Transistors are always very confusing to me. I read and try to understand the specsheet always but never confident about my understanding :( I have PCB (readymade), which gives me a 2.8V output when I switch on the PCB. The 2.8V is coming out from the MCU directly. Now if I drive a BC109 with that output and put a 75ohms relay in a common emitter fashion then will the transistor be able to supply enough current to turn on the relay? According to my calculation, the BC109 max collector current is 200mA and a 12V/75ohms relay will require around 160mA current. If I put an additional resistance in serial with the relay's coil, then I can bring down the current more. Can someone please tell me if I am theoretically all correct or not. I am not attaching any drawing as it is really a tiny circuit and can be visualized easily.
After reading the comments, here is the question in a more mathematical way. Please let me know if I understand it correct.
Transistor : BC337 Relay: 12V / 75 Ohms
Connection type: Common Emitter
Input Voltage (applied to base) = 2.8
Required Current to turn on the relay = 12V / 75 Ohms = 160 mA
Max Ic = 800mA (Safe to drive the relay)
Current Gain to hard on BC337 = 10
Base current required to switch on the transistor (thus the relay) = (Current Gain / Load Current) = 160/10 = 16mA
Vbe = 0.7
Effecting Base Voltage = 2.8 - 0.7 = 2.1
Required base resistor = Vb / Ib = 2.1 / .016 = 131 Ohms (approx)
Transistor's power dissipation: (Vce * Ic) + (Vbe * Ib) = 196mW Approx