0
\$\begingroup\$

Ignoring the modulation of the effective base width, why is the collector current independent of the reverse bias voltage across the Base-Collector P-N junction? On one hand, it is easy to see that KCL must be obeyed, so the Emitter current flowing into the base must exist through the collector; but how is that "obeyed" given the large electric field the carriers roll down when they reach the Base-Collector junction. Surely the electric field in the depletion region accelerates the carriers which by the Drift current equation: Drift Current Equation, increases the current as E can be very large?

\$\endgroup\$
  • \$\begingroup\$ In the saturation region this is not the case. However, as Vce reaches a certain point you have two opposing mechanisms (caused by the increase in Vce) as I see it; (1) an increase in Vce voltage ought to push more current and (2) a widening of the base-collector depletion layer ought to produce less current. Just my simple way of thinking telling me that when you move from saturation there are then two opposing things that tend to cancel each other and current remains largely the same. \$\endgroup\$ – Andy aka Dec 23 '15 at 11:36
  • \$\begingroup\$ the base collector junction does leak a little, so changing V_BC will change I_C a little, so there's no exactly 0 effect from collector voltage, just mostly zero effect \$\endgroup\$ – Jasen Dec 27 '15 at 7:02
  • \$\begingroup\$ Could you specify: which parameters are held constant while the collector–base voltage changes? \$\endgroup\$ – Incnis Mrsi Aug 28 '16 at 14:40
4
\$\begingroup\$

You seem to think that because the depletion area is large for the base collector junction that there must be a large field there. That is the opposite of how it works. The voltage difference between the emitter, base, and collector is dropped exclusively in the depletion region. Thus the larger the depletion region, the more spread out the field is and so the less strong the field is. In fact, back injection is one of the reasons for hetero junction transistors. In the non depletion regions, diffusion is the dominant mechanism. Thus by creating a large electric field that essentially acts like a reverse flow barrier, the BJT forces diffusion to occur. The reason the current flows from C->E (NPN) or E->C (PNP) is because the BJT uses different doping levels to create depletion widths for different purposes. The reverse biased C-B depletion region, being large, is dropped almost exclusively on the collector's side. That means there's very little length for diffusion. That small length means that the Collector becomes an excellent source of minority carriers for the base compared to the emitter, whose large electric field is almost insurmountable. When those minority carriers cross, they drive diffusion in the base. Once the base provides that minority carrier to the emitter, it can't get it back, so it must get another one from the collector.

In essence the collector current is dependent only on the amount that flows out of the emitter, not on the field, which is designed to drop almost entirely over the collector.

Please leave a comment on this if you don't understand, because I needed about 10 pictures to get this down in college.

Edit: The BJT picture to rule them all! enter image description here

\$\endgroup\$
  • \$\begingroup\$ I completely overlooked the fact that a larger depletion region = a weaker field!. I am not sure if I understand the rest: VCE, is designed to drop the largest at the base collector junction, this leads to a larger depletion region there, that ineffect decreases the electric field strength? So as VCE increases, its affect is compensated by a widening of the depletion region? However, current still increases in the active region because the effective base width also decreases thereby increasing the diffusion gradient? PS the pictures would be nice :) \$\endgroup\$ – Adil Malik Dec 26 '15 at 17:48
  • \$\begingroup\$ Okay so BJT's are a three terminal device and Vce is most of the time just a side effect of the more important Vbe, and Vcb. Because Vcb is a reverse biased pn junction, and because the collector is often doped at ~1/100th the concentration of the base, the majority of the field for Vcb only gets dropped on the collector side. The field from Vbe is used for a different reason. Since this is forward biased, the depletion region is small and the field is high. This high field makes it so that once a carrier has cross from b->e that it won't be able to go back, so it diffuses to metal contact \$\endgroup\$ – Dave Dec 26 '15 at 19:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.