I have designed a power supply that will supply a 5V and 800mA. my sensor requires 2.7 V and 3mA maximum.
Do you have any idea how to divide the voltage into 2.7 volt and to limit the current to 3mA only.
You can use the circuit below easily for your requirements. Maybe you see the circuit little bit difficult. There are three battery charger circuits. You will just use one of them. R2,R5 or R6,R7 or R11,R12 are for limiting the output current. Higher value of resistor, lower the output current. RV1, RV2 and RV3 are set the desired output voltage. Input and output capacitors are 1000uF and 100uF,respectively. Others can be same value. You can use this circuit for sensors. Also, you can use it for batteries for charging.
I dont know, but my power supply is only 5V and 800 mA. this power supply will supply my sensor and other circuits including( pic16f877a & max232). but the above circuit is not clear, and how can I calculate that it will be only 3mA?
ı have used proteus for simulating and results were accurate with that simulation but you re right, ı read the 3mA as 30mA =). Sorry for that. Your current limit is so low that can be easily ignored by these basic circuits.
That's no meaningful specification. What's the specified voltage range, minimum and maximum supply voltage? Presuming it's intended for constant voltage supply, does it require additional current limiting? Or is 3 mA just a maximum current consumption?
of course you need a resistor [in series] before the zener. It's value is [5 - 2.8] / [3 + 5]mA The zener needs current to engage - check the data-sheet.