Transistor-saturation-input characteristics

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The discussion focuses on the behavior of transistors in the saturation region, particularly how charge carriers move from the base to the collector. It is clarified that in saturation, the transistor behaves like two back-to-back diodes, with the base-emitter junction acting as a pn junction that does not change current unless influenced by the collector. The input characteristics of a BJT are analyzed, highlighting that in saturation, the base experiences a smaller depletion region, requiring less voltage to achieve the same current compared to the linear region. The conversation emphasizes the role of minority carriers and the importance of base width in controlling current flow. Overall, a deeper understanding of charge transport mechanisms in different operational regions of BJTs is sought.
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So I want to discuss a few things about transistors in saturation region.

First the saturation part.

Lets say the saturation occurs at 0.2 V and transistor cannot go below that voltage. We say that it is bottomed and its now in saturation region.

How do charges go from base to collector? Does transistor effect still occur?
Let me clarify what I mean.

I know when in linear polarization, charges, get injected into base-collector depletion region and get swept across it because of large electric field. (without much discussion, let's say that this the effect)

But when we have saturation region, base-collector is directly polarized, and charges don't have that depletion region to get swept across.

Do they then, pass over to collector normally like in diodes?

I generalized this pnp and npn. You may answer like so, too.

Now for the second part, which is closely related. http://pokit.org/get/c4446cd88893f30ed96ced2ec01d0b14.jpg I was told that this is input characteristic of a BJT. I believe these curves represent the diode characteristic of base-emitter.

I want to discuss the difference between 2 curves, when transistor is in linear and saturation region.

My explanation of the curves is:

When transistor is in saturation region, base only has depletion region from emitter-base junction. And indeed, this part really acts like a diode, and will have that characteristic. Peaks out at 0.7 VBut when transistor is in linear region, we have that depletion region from collector-base too. So base is being "squished" between 2 depletion regions.

Because of this, we will need a larger voltage drop, to achieve the same current as in saturation region case. In a nutshell, base "offers" smaller path for current to go through and ergo we need more voltage to get more current?

Am I thinking right here?
 
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Yes, you are very correct. In forward active the BC jn acts like controlled source. In saturation the transistor is just two back to back diode.
 
Kholdstare said:
Yes, you are very correct. In forward active the BC jn acts like controlled source. In saturation the transistor is just two back to back diode.

Thank you, makes sense. Any thoughts on the second part?
 
Base - emitter junction is essentially pn junction. Thus it will not change its current until and unless its properties are changed (by collector). Usually emitter is heavily doped n-type. Thus approximately all Ibe is carried by electrons. Now electrons in base is minority carriers. And its concentration gradient (this quantity is proportional to the base emitter current) depends on how close is the CB depletion region. Actually we deliberately keep base width small such that collector can easily control base carriers. Otherwise it won't work.
 
Kholdstare said:
Base - emitter junction is essentially pn junction. Thus it will not change its current until and unless its properties are changed (by collector). Usually emitter is heavily doped n-type. Thus approximately all Ibe is carried by electrons. Now electrons in base is minority carriers. And its concentration gradient (this quantity is proportional to the base emitter current) depends on how close is the CB depletion region. Actually we deliberately keep base width small such that collector can easily control base carriers. Otherwise it won't work.

Yes, I know that. But the curves, is the explanation right? You do see the difference between first and second curve?
 
If base offers smaller path for current flow and you consider drift transport, would not the current increase due to lower resistance of smaller base?

Diffusive transport cannot be thought of as a simple intuitive rule. Try doing that for a pn junction.

The point here is the bjt diffusive transport has a parameter minority carrier. In linear they have to vary very steep as base is small. In saturation they can relax and even store extra charge due to large base width. The greater the slope of the minority carrier the more "diffusion pressure" the more current.
 
Kholdstare said:
If base offers smaller path for current flow and you consider drift transport, would not the current increase due to lower resistance of smaller base?

Diffusive transport cannot be thought of as a simple intuitive rule. Try doing that for a pn junction.

The point here is the bjt diffusive transport has a parameter minority carrier. In linear they have to vary very steep as base is small. In saturation they can relax and even store extra charge due to large base width. The greater the slope of the minority carrier the more "diffusion pressure" the more current.

Ok thanks, I see I will have to dig a little more deep into this base and minority carrier thing.

Thank you very much, kind sir, you have been very helpful.
 
You're welcome. :smile:
 

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