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When i connect a NPN transistor with its collector to 12V and its base through a 10k resistor to a PIC to switch the transitor on, I only get 4,45V at the emitter when the transistor is turned on. How come I only get 4,45V and not full 12V?
Your circuit is a voltage follower amplifier, with grounded collector. The emitter must be connected to ground via resistor (for ex. 1kOHM). On the output will appear a voltage which is approaching equal to input (base) voltage. UE=UB-UBE when the UE is the emitter voltage to the GND, UB is the base voltage to GND and UBE is the base-emitter voltage (~0,5…0,7V). Therefore your circuit knows the theory: UE=5V-0,7V=4,4V. What would you like to drive with this transistor?
Base bias resistance is to high to drive the transistor into saturation, put a diode from the pic to the base to keep the current leaking back to a minimum. I would use it as a Com emitter.
It appears to have Vcc on the collector and GND on the emitter.
I am guessing that your pic is driving a 12 volt signal, then a drop is apparent at the resistor (6.85 volts) leaves 5.15 volts at the base, minus VBE .7v this yield at 4.45 volt dc at the emitter.
With Diode, driving a 12 volt signal, minus diode drop .7, this leave 11.3 volts at base minus VBE, which yields a 10.6 volt DC at the emitter.
To condition it to use low level signals you must use voltage dividers to drive a transistor into saturation at 12VCEcut, and figure out the resistor values needed to drive this way, acts like a clipping amp so to speak. It is better to use the collector to drive which also depends on what you are driving.
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