When i run the Simulation for the Voltage v2~ 0V, how come there are 12V at the base of Darlington transistor TIP127G,
View attachment 102844 Please Explain
Take a look at the TIP127 data sheet: Q2 (TIP127) is a Darlington transistor pair, but it also includes a resistor connected between the base and emitter of each transistor. These two resistors serve three functions:
(1) Conduct leakage current away which may turn the second TIP127 transistor on.
(2) Vastly speed up the turn-off time of the transistors, especially the second transistor, which is a high power type (the TIP127 is still pretty slow at 50KHz though).
(3) Increase the break-down voltage of the two transistors in the TIP127, especially the second transistor
In your circuit, the base of the TIP127 is connected to the 12V supply line by two resistors (8K Ohm and 220 Ohm) in the TIP127. This is why your simulator, quite correctly, says that, with Q1 turned off, the base of the TIP127 is at 12V.
When Q1 (BC547) collector current increases, the two resistors in the TIP127 will conduct current and Q2 base will drop linearly, until the TIP127 base voltage reaches around -0.5V from 12V supply line, when the first transistor of the TIP127 will conduct and turn the second transistor on. At that point, Q2 base will snap down to around -1.2V from the 12V supply line. After that, the base voltage of Q2 will not drop much further as the collector current of Q1, and hence the current through the TIP127, increase.
spec
(crossed posts Les
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DATASHEETS
https://www.onsemi.com/pub_link/Collateral/TIP120-D.PDF
https://www.onsemi.com/pub_link/Collateral/BC546-D.PDF