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In a transistor, we know that current amplification factor alpha (DC) for CB Configuration is given by :

Alpha = Ic / Ie

Where, Ic = collector current; Ie = Emitter current

This implies that :

        Ic = alpha × Ie ....................(1)

Also, the total current is given by :

  Ic = alpha × Ie + Icbo ................(2)

Where, Icbo = collector base current with open Emitter (leakage current)

From 1 and 2,

Ic = Ic + Icbo
Icbo = 0 ...................................(3)

This means for any numerical values of alpha, Ic and Ie, the leakage current is always going to be 0. But practically, this is not the case. A small current of the order of micro/nano amps flows as Leakage current. This contradicts the above equation. Does this mean to say that the above equation is faulty? Please explain.

check out this image from my textbook... This is where i got the second equation which I have highlighted in yellow. Also have a look at the image at the top right. A page from my textbook

Aditya DS
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  • Have you looked a current diagram of a BJT transistor? – Voltage Spike Nov 03 '16 at 17:19
  • Yes I have seen the diagram showing direction of currents – Aditya DS Nov 03 '16 at 17:21
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    Related: [ICEO, ICBO physical interpretation in BJT](http://electronics.stackexchange.com/questions/124668/) – The Photon Nov 03 '16 at 17:29
  • eq 1 and 2 contradict each other. – Andy aka Nov 03 '16 at 17:35
  • No, the first equation must be true by definition of alpha(current amplification factor). What Im not sure is the validity of the second equation. How true is the second equation is my question – Aditya DS Nov 03 '16 at 17:56
  • No, the first equation must be true by the definition of alpha(current amplification factor)... What Im not sure is the validity of the second equation. – Aditya DS Nov 03 '16 at 17:58
  • Since the second equation contradicts the first equation, which is true by definition, what can you conclude? – The Photon Nov 03 '16 at 18:13
  • I think you (OP) have a grossly oversimplified view on BJTs are are struggling to make sense of both incorrect as well as highly distorted views about BJTs. For a comprehensive DC model that does make sense, see the equations in my answer here for three equivalent DC perspectives: http://electronics.stackexchange.com/questions/252197/why-is-vbc-absent-from-bjt-equations/252199#252199 and for the currents see Figure 3-4 in my answer here: http://electronics.stackexchange.com/questions/254391/terminology-for-bjt-regions-of-operation/254397#254397 – jonk Nov 03 '16 at 18:47

1 Answers1

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Yes any real transistor does have some leakage. Usually a resistance Ro is added in the model between the collector and emitter to account for leakage.

The equation Ic = alpha * Ie is only an approximation based on a simplified model of a transistor.

It is no more faulty than say Newtons laws of motion, even though we know that they become inaccurate at relativistic speeds. In the same way the equation in your book gets very close to the real answer most of the time.

Most text books start with the simplified model, and then gradually build up their model of a transistor adding additional parts to more closely approximate the real behavior. Starting with the full model would probably hinder understanding for most people.

user4574
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