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EEET 201: ELECTRONIC CIRCUITS

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PHYS 162: Solid State Devices<br />

(Chapter 1) Introduction to Semiconductors<br />

HW 1 QUESTIONS<br />

DUE DATE: SATURDAY, 1 OCTOBER <strong>201</strong>1<br />

Conduction in Semiconductors<br />

1.1 What happens when heat is added to silicon?<br />

1.2 Name the two energy bands where current is produced in silicon.<br />

N-Type and P-Type Semiconductors<br />

1.3 Describe the process of doping and explain how it alters the atomic structure if<br />

silicon.<br />

The Diode<br />

1.4 How is the electric field in the pn junction created?<br />

1.5 Because of its barrier, can a diode be used as a voltage source? Explain.<br />

Biasing a Diode<br />

1.6 Explain why a series resistance is necessary when a diode is forward biased.<br />

Diode Models<br />

1.7 Determine whether each diode in Figure 1.1 is forward-biased or reversed-biased.<br />

1.8 Determine the voltage across each diode in Figure 1.1, assuming the practical model.<br />

1.9 Determine the voltage across each diode in Figure 1.1, assuming the ideal model.<br />

1.10 Determine the voltage across each diode in Figure 1.1, using the complete diode<br />

model with r’ d = 7Ω and r’ R = 100MΩ.<br />

Figure 1.1<br />

1


SOLUTION<br />

Conduction in Semiconductors<br />

1.1<br />

1.2<br />

N-Type and P-Type Semiconductors<br />

1.3<br />

The Diode<br />

1.4<br />

1.5<br />

Biasing a Diode<br />

1.6<br />

Diode Models<br />

1.7<br />

1.8<br />

1.9<br />

1.10<br />

Ignoring r′ R<br />

(a) V R ≈ 5V − 8V = −3V<br />

(b)<br />

100V − 0.7V<br />

I F =<br />

560Ω + 7Ω = 175.1mA<br />

2


(c)<br />

(d)<br />

V F = I F r ′ d + V B = 175.1mA 7Ω + 0.7V = 1.9V<br />

I tot = 30V<br />

R tot<br />

R tot = 1KΩ + 1.5KΩ + 4.7KΩ = 2.5KΩ + 2.35KΩ = 4.85KΩ<br />

2<br />

I tot =<br />

30V<br />

4.85KΩ = 6.19mA<br />

I F = 6.19mA = 3.1mA<br />

2<br />

V F = I F r ′ d + V B = 3.1mA 7Ω + 0.7V = 0.722V<br />

I F =<br />

20V − 0.7V<br />

10KΩ + 7Ω = 1.93mA<br />

V F = I F r ′ d + V B = 1.93mA 7Ω + 0.7V = 0.71V<br />

3

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