3.3 Doped Semiconductors

From the previous section we found that the number of charge carrying species is extremely small in the case of pure semiconductors -- it was of the order of 1016 #/m3. That is, only one out of one trillion Si atoms loses its electron to the conduction band. Doping is a process by which we can increase structurally the number of conduction electrons or conduction holes in the Si crystal. If we were to increase the number of electrons by say 1016 #/m3, conductivity will double. Although 1016 charges per m3 seems large, it turns out to be a very small concentration, of the order of one part in a trillion.

How do we then introduce conduction electrons or holes. It can be done by forcing into a silicon crystal another atom that has more number of valence electrons. An example of this is phosphorous. which is pentavalent. When we introduce trivalent atoms such as Boron, we create holes or positively disposed semiconductor, also called p-type semiconductor. When phosphorous is introduced we create negatively disposed semiconductor because of excess electrons and is called n-type semiconductor.

You will note that Group III elements are a source of p-type dopants and n-type semiconductors are created using Group V elements. Another term commonly used is donor and acceptor semiconductor. n-type is called donor because of its ability to "donate" electrons while p-type is called acceptor because of its propensity to "accept" electrons.


Group V elements = Forms Donor Semiconductor = n-type Semiconductor

Group III elements = Forms Acceptor Semiconductor = p-type Semiconductor

When we introduce dopants to an intrinsic semiconductor, practically all donor or acceptor atoms generate electrons or holes in proportion to their concentration in Si. This is because intrinsic concentration of free electrons or holes in pure Si crystal is so small, we can make the above approximation without any loss in accuracy. Let us consider an example.


Example 2: What concentration of Boron is needed to increase Silicon conductivity from 4.4 x 10-4 S/m to 100 S/m.

Solution: Since B is a Group III element, it will introduce holes into Si crystal. Let the concentration of holes be p. Mobility of p in Si is 0.048 m2/A-s. Use conductivity equation to determine desired value of p.


The number of negative charges in the silicon will be very small compared to holes introduced by Boron doping. Hence, we will set n = 0 in the above equation. Substituting values, we get:


That is, about 0.3 parts per million Boron is needed !


The above example shows that to increase conductivity by nearly a million fold, we need B concentration of 0.3 ppm! In Example 1, we found that resistance of pure silicon rod was 2.9 x 109 ohms. If it was modified with 0.3 ppm B, its resistance would decrease to 12,760 ohms. In the next section, we will learn how to determine conditions needed (temperature and time) to achieve the above desired level of Boron concentration.