0
\$\begingroup\$

While studying Grob's Basic Electronics chapter 28 I came across this statement,

Notice in Fig. 28–2 that the EB depletion layer is narrower than the CB depletion layer. The reason for the different widths can be attributed to the doping level of the emitter and collector regions. With heavy doping in the emitter region, the penetration into the n material is minimal due to the availability of many free electrons. On the collector side, however, there are fewer free electrons available due to the more moderate doping level in this region. Therefore, the depletion layer must penetrate deeper into the collector region to set up the barrier potential, VB , of 0.7 V. In Fig. 28–2, dash marks are used in the n-type emitter and collector to indicate the large number of free electrons in these regions. Small circles are used to indicate the holes in the p-type base region. (For an npn transistor, holes are the minority current carriers in the n-type emitter and collector regions, whereas free electrons are the minority current carriers in the p-type base.) enter image description here

Isn't it the converse? If there is heavy doping in the collector region and it has free electrons, will more electrons from the emitter not move into the base than the collector? The emitter has more electrons than the collector and will have much more repulsion to push many electrons to the base to form depletion layer.

The above all is about NPN transistors.

\$\endgroup\$
6
  • \$\begingroup\$ Are we discussing an npn or a pnp transitor? \$\endgroup\$
    – Syed
    Dec 2, 2021 at 15:50
  • \$\begingroup\$ NPN transistor... Sorry I should add this to the question. \$\endgroup\$ Dec 2, 2021 at 15:56
  • \$\begingroup\$ Think about the relevant voltages (b-e and c-b) \$\endgroup\$
    – user16324
    Dec 2, 2021 at 18:46
  • \$\begingroup\$ @mohsinraza Are you discussing an unbiased case? (Just a BJT sitting on the table, unconnected?) It looks as though that's the case. Just want to be sure. Also, do you feel comfortable with the separated ideas of diffusion and drift currents? (To test that in your mind, do you agree that in the case of a simple unbiased p-n junction that both diffusion and drift currents exist in the depletion region, but in such a way that the net result is in equilibrium?) \$\endgroup\$
    – jonk
    Dec 2, 2021 at 19:12
  • 1
    \$\begingroup\$ @mohsinraza I see you feel you have an answer. Great! If you have a high dopant concentration on one side (emitter) and very lightly so on the other (base, which is thin like a membrane, too) then the diffusion doesn't proceed deeply into the emitter region as it has far more available "slots" than the base can provide. Similar is true in the collector-base side because the collector is doped more than the base, too. If the emitter were lightly doped then the base charges could penetrate more deeply. But that's not the case. \$\endgroup\$
    – jonk
    Dec 3, 2021 at 20:41

1 Answer 1

1
\$\begingroup\$

When you say this: -

Why is the depletion layer near the emitter region narrower than the collector region?

I would point out that for a BJT, the normal operation of the base emitter is one of being forward biased and hence, the depletion region between base/emitter is much smaller than between base/collector because, the base/collector is working with a reverse bias for most applications in most of the time: -

enter image description here

The above is taken from here and shows the relative widths of the depletion regions for a PN junction where anode is the P material (equivalent to the base of an NPN transistor) and N is either collector or emitter.

With a forward bias (normal base/emitter situation) the depletion region is smaller. With a reverse bias (collector/base) the depletion region is bigger. Clearly, due to the very tiny base width of a BJT this is made slightly more complicated but, the general picture can be understood.

Regarding what the book says (and I'm very aware that such a limited quote might be easily misconstrued), heavier doping does mean a higher availability of carriers and given that in an NPN device the majority carriers (electrons) are in the emitter and collector, it makes total sense.

If there is heavy doping in the collector region and it has free electrons, will more electrons from the emitter not move into the base than the collector?

You don't want that to happen, you want the majority of electrons in the emitter to be swept across to the collector under control of the base.

\$\endgroup\$
4
  • \$\begingroup\$ I should update the statement and add the diagram to the text. I understand those biasing statements but the thing I am confused in is that the transistor is not connected to the battery yet.. \$\endgroup\$ Dec 2, 2021 at 16:58
  • \$\begingroup\$ It will still be because the emitter is more heavily doped than the collector. \$\endgroup\$
    – Andy aka
    Dec 2, 2021 at 17:04
  • \$\begingroup\$ @mohsinraza The original papers are excellent sources: "The Theory of p-n Junctions in Semiconductors and p-n Junction Transistors," Shockley, 1949; "Electrons and Holes in Semiconductors," Shockley, 1950; "p-n Junction Transistors," Shockley & Sparks & Teal, 1951; "Some circuit properties and applications of n-p-n transistors," Wallace & Pietenpol, 1951; "Effects of Space-Charge Layer Widening in Junction Transistors," Early, 1952; "Properties of Junction Transistors," Smith, 1953; and "Large-Signal Behavior of Junction Transistors," Ebers & Moll, 1954? These are sufficient. \$\endgroup\$
    – jonk
    Dec 2, 2021 at 19:03
  • \$\begingroup\$ @mohsinraza please take the 2 minute tour to understand the motivation behind giving free help. \$\endgroup\$
    – Andy aka
    Dec 3, 2021 at 13:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.