow.
Note that M2 cannot disconnect the battery during normal operation due to it's internal parasitic body diode, and this is the reason why M1 is required in this orientation (to block "forward" flow when off.)
In normal operation, both 8205A MOSFETS have their \$VGS_{TH}\$ met (usually far exceeded to "really turn them on hard"), so each device imposes their minimum \$RDS_{ON}\$ resistance (about 0.03 Ohms for this device.) So the voltage at the combined drains will be a resistor-divider's-worth above ground, say perhaps 0.05v, which of course will vary with current consumption. This is assuming you measure in a "forward" direction from the BATT- terminal to the drains.
Unlike BJT devices, there are no parasitic voltage drops (aside for \$VGS_{TH}\$) in MOSFETS, and the gate is not operated via current but voltage. The steady-state gate current \$I_{GSS}\$ is very small (100nA) as opposed to mA's for BJT's. However a MOSFET gate can have significant capacitance (\$C_{ISS}=600pf\$) especially for power devices. 600pF isn't a lot; some power MOSFETS are 10x that or more, but this makes it difficult when trying to switch them on and off very quickly.