0
\$\begingroup\$

Regarding the active mode of operation: Icbo is leakage of the reverse biased C-B junction. It flows whether or not the emitter is floating, as long as there is a complete C-B circuit. The emitter just has to be floating in order to measure it. But It always flows whether or not the emitter is floating. So the conclusion Icbo always flows. The question is does it always(CB and CE) flow through the emitter base junction. It seems to me that it doesn't flow through emitter base junction in CB but it does in CE (?). Now to express my confusion I start from the basic equations and develop the question as follows:

Now the following equations are valid for a BJT in any configuration CB or CE ect.:

Ic = α×Ie + Icbo

Ie = Ic + Ib

So from those above the following can be derived:

Ic = (α/(1-α))×Ib + (1/(1-α))×Icbo

Since α/(1-α) = β

Ic = β×Ib + (β+1)×Icbo

Now above the second part of the equation is (β+1)×Icbo and it is called Iceo.

This means Icbo is amplified and became Iceo.

But here my question is: The above equation are valid both for CB and CE configurations. But in common base configuration Icbo is not flowing through emitter base junction yet it appears in the formula as if it is amplified as being (β+1)×Icbo.

This is confusing, is my knowledge is wrong or how should we interpret this amplification of Icbo appearing in the formula of CB configuration.

Edit:

Here is an excerpt from a text:

enter image description here

I saw many paragraphs like above which states "However, when the transistor is in the common-emitter connection...". Obviously these are telling that this Icbo is more a problem in CE because Icbo flows through emitter base junction and gets amplified.

But again in CB we still have the same equitation with (β+1)×Icbo which indicates Icbo multiplied by factor of β. This is very confusing why they relate the issue only with CE but not CB. I hope I made the question more clear now.

Another example:

enter image description here

It says "The leakage current in CE is larger than that in CB" and the reason is given by Iceo = (β+1)×Icbo.

But this same equation appears in CB as well

Ic = β×Ib + (β+1)×Icbo

Where am I wrong? Why is the leakage current in CE is larger than that in CB? Can we show it explicitly by an equation.

\$\endgroup\$
4
  • \$\begingroup\$ If the base is tied to some "ground", so most of the Icbo can exit thru the base, then the "amplification" is only 1, or nearly 1 (alpha, or 1/alpha). Is alpha the ratio of collector-to-emitter current? thus is about 1.0? \$\endgroup\$
    – analogsystemsrf
    May 14, 2019 at 13:20
  • 1
    \$\begingroup\$ The current relations inside the transistor still hold whatever the configuration (CB, CE) you use the BJT in. In CB the Icbo current can flow out of the base. However, in a CB configuration I expect that a (much larger) Ib is flowing into the base. That much larger Ib will "eat up" the small Icbo, the net resulting Ib will then be (Ib - Icbo). \$\endgroup\$
    – Bimpelrekkie
    May 14, 2019 at 13:24
  • \$\begingroup\$ @Bimpelrekkie (β+1)×Icbo = β×Icbo + Icbo. This indicates at least in the formula Icbo is amplified by factor β. I'm wondering how this happens because transistor amplification action only as far as I know is due to charge carries passing through emitter base junction. Do we at least agree Icbo does not flow through emitter base junction in a properly biased CB? \$\endgroup\$
    – user1245
    May 14, 2019 at 13:33
  • \$\begingroup\$ I know that Ib - Icbo is the net base current but that is not my question. \$\endgroup\$
    – user1245
    May 14, 2019 at 14:18

1 Answer 1

Reset to default
1
\$\begingroup\$

I think I found where it was hidden when I draw the charge carriers loops:

enter image description here

where Ib’ = Ib-Icbo

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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