4
\$\begingroup\$

enter image description here

For the given circuit, a statement has been given in my book, saying

It is obvious that if Vbe is increased by a small amount, both hole current from emitter region and the electron current from the base region will increase. As a consequence both Ib and Ic will increase proportionally

I believe there is a mistake and it should be rather,

electron current from the emitter region

Because BE seems to be forward biased and thus the majority carriers, that is, electrons from Emitter should increase and go into the base and further in collector.

Am I right or wrong?

\$\endgroup\$
2
  • \$\begingroup\$ @WhatRoughBeast, you are saying that the book would be "wrong". It certainly would help if you would be more specific: Which statement in the book is wrong? And - who has mentioned a current source ? Neither the book nor the questioner? \$\endgroup\$
    – LvW
    Mar 16 at 11:47
  • \$\begingroup\$ @LvW - In looking at the figure more closely, and thinking about your comment, I realized that I was mistaken. I interpreted a current meter on the base as a current source, which is clearly what is technically called a brainfart. I have deleted my comment. Thanks for drawing it to my attention. \$\endgroup\$ Mar 17 at 16:15
1
\$\begingroup\$

You are correct and the book is wrong if interpreted as follows. The emitter majority carriers in an NPN BJT are electrons that are injected into the base. Holes from the base cross the b-e junction and recombine as minority carriers with electrons in the emitter just as electrons recombine as minority carriers in the base with the majority-carrier holes, resulting in base current and finite beta.

However, the book is referring to the minority carriers in both emitter and base and as such, they are holes in the emitter and electrons in the base. So both you and the book are right if each is referring to the right currents, minority or majority, in base and emitter.

\$\endgroup\$
0
\$\begingroup\$

The statement of the book is correct. When you increase the base-emitter junction voltage the electron current injected into the base (and thus the collector current) increases but also the hole current injected from the base into the emitter increases similar to an isolated pn-junction. This hole current is usually the dominant cause for the base current flowing into the base. Thus the collector current and the base current increase proportionally to each other when you increase the base emitter-voltage.

\$\endgroup\$
3
  • \$\begingroup\$ But then wouldn't that be hole current from the base region rather than the emitter region? And why would that be more significant? Base is slimmest and least doped. While the emitter competitively thicker and more doped \$\endgroup\$
    – Daksh Shah
    Mar 10 '17 at 16:26
  • \$\begingroup\$ That's what I said. The holes injected from the base into the emitter are the main cause for the base current. This hole current is, of course, much smaller than the electron current injected from the emitter into the base which is practically the same as the collector current. \$\endgroup\$
    – freecharly
    Mar 10 '17 at 17:33
  • \$\begingroup\$ Hole current... that always made me laugh... Reminds me of the gardener moving a hole from one end of the garden to the other by digging a hole next to it an filling in the previous. \$\endgroup\$
    – Trevor_G
    Mar 10 '17 at 17:49
0
\$\begingroup\$

You can take into consideration the formula Ic=Isexp(Vbe/Vt), so obviously if Vbe rises the collector current rises, and if transistor is directly active, you can use the formula Ic=betaIb, so obviously if Ic rises, Ib also rises.

\$\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.