Im stuck with figuring out what the following circuit does. The 3v3A feeds the AVDD of an MCU and 3v3D feeds the IOVDD of the mcu. The 3v3_protect comes externally to introduce to the circuit. is this some protection type circuit? can anybody help me figure out whats going on here?
R15 and R6 form a voltage divider. The voltage at C1 would be 10k/(10k + 2k) = 5/6 of the supply voltage.
If the supply voltage is 3V, then 5/6 of 3V is 2.5V.
Z1, LM4040C25Q is a precision 2.5V voltage reference.
Hence the comparator circuit is designed to change state when the supply voltage (+3V3) crosses 3V.
I have not examined the rest of the control circuitry, i.e. the two MOSFETs.
low output from comparator would turn mosfets on. that causes 3.3V to appear on the +3V3D.
it seem to kill supply to MCU when protected voltage falls below threshold.
This is an undervoltage lockout circuit. The +3V3D output is disabled until the +3V3A input is high enough to enable it.
PROTECT is divided down by R15 and R6, and filtered by C1. Z1 provides a reference voltage to compare against. When PROTECT is high enough, U7pin1 goes low, turning on both halves of Q3.
There is something strange about how Q3 is implemented. That is not the normal back-to-back MOSFET power switch configuration.
not sure what U7 is (and what level) but this is boiled down version. depending on U7 level diode may be needed... most likely U7 and R18 can also be removed.
R15/C1/R6 takes about 3.8mS after 3v3Protect hits 3.3v and for the comparator to turn on Q3a and Q3b.
I originally thought C3/R21 was some form of soft-start, but the values are so low all it seems to do is inject a spike that has no obvious effect.
Here's a simulation with near-equivalent MOSFETs and comparator and a resistive load with some bulk capacitance. Q3b is the main power switch, Q3a seems to do very little...