I have the VIN pin at 3.0V and the CE pin at 3.3V, however the part is still enabled.
Does anyone know why?
Kalin,
From the datasheet.
The chip enable works by comparing the CE pin voltage to the input voltage. When the CE pin voltage is higher
than VIN by 80 mV, the device is disabled and the MOSFET is off. When the CE pin voltage is lower than VIN by
250 mV, the MOSFET is on. The LM66100 also comes with reverse polarity protection (RPP) that can protect the
device from a miswired input, such as a reversed battery.
A snippit of the circuit.
The RED LED is always on and should not be if the device is disabled, there would not be enough current to drive the LED.
Regards,
Rick
Hello Rick,
See the below image from figure 11 of the datasheet for expected Vout.
Meaning Vout when the device is disabled should be 3.0V - Vf. This could still drive your LEDs. The device is an ideal diode and not a switch. It is meant to be used in a dual ORING configuration or used as a diode.
If you need a device with reverse current protection that will act as a switch, not a diode, you can check out our load switch portfolio below and sort by the reverse current protection feature.
Regards,
Kalin Burnside
