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Why does a linear charge profile in the base imply no recombination of charges at the base, but in case exponential decay it implies recombination at the base?

bjt charge profiles

Why does charge density at the base decrease with distance reaches zero at junction of collector? I read some answers saying that it is because of the electric field but the E field has no effect outside the junction due to Gauss' law.

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When the B-E junction is forward biased, carriers are injected from the emitter into the edge of the base near emitter. These are electrons (e-) in the case of an NPN transistor. Carriers (holes) are also injected from the base into the emitter, but we can ignore that for now.

If the B-C junction is reverse biased, then the minority carrier (e- for NPNs) concentration at that junction edge is 0.

Now we have a situation with a high density of carriers at the B-E edge, and a 0 density at the B-C edge. Ignoring all other effects and non-idealities, the carriers will diffuse from the high concentration to the low concentration. If none get 'lost' along the way, then the concentration gradient must be constant (because the diffusion flow depends only on the concentration gradient). This will give a linear charge profile.

However there can be some significant non-idealities. These carriers are minority carriers in the region of interest (e- in a p-type region). Thus they are subject to recombination -- some fraction of the carriers will recombine. The rate of recombination depends on the local minority concentration (ignoring many non-idealities and other effects).

Because the rate of recombination depends on the concentration, this leads to an exponential decay in the concentration with distance.

In a useful BJT, the base is lightly doped and so the recombination rate is relatively slow. Most carriers 'make it through'.

The original (linear) slope and the additional exponential decay combine to add some curvature to the carrier concentration.

Other things that affect this are:

  1. Non-uniform base doping. This means that recombination rates change along the base.
  2. The base recombination (and also the injected carriers from the base to the emitter) creates a base current. The I*R drop from this creates additional fields in the base which affects the carrier distribution.

In useful BJTs - at low frequencies - these other effects are negligible, and recombination and injection efficiency become the dominant non-idealities, leading to hFE parameter beta being in the range of 100 instead of infinite if everything was ideal.

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  • \$\begingroup\$ I dont get why none of charges gets lost along the way the profile has to be linear \$\endgroup\$ Commented Aug 31, 2021 at 11:28
  • \$\begingroup\$ Because the carrier current is constant from one edge of the base to the other, and the current is driven by a concentration gradient (diffusion). The current is J = q.D.d(concentration)/dx \$\endgroup\$
    – jp314
    Commented Sep 1, 2021 at 0:34

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