I have a problem: I have a circuit with a MSP430 MCU. I have a sensor which is supposed to be powered in 3.3V. My main power rail in my circuit is 3.3V. I need to power on and power off this sensor using a GPIO to pilot the power, so I suppose I need to use a transistor BJT being controlled by GPIO to let go the current in on/off mode into the sensor. Problem is power rail value = sensor value and my transistor got VCEsat = 0.7V (or I can also change it to -0.7V) So how do I manage with the Resistor value (R208) ?
With a collector current of 100 mA, the BC807 would have a typical saturation voltage of about 50 mV:
However, even if you can live with this 50 mV, this would require a base current of about 10 mA, which is too much for your MSP430FR2311.
A P-channel MOSFET can give you a smaller voltage drop:
This requires that Q1 is a FET with an RDS(on) value of much less than 500 mΩ; some widely-used models would be DMP2066, IRLML6401, or FDN338P.
The sensor will always be connected to ground on one end. The power supply can be enabled or disabled by the MCU. this is just the basic FET based switch. Advantage here is that, the sensor can always have ground reference. FETs can be from TI. the threshold voltage is well with in 2 V to turn on the FET.
Correct me if i am wrong.
Get rid of R208 and use a low on-resistance N channel MOSFET instead of the BC817. Make sure the MOSFET is turned on enough with a 3V3 IO line - look at Alpha and Omega - they have some low gate threshold devices.