# BJT dc analysis help

Discussion in 'Homework Help' started by kaiosama, Aug 6, 2013.

1. ### kaiosama Thread Starter New Member

Dec 6, 2010
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Sedra & Smith Microelectronics, 6th edition, page 384. Example 6.6. Here is the schematic:

The solution says:
"The collector-base junction cannot conduct, since the n-type collector is connected through Rc to the positive power supply while the p-type base is at ground. It follows that the collector current will be zero."

I don't really understand this explanation. I was under the impression that for a transistor in active mode, Vc > Vb so CB junction is reverse biased, which is the case in this example.

What am I missing? Thank you.

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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What was the question asked in relation to the schematic in the text. You only gave us the proffered answer.

3. ### WBahn Moderator

Mar 31, 2012
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I suspect that the quote is either an answer to a very narrowly asked question or only a part of a more complete answer.

An example of a narrow questin might be: Because the base-emitter junction is not forward biased, no collector-emitter current flows. Why doesn't any collector-base current flow?

This is not only wrong, but it is a case of apples and oranges. You are talking about a transistor being in active mode while the transistor in the example problem is in cutoff.

Even in active mode, Vc does not have be greater than Vb. Think of a transistor that is almost in saturation but still in the active region. Since Vcesat < Vbe, that means that Vc<Vb, though not by much.

Finally, the question was asking about the collector current. Even in a transistor that is cutoff and has no emitter current, you can still get base-collector current if Vc is lower than Vb by a diode drop (i.e., if it is forward biased sufficiently to turn it on).

4. ### Brownout Well-Known Member

Jan 10, 2012
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That's what makes a transistor special. In cutoff, as in the drawing, current cannot flow through the reversed-biased C-B junction. So, the answer given is correct. However, in the active region, minority carriers are injected into the transistor base, lowering the barrier to C-B current. In active mode, IB=IC/β and IE=IC/α. So, while the transistor is in cutoff, the C-B junction is a norman P-N junction. But in the active region, the junction acts very differently.

5. ### screen1988 Member

Mar 7, 2013
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Is this case transistor operating in reverse-active mode?

6. ### LvW Active Member

Jun 13, 2013
674
100
Kaiosama,
Let me try a rather short answer: The base-emitter diode is not forward biased. That`s all.
As for the classical pn-diode, you need approx. +0.7 volts between B and E.
Remember: The bipolar transistor can be seen as a voltage-controlled current source and the controlling path is the pn junction between base and emitter.

7. ### WBahn Moderator

Mar 31, 2012
17,715
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Did you mean to say base-collector diode?

8. ### hobbyist Distinguished Member

Aug 10, 2008
764
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The point the book is trying to make, is that the transitor is not being used as a transitor, in this case, it is hooked up as a reverse biased diode, through the base collector junction, any diode reversed biased will not conduct current.

However when the base emitter diode is forward biased, then the reversed biased collector base diode, will conduct due to transitor action.

The base collector diode is not simply a reversed diode anymore, its taking on another characterstic, because the base substrate, is connected to both the emitter and collector, so when the emitter is forward biased, with the base substrate, then the base collector junction is no longer just a reversed biased diode alone, it has a new characteristic due to its same substrate connected to the emitter.

Its all in the design of a transitor, verses a simple diode alone.

9. ### LvW Active Member

Jun 13, 2013
674
100
No - I didn`t.
For my opinion, the main question was: Why no collector current?
Ans my answer was: Because the B-E junction is not forward-biased.
Anything wrong?

10. ### WBahn Moderator

Mar 31, 2012
17,715
4,788
Okay, I see where you are coming from.