About John Penton Expertise I am happy to answer questions on semiconductor devices - in particular high-voltage devices.
Also any electronics or physics homework questions.
I`m not a semiconductors professional (at the moment), so no really in-depth questions please.
Experience Degree level education in Natural Sciences and Electrical and Information Sciences.
My final-year project was in high-voltage power MOSFETs.
Currently work for ARM Ltd, designing microprocessors.
Answer First point to make: you show Vbe as being 2V - in practice this will never be possible. If Vbe was 2V, a huge current would flow which would burn out the transistor. Always make sure the base is fed from a high impedence source (so the base voltage falls as the current rises - this is a current-switching configuration) or the emitter is driving a high impedance load (so the emitter voltage rises as the current rises - this is is emitter-followered configuration).
Even given that there will never be more than about 0.7V Vbe, your question is valid. I've been thinking about it for a while and suddenly realised what the answer is. Current will flow from the base to the collector, as well as from the base to the emitter. The later will be greater than the former - but in general (I believe) the device will start to act like two diodes in parallel.
(Phew! I'm pleased I figured that out)
HTH
John
