Given this circuit:


simulate this circuit – Schematic created using CircuitLab

I don't understand why, when simulating it by increasing parametrially the V2 from 0V to 12V keeping V1 constant, the base current decreases exponentially until V2 reaches the V1. I understand that when V2 > V1 the bjt enters in the direct active region and then the Vce increases, keeping the collector current almost constant. But I don't see why the base current should decrease until V1 <= V2. This reduction is more higher than the Vbe variation. Sorry if it is a stupid question.

I've noticed that when simulating an emitter follower, and I thought that was due to the emitter drop that this base current was reducing. But that happens also in that circuit above.

EDIT: I've discovered that this decrease is coherent with the increase of the Vbe voltage, with a given base resistor. So the question is: why the Vbe increase while the BJT exits from its saturation state? Perhaps that answer is more easy, I think.

  • \$\begingroup\$ Where did you simulate it ? If you simulate the circuit above, I do not see any decrease in base voltage or current. I just ran it through CircuitLab and it behaves as expected. \$\endgroup\$
    – efox29
    Commented Dec 19, 2014 at 14:56
  • \$\begingroup\$ PSpice using OrCad. Moreover, the Vbe voltage varies in a right way, because the BJT exits from saturation. The problem is the current. The 3,3V are kept constant. \$\endgroup\$
    – thexeno
    Commented Dec 19, 2014 at 15:05
  • \$\begingroup\$ Post your plots. I ran the simulation in LTSPICE and I see your exponential change (albeit very small change). I don't have a concrete factual answer - interesting question though \$\endgroup\$
    – efox29
    Commented Dec 19, 2014 at 15:30
  • \$\begingroup\$ So, it is not a normal thing of those things that "everyone should know". One point is solved. What if it is just a simulator issue? Like mine at a certain V2 (like more than 6V) start to oscillate: that of course is a simulator issue that put me to make a lot of asks, right here on stackexchange. \$\endgroup\$
    – thexeno
    Commented Dec 19, 2014 at 16:05
  • \$\begingroup\$ @thexeno: When you have the components there, you could just build the circuit and compare to simulations. But also perhaps post your plots, I don't think that I see what you see. \$\endgroup\$
    – PlasmaHH
    Commented Dec 19, 2014 at 16:58

2 Answers 2


The simulation result is correct. What you are seeing is Ibc: Current flowing from base to collector. Remember that the base collector junction is forward-biased.

Normally we don't think about the current flowing in that direction. But it can/does.

In saturation, some base current flows to the collector. The net flow may be from outside of the transistor into the collector, but that does not mean that Ibc = 0. This is something I did not fully grasp until quite recently, and I even down-voted someone over it. I mean, I am sure I knew it at one time when I was in school. But I must have forgotten it long ago.

  • \$\begingroup\$ There was a guy who told the same, but has canceled his answer, don't know why. He told almost the same: but I replied that the Ib decrease is the same even with a diode put in the collector, the only difference is that the diode block the higher value of current that should exist with that base resistor. In other words: I've considered that BC junction current, what I do not understand is why the Vbe is varying. I put an EDIT part on the original answer. \$\endgroup\$
    – thexeno
    Commented Dec 19, 2014 at 17:25
  • \$\begingroup\$ To me it seems intuitive. When Vc = Ve = 0 < Vb, you basically have two diodes in parallel. As Vc goes up, at some point, the base-collector junction comes under reverse bias. Now you no longer have two diodes in parallel. This causes the base voltage to go up slightly. \$\endgroup\$
    – user57037
    Commented Dec 19, 2014 at 18:00
  • \$\begingroup\$ You mean that was like seeing a voltage on a node with two parallel voltage generators with different values? \$\endgroup\$
    – thexeno
    Commented Dec 19, 2014 at 19:10
  • 1
    \$\begingroup\$ No. I mean it is like having a parallel load, where one of the loads gradually has its impedance increased to infinity. Don't you expect the voltage to go up slightly? Here is another way to look at it. When the transistor comes out of saturation, the effective beta goes up dramatically, which means the collector and emitter current goes up dramatically. Re, the intrinsic emitter resistor goes up as Ie goes up. When Re goes up, Vbe must also go up. How is that? \$\endgroup\$
    – user57037
    Commented Dec 19, 2014 at 20:03
  • \$\begingroup\$ Yes, could be intuitive. But only if beta (Hfe) factor increases more than Ic, and seems to be realistic, since if Ic is very high we can be again in saturation if Ib is high enough and Vbe can be a bit lower again. And also, in this case I presume beta increase far away more than the Ic. It's correct? \$\endgroup\$
    – thexeno
    Commented Dec 21, 2014 at 11:38

Brian has the answer, but I'll add this simulation showing the base current vs. Vce

enter image description here

The collector voltage will influence the base voltage (and, in this circuit, thus the base current) somewhat whenever it starts to approach a diode drop less than the base voltage.

  • \$\begingroup\$ Is this Brian mkeith? \$\endgroup\$
    – thexeno
    Commented Dec 21, 2014 at 11:39
  • \$\begingroup\$ @thexeno no another Brian- he deleted his answer for some reason .. I can still see it.. I'll have to augment mine when I get a chance. \$\endgroup\$ Commented Dec 21, 2014 at 13:45
  • \$\begingroup\$ My answer was only part of the story. When the question was edited (a diode suddenly appeared, but not in the schematic) I edited it further. The diode's disappeared now. I didn't have time to keep the answer up to date and add consideration of the b-e junction and their interaction, leaving my answer partially obsolete and misleading. \$\endgroup\$
    – user16324
    Commented Dec 22, 2014 at 15:03

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