The voltage of the base is the voltage of the B node with respect to the reference node, which by connection is the same voltage across R2. Because IB is zero, that voltage can be found quickly with a voltage divider VB=10*R2/(R1+R2). Or, alternatively but more laboriously, you can compute the voltage drop across R1, which tells you how much lower B is than the 10V source node, and subtract this drop from 10V. ... (to be continued)
(.continuation) ...You'll need, however, to compute the current in R1, which is I=10/(R1+R2), the drop in R1 is V1=10R1/(R1+R2), so VB = 10 - V1, do the math and you get exactly the same as with the voltage divider. Cheers.
Im kindda confused here. When Ib is negative, the circuit is in cut off mode, however on our book when Ib is zero it is in cutoff. Why are they different? Or am I just misunderstanding it? Help please. Thank you
+Jackielyn Jane Yogawin In reality, the base current cannot be negative (it is the current through a PN junction, that between the base and the emitter). That is the physical reality. In the equivalent circuit on paper, where we are modelling the BJT with linear ideal components, when we solve the circuit for Ib and we find it as a negative value we reason "Aha! This is impossible, the actual current cannot be negative ... so it has to be zero ... and the BJT is in cut-off." But the procedural short path is " ... in the llinear model, if the base current is computed as a negative value, it's because the BJT is in cut-off ..." (The actual Ib is zero in the real, physical non linear device.)
So far you are the only one in UA-cam who explains very well the electric circuits problems...Thanks Mr.rolinychupetin.
Iv been looking for this. Thanks a lot
The voltage of the base is the voltage of the B node with respect to the reference node, which by connection is the same voltage across R2. Because IB is zero, that voltage can be found quickly with a voltage divider VB=10*R2/(R1+R2). Or, alternatively but more laboriously, you can compute the voltage drop across R1, which tells you how much lower B is than the 10V source node, and subtract this drop from 10V. ... (to be continued)
You explain very well! Thanks for the videos!
(.continuation) ...You'll need, however, to compute the current in R1, which is I=10/(R1+R2), the drop in R1 is
V1=10R1/(R1+R2), so VB = 10 - V1, do the math and you get exactly the same as with the voltage divider. Cheers.
Heh. I'm one day too late in watching this. The MITx online circuits course had this circuit on the final exam(never having shown us a BJT previously)
Im kindda confused here. When Ib is negative, the circuit is in cut off mode, however on our book when Ib is zero it is in cutoff. Why are they different? Or am I just misunderstanding it? Help please. Thank you
+Jackielyn Jane Yogawin In reality, the base current cannot be negative (it is the current through a PN junction, that between the base and the emitter). That is the physical reality. In the equivalent circuit on paper, where we are modelling the BJT with linear ideal components, when we solve the circuit for Ib and we find it as a negative value we reason "Aha! This is impossible, the actual current cannot be negative ... so it has to be zero ... and the BJT is in cut-off." But the procedural short path is " ... in the llinear model, if the base current is computed as a negative value, it's because the BJT is in cut-off ..." (The actual Ib is zero in the real, physical non linear device.)