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TPS2121: Fast Switchover design review requested

Part Number: TPS2121


Tool/software:

Hoping you can assist me in verifying the below implementation:

I have 2 supplies VUSB @ 5V and VBAT @ 9V, i wish to use 5V USB when it is present, otherwise i want to use VBAT 9V.

Previously when both supplies were present and unplugging the 5V USB supply the system was crashing, so i have made the following changes to implement "fast switchover".

I am looking for confirmation that i have understood correctly how this is to be setup.

Circuit:



VPR1 = VUSB * (R10 / (R10 + R9))
VPR1 = 5V * (20 / (20 + 5))
VPR1 = 4V

VCP2 = VBAT * (R8 / (R8 + R7))
VCP2 = 9V * (10 / (10 + 20))
VCP2 = 3V


FACTS

If VPR1 > VCP2 IN1(VUSB) is powering output.

If VPR1 < VCP2 IN2(VBAT) is powering output.

If VCP2 > VREF(1.06V) fast switchover mode is enabled.


SCENARIO 1 (VBAT & VUSB PRESENT)

VUSB (IN1) = 5v
VBAT (IN2) = 9v (fully charged)

VPR1 = (5 * (20/(20 +5)) = 4V
VCP2 = (9 * (10/(10 + 20)) = 3V

VPR1 (4V) > VCP2 (3V) so IN1(USB) is powering output.
VCP2 (3V) > VREF (1.06V) so FAST SWITCHOVER MODE

Low battery voltage makes no difference as VPR1 is always greater than VCP2
as VCP2 drops with battery depletion.


SCENARIO 2 (VBAT PRESENT & VUSB DISCONNECTED)

VUSB (IN1) = 0v (Disconnected)
VBAT (IN2) = 9v (fully charged)

VPR1 = (0 * (20/(20 +5)) = 0V
VCP2 = (9 * (10/(10 + 20)) = 3V

VPR1 (0V) < VCP2 (3V) so IN2(VBAT) is powering output.
VCP2 (3V) > VREF (1.06V) so FAST SWITCHOVER MODE


SCENARIO 3 (VBAT DISCONNECTED & VUSB PRESENT)

VUSB (IN1) = 5v
VBAT (IN2) = 0v (Disconnected)

VPR1 = (5 * (20/(20 +5)) = 4V
VCP2 = (0 * (10/(10 + 20)) = 0V

VPR1 (4V) > VCP2 (0V) so IN1(VUSB) is powering output.
VCP2 (0V) < VREF (1.06V) FAST SWITCHOVER DISABLED

SCENARIO 4 (VBAT & VUSB PRESENT then VUSB DISCONNECTS)

This starts as scenario 1 with
- IN1 (VUSB) is powering the output.
- Fast Switchover mode is enabled.

When VUSB is removed then we are in scenario 2
- IN2 (VBAT) is powering the output.

Because fast switchover was enabled then this should have
minimal voltage drop, especially with C16 100uf COUT cap.



Have i understood correctly? Anything i have missed or miscalculated?  Many thanks in advance.


Layout for reference: