0
\$\begingroup\$

Regarding BJT, forward active mode of operation i don't seem to get something. I will use NPN type. In the forward active regime, Vbe>= 0.75 V, so electrons flow from the emitter through the base .Ok, what i do not understand is why if Vbc <= 0.75 V there is a current Ic = Beta*Ib, how come electrons flow from emitter to collector if that part of the transistor is "closed" (Vbc <= 0.75)

schematic

simulate this circuit – Schematic created using CircuitLab

\$\endgroup\$
1
\$\begingroup\$

The thing you need to know is that the base is very thin/short.

So the electrons travel from emitter to base "thinking" it's a standard forward PN junction. However, like I said, the base is very small so many of the electrons end up at the collector side of the base and suddenly find themselves inside a big-#ss collector :-)

So there they all combine and form the collector current. Only few electrons make it out of the base without ending up in the collector, this explains the small base current.

So the base-collector junction itself isn't operating as if it is in reverse mode. It's not the junction causing the current, it is the base that is injecting carriers into the collector.

\$\endgroup\$
  • \$\begingroup\$ ..just to avoid misunderstandings: FakeMoustache is right (it is the base that is injecting carriers into the collector), however, I am sure he is referring to the base region and NOT to the base terminal or the base voltage. \$\endgroup\$ – LvW Jan 12 '16 at 17:17
  • \$\begingroup\$ That is correct, when I wrote "base" I mean the base area inside the silicon. The electrons that do not leave the base through the collector exit via the base terminal which is the conductive contact to the base area. \$\endgroup\$ – Bimpelrekkie Jan 12 '16 at 18:30
2
\$\begingroup\$

It's a bit unclear what you are really asking, but it seems the answer is "Because that's what transistors do". For more details, see semiconductor physics.

Also, you should be thinking about the current thru the base, not so much the B-E voltage as driving things. B to E looks like a diode to the external circuit, so will have about 700 mV drop depending on current. But, the function of voltage to current is quite steep, and current is ultimately what matters anyway.

V2 in your schematic should be a current source, not a voltage source. Let's say you set it for 100 µA and the transistor has a gain of 50. That means the collector can sink up to (100 µA)50 = 5 mA. If R1 were 300 Ω, then it would drop (5 mA)(300 Ω) = 1.5 V. The collector would therefore be at 3.5 V.

The above example was assuming the transistor had a gain of exactly 50 at the operating point we chose. Real transistor gains vary widely. The datasheet generally specifies a minimum, but no maximum. It's not unheard of for actual parts to have 10x more gain than the minimum guaranteed value.

Unpredictable gain is one reason transistors used in linear operation are biased with some feedback. The actual C-E voltage is used to adjust the bias current to keep the collector voltage within a useful range over the possible range of gains from part to part.

\$\endgroup\$
  • \$\begingroup\$ from what i understand there is a depletion layer between the Base and the Collector, a depletion layer that is supposed to stop electrons from flowing in there..that is what confuses me \$\endgroup\$ – user3682109 Jan 12 '16 at 12:34
  • 1
    \$\begingroup\$ @user: What are you trying to learn, semiconductor physics or electronics? If electronics, don't get hung up on how the details work inside. At first approximation, consider the B-E voltage fixed at roughly 700 mV, and the C-E current the gain times the B-E current. There are of course more details, but you can get a lot done with that simple view while you're learning. \$\endgroup\$ – Olin Lathrop Jan 12 '16 at 12:38
  • \$\begingroup\$ @user3682109, the depletion layer of the Base-Collector junction is more for shrinking the distance minority carriers have to travel across in the collector to get to the base. Then, once they reach the depletion layer boundary, they get swept across without any time to recombine. Once they've been move across another must appear to maintain charge neutrality. (note that i was talking about holes this hole time as NPN transistors are easiest to understand when talking about holes rather than electrons) \$\endgroup\$ – Dave Jan 12 '16 at 17:08
1
\$\begingroup\$

Forward active region biases the device so that it doesn't matter what value Vbc is so long as it's within electrical tolerances and less that 0. The Vbc voltage has nothing to do with determining current through Ic in forward active.

As for "close" I don't understand what you mean. Current flows like that because we forced it to do so. The BJT is constructed so that in an NPN with the Vbe forward biased and the Vbc reverse biased that current flows from the collect through to the emitter with a small leakage current added by the base.

\$\endgroup\$

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.