That's not the point... what is driving the voltage down at the collector.

As I understand it, the question is, why does current continue to flow in to the collector when $V_{cb}$ is negative?

@Trevor_G The voltage you are applying defines the collector emitter voltage! There is no reason why you should not be able to apply any arbitrary voltage to the collector with reference to the emitter. That the collector is drawn above the base doesn't mean that the voltage cannot be lower than the base.

@freecharly The diagram may not be clear.. the voltage at the collector is not applied, it is generated by the transistor.

@Trevor_G - The transistor doesn't "generate" any voltage. Usually, the collector-emitter voltage is defined by the applied voltage or equivalently by the external circuit. If your problem is the direction of the collector current, of course it can change direction when the collector-base voltage becomes negative and high enough, when the electron injection current from the collector into the base becomes larger than the electron injection current from the emitter into the base.

@freecharly you are missing the point of the question. How can/does 0.7V on the base pull down the load above the transistor to a voltage less than that. Or I am missing yours...

@Trevor_G - So you assume that there is a load resistance connected to the collector which is not shown in the diagram? Then the question is circuit related and you should draw the whole circuit to make clear what your question actually is.

If the transistor's in common emitter configuration, then the current flowing in to the collector, as it passes through the collector resistor, results in a collector voltage lower than the base voltage. So the question is, why does current continue to flow in to the collector when $V_{cb}$ goes negative?

@freecharly so done ;)

@freecharly, Trevor is a high-rep user on EE who knows his circuits. He isn't just asking for an explanation of KCL and KVL.

@ThePhoton - So he doesn't formulate the question in an understandable way. I have been teaching this subject for decades in EE courses as a professor at a university.

@ThePhoton - Let's talk about the electron current (which is opposite the usual current arrow). When $V_{CB}$ becomes negative, the CB junction becomes forward biased so that an electron injection current flows from the collector into the base (and emitter) which is superimposed opposite to the electron injection current from the emitter into the base and collector. For not too large negative $V_{CB}$, this electron current injected from the collector into the base is usually smaller than the usual electron current injected from the emitter reaching the collector ($n_E >> n_C$).

@ThePhoton - Thus you can have a positive electric current entering the collector terminal in spite of the negative $V_{CB}$, or equivalently, for $V_{CE} \lt V_{BE}$

@freecharly, that should be posted as part of your answer.

@Trevor_G - I see your circuit diagram now. You should have included this from the beginning to avoid guessing what your actual question is. With large enough positive collector current (flowing into the collector terminal), the voltage drop on the resistor can become easily so high that $$V_{CE}=V_2 - I_{C} R_1 \lt V_{BE}$$ This is simple circuit theory.

@Trevor_G You have also keep in mind that the CB junction is not a simple lumped resistance in this case.

@ThePhoton - I am not sure whether the OP really wanted to hear this answer considering the inner working of the bipolar transistor. Actually his question was based on a certain circuit which he entered only in a later editing. Thus the original question was completely unclear and one had to make intelligent guesses, as yours, which was very founded, what the question was actually about. Now it turns out that the voltage drop in the collector circuit plays a major role in his thinking.

Again, you are missing the point, Yes obviously, if sufficient current is flowing through the collector the voltage at the collector will be whatever the collector resistor dictates, I'm not an idiot. What does not make sense is the current must flow down through the collector material the base material and the emitter material. If current is flowing in one direction... how can the emitter be lower than the base. How can the base create so much current that that can happen.

Continued: Answering that with " is not a simple lumped resistance in this case" is like telling a five year old "because".. and quite frankly is I find rather insulting.

@Trevor_G - I am really sorry, it was definitely not my intention to insult you! I sincerely apologize if you got this impression. I am not thinking that you are an idiot! Communication of technical issues is sometimes not easy. The whole problem obviously stems from the unclear formulation of the question. You were assuming a load circuit which was not at all obvious to the reader before you added the circuit diagram. My honest intention is to understand your problem and to give you a satisfactory answer.

@Trevor_G - The base is not creating so much current. It is a current that flows through the base mainly by diffusion and in the CB junction for $V_{CB} \gt 0$ against the applied voltage. Therefore, in a circuit diagram the output circuit usually includes a controlled current source.

@Trevor_G - In the CB junction, for $V_{CB} \lt 0$, the current coming through the base by diffusion flows against the applied potential difference. But, due to the built-in voltage being larger than the opposing applied $V_{CB}$ , the electrical field in the junction is still directed towards the base so that electrons are transported to the collector producing a positive electrical current flowing into the collector terminal.