Equilibrium Electron and Hole Concentrations

4.5 | CHARGE NEUTRALITY

4.5.2 Equilibrium Electron and Hole Concentrations

Figure 4.14 shows the energy-band diagram of a semiconductor when both donor and acceptor impurity atoms are added to the same region to form a compensated semiconductor. The fi gure shows how the electrons and holes can be distributed among the various states.

The charge neutrality condition is expressed by equating the density of negative charges to the density of positive charges. We then have

n0 N _a p0 N _d (4.56) or

n0 (Na pa) p0 (Nd nd) (4.57) where n0 and p0 are the thermal-equilibrium concentrations of electrons and holes in the conduction band and valence band, respectively. The parameter nd is the concen- tration of electrons in the donor energy states, so N _d Nd nd is the concentration of positively charged donor states. Similarly, pa is the concentration of holes in the acceptor states, so N _a Na pa is the concentration of negatively charged acceptor states. We have expressions for n0, p0, nd, and pa in terms of the Fermi energy and temperature.

Total electron concentration Thermal

electrons ⴚ ⴚ ⴚ

ⴚ ⴚ ⴚ ⴚ ⴚ ⴚ ⴚ ⴚ ⴚ ⴚ

ⴙ ⴙ ⴙ ⴙ ⴙ ⴙ ⴙ

ⴙ ⴙ ⴙ ⴙ

ⴙ ⴙ Un-ionized

donors Un-ionized

acceptors

Donor electrons

Thermal holes

Acceptor holes n₀

E_c E_d

E_a E_v E_Fi N_d (N_d n_d)

Ionized donors N_a (N_a p_a) Ionized acceptors

Total hole concentration

p₀

Figure 4.14 | Energy-band diagram of a compensated semiconductor showing ionized and un-ionized donors and acceptors.

Thermal-Equilibrium Electron Concentration If we assume complete ioniza- tion, nd and pa are both zero, and Equation (4.57) becomes

n0 Na p0 nd (4.58)

If we express p0 as n _i² n0, then Equation (4.58) can be written as n0 Na n _i²

_ n0 Nd (4.59a)

which in turn can be written as

n ₀² (Nd Na)n0 n _i² 0 (4.59b) The electron concentration n0 can be determined using the quadratic formula, or

n0 (Nd Na)

__ 2

______________

__ Nd Na

2

^{2 n i2} (4.60) The positive sign in the quadratic formula must be used, since, in the limit of an intrinsic semiconductor when Na Nd 0, the electron concentration must be a positive quantity, or n0 ni.

Equation (4.60) is used to calculate the electron concentration in an n-type semi- conductor, or when Nd Na. Although Equation (4.60) was derived for a compen- sated semiconductor, the equation is also valid for Na 0.

EXAMPLE 4.9

Objective: Determine the thermal-equilibrium electron and hole concentrations in silicon at T 300 K for given doping concentrations. (a) Let Nd 10¹⁶ cm³ and Na 0. (b) Let Nd 5 10¹⁵ cm³ and Na 2 10¹⁵ cm³.

Recall that ni 1.5 10¹⁰ cm³ in silicon at T 300 K.

■ Solution

(a) From Equation (4.60), the majority carrier electron concentration is n0 10_ ¹⁶

2

____________________

10_ ¹⁶

2

^{2 1.5 1010 2 1016 cm3}

The minority carrier hole concentration is found to be p0 n _i²

_ n0 _{__} (1.5 10¹⁰) 10¹⁶

2

2.25 10⁴ cm³

(b) Again, from Equation (4.60), the majority carrier electron concentration is n0 5 _{____} 10¹⁵ 2 10¹⁵

2

________________________________

5 _{____} 10¹⁵ 2 10¹⁵

2

^{2 (1.5 1010) 2 3 1015 cm3}

The minority carrier hole concentration is p0 n_i²

_ n0 (1.5 10¹⁰)

___ 3 10¹⁵

2

7.5 10⁴ cm³

4 . 5 Charge Neutrality 137

We have argued in our discussion and we may note from the results of Ex- ample 4.9 that the concentration of electrons in the conduction band increases above the intrinsic carrier concentration as we add donor impurity atoms. At the same time, the minority carrier hole concentration decreases below the intrinsic carrier concen- tration as we add donor atoms. We must keep in mind that as we add donor impurity atoms and the corresponding donor electrons, there is a redistribution of electrons among available energy states. Figure 4.15 shows a schematic of this physical redis- tribution. A few of the donor electrons will fall into the empty states in the valence band and, in doing so, will annihilate some of the intrinsic holes. The minority carrier hole concentration will therefore decrease as we have seen in Example 4.9. At the

■ Comment

In both parts of this example, (Nd Na) ni, so the thermal-equilibrium majority carrier electron concentration is essentially equal to the difference between the donor and acceptor concentrations. Also, in both cases, the majority carrier electron concentration is orders of magnitude larger than the minority carrier hole concentration.

■ EXERCISE PROBLEM

Ex 4.9 Find the thermal-equilibrium electron and hole concentrations in silicon with doping concentrations of Na 7 10¹⁵ cm³ and Na 3 10¹⁵ cm³ for (a) T 250 K and (b) T 400 K.

[Ans. (a) n 0

4

15 10

3 cm

, p 0

1.225

3 cm

; (b) n 0

4

15 10

3 cm

, p 0

1.416

9 10

3 cm

]

ⴚ

ⴙ ⴙ ⴙ ⴙ ⴙ ⴙ ⴙ ⴙ ⴙ ⴚ ⴚ ⴚ ⴚ ⴚ

ⴙ ⴙ ⴙ ⴙ

Ionized donors Intrinsic

electrons

Intrinsic holes

Un-ionized donors

Net p₀

E_c E_d

E_v E_Fi

n_i² n₀

A few donor electrons annihilate some intrinsic holes ⴚⴚⴚⴚⴚ

Figure 4.15 | Energy-band diagram showing the redistribution of electrons when donors are added.

same time, because of this redistribution, the net electron concentration in the con- duction band is not simply equal to the donor concentration plus the intrinsic electron concentration.

EXAMPLE 4.10

Objective: Calculate the thermal-equilibrium electron and hole concentrations in germanium for a given doping concentration.

Consider a germanium sample at T 300 K in which Nd 2 10¹⁴ cm³ and Na 0.

Assume that ni 2.4 10¹³ cm³.

■ Solution

Again, from Equation (4.60), the majority carrier electron concentration is n0 2 _{__} 10¹⁴

2

________________________

2 _{__} 10¹⁴

2

^{2 2.4 1013 2 2.028 1014 cm3}

The minority carrier hole concentration is p0 _{_} n _i²

n0 ___ 2.4 10¹³²

2.028 10¹⁴ 2.84 10¹² cm³

■ Comment

If the donor impurity concentration is not too different in magnitude from the intrinsic carrier concentration, then the thermal-equilibrium majority carrier electron concentration is infl u- enced by the intrinsic concentration.

■ EXERCISE PROBLEM

Ex 4.10 Repeat Example 4.10 for (a) T 250 K and (b) T 350 K. (c) What can be said about a very low-doped material as the temperature increases?

p 0

1.059

14 10

3 cm

; (c) Material approaches an intrinsic semiconductor]

[Ans. (a) n 2 0

14 10

3 cm

, p 0

9.47

9 10

3 cm

; (b) n 0

3.059

14 10

3 cm

,

We have seen that the intrinsic carrier concentration ni is a very strong function of temperature. As the temperature increases, additional electron–hole pairs are ther- mally generated so that the n _i² term in Equation (4.60) may begin to dominate. The semiconductor will eventually lose its extrinsic characteristics. Figure 4.16 shows the electron concentration versus temperature in silicon doped with 5 10¹⁴ donors per cm³ . As the temperature increases, we can see where the intrinsic concentration begins to dominate. Also shown is the partial ionization, or the onset of freeze-out, at the low temperature.

Thermal-Equilibrium Hole Concentration If we reconsider Equation (4.58) and express n0 as n _i² p0, then we have

n _i²

_ p0 Na p0 Nd (4.61a) which we can write as

p ₀² (Na Nd)p0 n _i² 0 (4.61b)

4 . 5 Charge Neutrality 139

Using the quadratic formula, the hole concentration is given by p0 __ Na Nd

2

______________

__ Na Nd

2

^{2 n i2} (4.62) where the positive sign, again, must be used. Equation (4.62) is used to calculate the thermal-equilibrium majority carrier hole concentration in a p-type semiconductor, or when Na Nd. This equation also applies for Nd 0.

EXAMPLE 4.11 Objective: Calculate the thermal-equilibrium electron and hole concentrations in a compensated p-type semiconductor.

Consider a silicon semiconductor at T 300 K in which Na 10¹⁶ cm³ and Nd 3 10¹⁵ cm³. Assume ni 1.5 10¹⁰ cm³.

■ Solution

Since Na Nd, the compensated semiconductor is p-type and the thermal-equilibrium majority carrier hole concentration is given by Equation (4.62) as

p0 10_{___} ¹⁶ 3 10¹⁵

2

_____________________________

10_{___} ¹⁶ 3 10¹⁵

2

^{2 (1.5 1010)2}

so that

p0 7 10¹⁵ cm³ The minority carrier electron concentration is

n0 n _i²

_ p0 _{___} (1.5 10¹⁰)²

7 10¹⁵ 3.21 10⁴ cm³

■ Comment

If we assume complete ionization and if (Na Nd) ni, then the majority carrier hole concen- tration is, to a very good approximation, just the difference between the acceptor and donor concentrations.

Figure 4.16 | Electron concentration versus temperature showing the three regions: partial ionization, extrinsic, and intrinsic.

Partial ionization

Extrinsic

T (K) n0(cm3)

n_i

Intrinsic

10¹³ 10¹⁴ 10¹⁵

5

10¹⁶

10¹²

100 200 300 400 500 600 700