Features: Transistor NPN common emitter Assembly positive polarization (living positive) signal line of input: 0. 05VPP output signal: Vcc = 12V Recor 3VPP demos (Vb) the base voltage is obtained through a couple of resistors (R1-R2) forming a divider. (It is not rule) Emitter voltage is the product obtained from It x Re while the collector voltage is not more than the subtraction of the Rc drop to the mains voltage which in this case would be a 3V (12V-3V = 9V) (3V that are the product of the multiplication of Ic x Rc.) Values that they would be represented in a diagram and serious the reference that we as technicians for service. (All values represented are mere reference to exercise) Below were to cause a series of possible failures registering the voltages obtained value and that will serve us for analysis. In each case, such an analysis based on what was written above will be realized. 1st. example: Vc = 12V Vb = 0V Ve = 0V the Vc is equal to the food (if Vc = Vcc – fall of Rc) that step with the fall of Rc? In this case the product of Rc x Ic = 0V for that is this outcome it is necessary that one factors is equal to 0 and know resistances could ever give that value by what remains that there is no Ic (Ic = 0) Ic is the result of the Ib and for that flow are IB the dollarization of base which are observed 0V is necessary.
Which implies a problem with the divider and more specifically with R1 who connects to the polarization. Result open R1. Do you find difficult? Do we expect that not to review the second example Vc = 12VVb = 2.4VVe = 1.8V follow the collector 12V, even having Vb? By what becomes to make the case that there is no Ic therefore there is no fall in Rc and hence the 12V in collector must remember that Ic is part of It flowing through transmitter is a similar problem to which provokes a transistor with its open collector electrode but actually what opened was Re draw the first conclusion: The increase in Vc means no driving the transistor.
