I am having problem understanding how this MOSFET configuration (used in battery charger) works?
In this slide,
what does it mean "blocking voltages in both directions while allowing bidirectional current"?
Electronic – Understanding Two MOSFET with sources connected
battery-chargingmosfetswitches
Related Solutions
The datasheet says this about the "PACK" (2) pin:
Battery pack input voltage sense input. It also serves as device wake up when the device is in shutdown mode
It is also marked as "analog input" and "power".
So, when you connect the charger, the chip gets power from the PACK pin then it turns on the relevant FET. After that it stays on until the batteries discharge completely (after that you need to connect the charger to turn the chip back on).
Point 1 and point 2 are quite accurate I would say. For point 3, I always search for the following graph when I'm looking for the likely conduction capabilities of FETs: -
It tells me that if I put 3.5V on the gate and I want to pass 10A drain current there is likely to be 0.5V dropped across the device leading to a power dissipation of 5W. This is an on-resistance of 50 m\$\Omega\$. However, if i used a gate drive of 10V, I can expect the voltage dropped by the device to be about 0.25V i.e. a power dissipation of 2.5W or an on-resistance of 25 m\$\Omega\$.
Regarding the state of "fully-on" you can see that there is no magic state that is reached at some arbitrary gate voltage but I'd say that at about 4.5V gate drive, the benefits of driving with a higher gate voltage are diminishing.
Figure 3 in the data sheet (which shows the graph which you deduced the on resistance variation of 10 m\$\Omega\$) is a derivation of the graph (figure 1) in my answer. Obviously, Figure 1 carries more information because it covers a wider range of gate voltages.
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Best Answer
The below assumes NMOS, just for simplicity. Also note that I'm describing which directions current can flow, not which way it will flow. Which way it will flow depends on the voltages the circuit is connected to.
A single MOSFET has two states: ON and OFF.
ON: current can flow both directions
OFF: current can flow one direction (source-drain, due to the body diode) and not the other direction (drain-source, because the FET is off)
A single MOSFET in series with a diode:
also has two states, ON and OFF.
ON: current flows one direction (drain-source, through the FET and the second diode), but not the other direction (source-drain, because the second diode is pointing the wrong way)
OFF: current won't flow either way, because whichever way you look there's a diode opposing the flow of current.
Two MOSFETs in series, pointing opposite directions:
have four possible states. ON-ON, ON-OFF, OFF-ON, and OFF-OFF. For this example, I'll describe current flow with directions as seen in the picture above.
ON-ON: current can flow both directions through the circuit.
ON-OFF: current can flow from left to right (through the left FET and the right diode) but not from right to left (because of the right diode)
OFF-ON: current can flow from right to left (through the right FET and the left diode) but not from left to right (because of the left diode)
OFF-OFF: current can not flow in either direction, because with both FETs off all you have is two diodes blocking current flow in either direction.
So the third topology has the option of either blocking currents in both directions, or allowing current flow in either direction, depending on how it's gated.