(iT MIGHT BE bATTERY mANAGEMENT FORUM QUESTION BUT I ASKED IT HERE AS IT IS RELATED TO FETS)
1. ON PAGE 13 OF SLUU474, THERE IS PARAGRAPH GIVEN
The high input capacitance of the Discharge and Charge MOSFETs can lead to a slow turn off time if
proper drive circuitry is not used. In this EVM, the drive circuit employs a JFET that has a low output
resistance compared to the 1-MΩ resistor across the MOSFET gate when the MOSFET control signal is
turned OFF. This low resistance quickly pulls the charge off of the gate of the MOSFET. Fast shutoff time
is desirable in battery management because of the high energy in the cells. Turnoff times of less than one
millisecond can be expected for these 10-mΩ switches (two 0.020 Ω in parallel).
2. NOW ON PAGE 43 OF SLUU474, IN CKT DIAGRAM
JFET ARE - q5 & q6 & THEY ARE PULLING CHARGE FROM q13 & q15 RESPECTIVEY.
3. HOW DOES JFET AFFECTS THE CKT, IT IS ON IN BOTH THE STATES WHETHER Q3 & Q4 ARE ON FOR TURING ON Q13 & Q15.
AS TO TURN OFF JFET, -VE VOLTAGE HAS TO BE APPLIED. BUT HERE VGS>=0.
4. WHEN ONE SAY JFET HAS LOW OUTPUT IMPEDENCE, WHAT DOES THAT MEAN IN THE CKT. WHICH PATH HAS LOW IMPEDENCE- DRAIN TO SOURCE ?
BUT HERE FET HAS ALWAYS ON.
5. WHICH PIN SHOULD BE DRAIN OR SOURCE. I THINK BOTH ARE INTERCHANGEABLE.
6. DO TI HAVE EQUIVALENT PART FOR IT.
