EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: Nikos A. on June 03, 2020, 12:24:21 pm
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Hi everyone,
Could someone explain me how the P channel Mosfet at this circuit works?
When it is supposed to be ON and when it is supposed to be OFF?
I suppose that this is to distinguish and prioritize the two voltage domains VBUS and VBAT but I cannot see how it is works.
(https://i.ibb.co/G0mYrV6/Mosfet.png) (https://ibb.co/mGpZPHj)
Thanks in advance
Nick
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First of all the mosfet is a diode, so the input can receive current from both the bus and the battery. When the bus has no/low voltage, the p-channel mosfet will turn on and shorten out the diode voltage drop. When the bus has voltage comparable or higher than battery, the mosfet and its diode will both be off.
Regards, Dieter
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This is a standard "load sharing" circuit. If USB is not plugged in, the gate pulldown resistor will bring the gate low, and that turns on the mosfet so the battery can provide power to the regulator. But if USB is active, it will bring the gate high, which turns off the mosfet. The orientation of the mosfet permits its body diode to block current from flowing from VBUS back into the battery, which would overcharge it. D1 is there to block the battery from trying to power whatever else might be connected to VBUS, such as a battery charger. In fact, you could have an upstream battery charger powered by VBUS, and this circuit would let VBUS charge the battery to termination while also powering the load.
When VBUS is active the mosfet's body diode will not pass current from the battery because the voltage at its source terminal is greater than the battery's voltage, so the diode is reverse biased.
With this circuit, VBUS will always power the regulator when it's plugged in. When it isn't, the battery will take over, and the voltage drop though the mosfet will be very small, which maximizes battery life. You could just have another diode there in place of the mosfet, and that would work the same way (the highest voltage input source would always power the load), but there would be a voltage drop across the battery diode. The mosfet just eliminates almost all of the voltage drop so you don't waste battery power generating heat.
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Thank you guys!! Now I can see clearly how it works!!
This is a standard "load sharing" circuit....
Great explanation, thanks!!