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I was studying Millman and Halkias’ ‘Integrated Electronics’ (Tata McGraw Hill, 1991) when I hit this paragraph and couldn’t proceed any further. I’d really appreciate suggestions of any kind.

I will first reproduce here the above mentioned paragraph, then write down the assumptions which I believe can be made and then proceed to list my questions.

Pg 27,

“2-4 Charge Densities in a Semiconductor

Equation (2-10) namely np = n(sub i)^2, gives one relationship between the electronnand the holepconcentrations. These densities are further interrelated by the law of electrical neutrality, which we shall now state in algebraic form: LetNdequal the concentration of donor atoms. Since, as mentioned above, these are practically all ionized (ψ),Ndpositive charges per cubic meter are contributed by the donor ions a. Hence the total positive-charge density isNd+p. Similarly, ifNais the concentration of acceptor ions, these contributeNanegative charges per cubic meter. The total negative-charge density isNa+n(b). Since the semiconductor is electrically neutral, the magnitude of the positive-charge density must equal that of the negative concentration, or

Nd+p=Na+n(Eq: A)”

Assumptions:

1.nrefers to the conc of free electrons.

2. When a donor atom is ionized, it loses one free electron and when an acceptor is ionized it gains one electron.

3. ALL the donor and acceptor atoms are ionized (from ψ).

Questions:

From a: If Nd donor atoms get ionized producing Nd positive ions, wouldn’t that also produce Nd free electrons, thereby increasing the RHS of equation A by Nd?

Wouldn’t some of these free electrons also fill up some of the p holes, reducing the LHS?

From b: Since the Na atoms become ionized by absorbing some of the free electrons from n, wouldn’t the total negative charge remain unchanged?

Thanks

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# Electron and hole concentrations - Long One.

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