Some fundamental questions about pn junctions

[cnh1995]

I am revising some topics in analog electronics and I am currently working on the BJT. I know the applications of these devices (diodes, transistors) and know the necessary formulae to work out problems.
But there are a few fundamental things which are still unclear to me and I am not very confident about my overall understanding of semiconductor physics.
So,
1)When we forward bias the B-E junction in a BJT, does emitter-base current start before the collector current? I mean the explanation I read in my books is like this:
"when the emitted electrons enter the base region, very few electrons recombine with holes and the rest are pulled in the collector region." Does this mean emitter injects electrons first irrespective of collector parameters (like biasing, doping etc) and then the collector pulls whatever electrons are injected in the base?

2)If a diode is having a very thin, lightly doped p region and wide, heavily doped n region and it is forward biased, will the forward diode current be very small due to fewer recombinations in the p-region or will the current be large due to large number of majority carriers in n-region that diffuse in the p region? In other words, can the electrons exiting p-region be conduction electrons (in which case the current will be large) or do they have to be only valence electrons (in which case this current will be small)?

Thanks in advance.

 

[Tom.G]

For 1) see http://www.fairchildsemi.com/ds/PN/PN2222A.pdf
Page 3, bottom table, Switching Characteristics, shows t_d (delay time) of 10nS

Cheers,
Tom

 

[cnh1995]

Thanks @Tom.G for the link , I looked up more stuff about transistor delay time and I think I have a satisfactory answer to 1).

What do you think about 2)?
I know the forward diode current I_d=I_electron+I_hole . Since the p-region is narrow and lightly doped, I believe I_hole should be very small but I_electron should not be affected, and a large number of conduction electrons in n-region should diffuse in the p-region (although only a few of them would recombine with holes and the rest will just move through the p-region as conduction electrons). Is this correct or does this violate any law in semiconductor physics?

 

[Tom.G]

Beats me. I posted a request for help in the Advisors Lounge thread asking for someone with Semiconductor Physics knowledge to jump in. It sometimes takes a few days, but people do tend to respond.

Cheers,
Tom

 

[jedishrfu]

Im no expert in semiconductor but did a search on some of your text and found this solution reference where they mention similar characteristics and some commentary about what they provide:

http://www.eecs.umich.edu/courses/eecs320/f00/solhw5.pdf

and this wikipedia article on depletion region:

https://en.wikipedia.org/wiki/Depletion_region

 

[cnh1995]

Thanks @Tom.G and @jedishrfu for your inputs. It turns out that there are two types of diode, long diode and short diode. The answer to 2) lies in the concept of short diode IMO. Need to study that tonight!

Thanks again!

 

[eq1]

For question #1 I think what you're really asking is what is the base-collector transit time. Or maybe to put it another way, what is the bandwidth of the transistor. I think this handout from OCW explains it pretty well.

https://ocw.mit.edu/courses/electri...vices-spring-2007/lecture-notes/lecture37.pdf

For question #2 the answer is; it depends on the biasing, I think. To be honest I am not sure I totally understand the question. I think what you're really asking is how does change in width and doping affect the current gain in a typically derived 1-dimensional forward biased NPN. If so, see the link below. Specifically equation 5.3.11.

https://ecee.colorado.edu/~bart/book/book/chapter5/ch5_3.htm

A book on these topics, which I highly recommend as it is written for the undergraduate, is very practical, and doesn't assume a lot of prior modern physics or math background, is Principles of Semiconductor Physics by Sima Dimitrijev.

 

[dlgoff]

long diode and short diode

I guess I never heard diodes described as long or short before so I checked one of my favorite online books; http://ecee.colorado.edu/~bart/book/book/contentc.htm. The section on https://ecee.colorado.edu/~bart/book/book/chapter4/ch4_4.htm gives this description:

The "long" diode expression applies to p-n diodes in which recombination/generation occurs in the quasi-neutral region only. This is the case if the quasi-neutral region is much larger than the carrier diffusion length. The "short" diode expression applies to p-n diodes in which recombination/generation occurs at the contacts only.

Thanks Electrical, Computer & Energy Engineering University of Colorado Boulder.

edit: @eq1 mentioned ecee first but a different chapter. :)

 

[cnh1995]

Excellent book!
Thanks a lot @eq1 and @dlgoff!

 

[jsgruszynski]

I am revising some topics in analog electronics and I am currently working on the BJT. I know the applications of these devices (diodes, transistors) and know the necessary formulae to work out problems.
But there are a few fundamental things which are still unclear to me and I am not very confident about my overall understanding of semiconductor physics.
So,
1)When we forward bias the B-E junction in a BJT, does emitter-base current start before the collector current? I mean the explanation I read in my books is like this:
"when the emitted electrons enter the base region, very few electrons recombine with holes and the rest are pulled in the collector region." Does this mean emitter injects electrons first irrespective of collector parameters (like biasing, doping etc) and then the collector pulls whatever electrons are injected in the base?

2)If a diode is having a very thin, lightly doped p region and wide, heavily doped n region and it is forward biased, will the forward diode current be very small due to fewer recombinations in the p-region or will the current be large due to large number of majority carriers in n-region that diffuse in the p region? In other words, can the electrons exiting p-region be conduction electrons (in which case the current will be large) or do they have to be only valence electrons (in which case this current will be small)?

Thanks in advance.

Minority carrier injection in on both sides of any PN junction and only requires base junction bias. This is a "fundamental" property of the PN junction itself. In the case of metal-semi Schottky junctions, you get injection on both sides also but the metal's carrier concentration is so much higher than any semiconductor that the injection distance is practically zero (or maybe on or two atoms only). So you are getting emitter injection into the base regardless of the collector bias.

And you are getting injection into the emitter from the base as well! This is a key aspect of "emitter efficiency" which is tuned up with a heterojunction BJT: the reverse injection of a heterojunction is pretty much zero so the injection efficiency is very high. Consider that the reverse injection is canceling some of the forward injection for normal forward-linear operation.

These "decay" exponentially with distance as long as their is a current path to suck up the recombination current that must be supplied to enable recombination. This is also related to base transit and base width: BJTs ONLY work when the depletion thickness is wider than the physical base width (based on including the base-collector depletion layer width from the metallurgical junction).

It's this minority carrier injection that leads to the extra diffusion capacitor seen in BJTs - if you pulse the base-emitter voltage and measure the base current, you'll get a different answer for the collector being open vs. being shorted to the base or emitter because of diffusion capacitance being caused by injected minority carriers NOT being instantly recombined. When you short the collector, you have a current path for that recombination current.

Per the diode depletion layer equation, the answer for this is pretty obvious. Intuitively you get a smaller depletion layer with higher doping because you "cancel" the charge in a shorter distance. The current you see in either case (light or heavy doping) changes the Is (saturation current) value though it's as a ratio with the intrinsic doping.

https://en.wikipedia.org/wiki/Saturation_current

There's never a current that is valance band electrons. All the electrons (or holes) that conduct current are 100% from dopant atoms near the conduction (or valance) band, respectively that have been thermalized into the conduction (or valance band) by heat. This is why semiconductors don't work at cryogenic temperatures: there are no longer free carriers available - for silicon it becomes like diamond or silicon carbide - an apparent insulator. This is also why, with enough heat, you can make the latter two into semiconductors.

 

[Chris Riccard]

Can you explain light and heavy doping, I always thought it was based solely on the electron configuration of silicon minus or plus an electron in the crystal lattice.