Transistors and Diode Test

[TFM]

Homework Statement

Explain why the BC junction will give a smaller voltage drop, and will thus have a higher voltage measured then a BE Junction.

Homework Equations

N/A

The Attempt at a Solution

Okay, hopefully this is the right section to post this in. See, I have read the book, 'The Art of Electronics', and from this I seemed to have gathered the following:

'The difference in voltages occurs because the BC requires more energy, because it is flowing anti-biased, whilst the BE will be conductive, and allows the current to flow more easily,'

However, what I wrote seems to go against the above, it would indicate that the BE should have the smaller drop/higher voltage measured?

Thanks,

TFM

 

[Defennder]

Look at the equation for the built-in potential barrier $$V_{bi}$$. That's all you need to answer this question.

 

[TFM]

Does it have any other names, because I am looking through 'The Art of Electronics', and cannot find it in there? and this is a big book!

TFM

 

[Defennder]

I don't have a copy of that book, but Vbi is given as a function of Na and Nd, the acceptor and donor ion concentration of the pn junction respectively.

This equation:
http://ece-www.colorado.edu/~bart/book/builtin.htm

 

[TFM]

There isn't anything like that on the transistor section

This is a physics course, but this section, practical electronics, is more about cicuits. No mention of ions in the Transistor section?

TFM

 

[Defennder]

? Are you sure about that? But surely that equation is present somewhere? The acceptor and donor ion concentrations refer to the concentration of acceptor and donor atoms in the doped semiconductor.

 

[TFM]

Pretty sure - the course really is about setting up circuits, despite being in a physics course, and the book is for electronics students really.

 

[MATLABdude]

Pretty sure - the course really is about setting up circuits, despite being in a physics course, and the book is for electronics students really.

The Art of Electronics is a good reference / review book (or for getting up and running with practical--and simpler electronics), but not so great for really learning about semiconductor operation. Try Solid State Electronic Devices (Streetman and Bannerjee) or Microelectronic Circuits (Sedra and Smith)--generic EE semiconductor physics and active device books, for a better introduction.

Then again, the answer to your problem is (partially) answered at the Wikipedia page on BJTs under structure (along with what Defennder says):
http://en.wikipedia.org/wiki/Bjt#Structure

 

[TFM]

Looking at Wikipedia, is this a useful reference:

The bipolar junction transistor, unlike other transistors, is usually not a symmetrical device. This means that interchanging the collector and the emitter makes the transistor leave the forward active mode and start to operate in reverse mode. Because the transistor's internal structure is usually optimized to forward-mode operation, interchanging the collector and the emitter makes the values of α and β in reverse operation much smaller than those found in forward operation; often the α of the reverse mode is lower than 0.5. The lack of symmetry is primarily due to the doping ratios of the emitter and the collector. The emitter is heavily doped, while the collector is lightly doped, allowing a large reverse bias voltage to be applied before the collector–base junction breaks down. The collector–base junction is reverse biased in normal operation. The reason the emitter is heavily doped is to increase the emitter injection efficiency: the ratio of carriers injected by the emitter to those injected by the base. For high current gain, most of the carriers injected into the emitter–base junction must come from the emitter.

Looks useful,

TFM

 

[Defennder]

Yes it looks correct to my knowledge. Except I don't know enough to verify that "a of the reverse mode is lower than 0.5". Clearly the bulk of semiconductor device physics focuses on the forward active region of the BJT so I didn't study the three other less used modes.