9.7 Semiconductor Devices

Semiconductor technology has been a driver of the computing, information technology, mobile device, and related industries. The transistor — a device that can amplify or switch electrical signals — is the key component. In the late 1940s and early 1950s, a number of discoveries and inventions demonstrated the promise of this technology. Researchers William Shockley, John Bardeen, and Walter Brattain invented the first working transistor, and followed up with several other advances. Soon after, their colleague Mohammad M. Atalla developed a surface process that significantly improved transistor performance, and enabled researchers (and eventually manufacturers) to utilize silicon as the primary material. With Dawon Kahng, Atalla went on to invent a specific type of transistor, the MOSFET, which could be miniaturized and therefore incorporated into a high-density integrated circuit (abbreviated ic). Chih-Tang Sah and Frank Wanlass built on this work by developing a MOSFET circuit fabrication method that would become (and still is) the dominant way to manufacture microprocessors.

The results of these efforts were widespread and immediate. Smaller and more efficient computing devices led to advances in arenas ranging from spaceflight (where every gram counts) to wristwatches to radios. Computers became smaller and smaller, enabling businesses and everyday consumers to own them and eventually carry them in their pockets in the form of smartphones. These changes have all been driven by silicon semiconductors, leading a region known for electronics to be renamed Silicon Valley.

We describe two basic semiconductor devices below.

Diodes

Perhaps the simplest device that can be created with a semiconductor is a diode. A diode is a circuit element that allows electric current to flow in only one direction, like a one-way valve (see Model of Conduction in Metals). A diode is created by joining a p-type semiconductor to an n-type semiconductor (Figure 9.23). The junction between these materials is called a p-n junction. A comparison of the energy bands of a silicon-based diode is shown in Figure 9.23(b). The positions of the valence and conduction bands are the same, but the impurity levels are quite different. When a p-n junction is formed, electrons from the conduction band of the n-type material diffuse to the p-side, where they combine with holes in the valence band. This migration of charge leaves positive ionized donor ions on the n-side and negative ionized acceptor ions on the p-side, producing a narrow double layer of charge at the p-n junction called the depletion layer. The electric field associated with the depletion layer prevents further diffusion. The potential energy for electrons across the p-n junction is given by Figure 9.24.

Figure a shows two blocks place side by side, in contact. The left one is labeled p and the right one is labeled n. Figure b shows a valence band at the bottom and a conduction band at the top. There are holes within the valance band on the left, labeled holes at the top of the valence band. There are electrons above the conduction line on the right, labeled electrons at the bottom of the conduction band. Impurity bands are shown above the holes and below the electrons. — **Figure 9.23** (a) Representation of a *p-n* junction. (b) A comparison of the energy bands of p-type and n-type silicon prior to equilibrium.

Figure a shows two blocks place side by side, in contact. The left one is labeled p and the right one is labeled n. Minus signs are shown in the p block near the side in contact. Plus signs are shown in the n block near the side in contact. Figure b shows a valence band at the bottom and a conduction line at the top. The valence band is higher on the left side almost reaching the central line between the two bands. There are holes with the valence band at the top, on the left. The conduction line is lower on the right, almost reaching the central line between the two bands. There are electrons just above the line, on the right. The displacement of the bands is labeled eV subscript 0, potential difference prevents diffusion of electrons from n side to p side. — **Figure 9.24** At equilibrium, (a) excess charge resides near the interface and the net current is zero, and (b) the potential energy difference for electrons (in light blue) prevents further diffusion of electrons into the p-side.

The behavior of a semiconductor diode can now be understood. If the positive side of the battery is connected to the n-type material, the depletion layer is widened, and the potential energy difference across the p-n junction is increased. Few or none of the electrons (holes) have enough energy to climb the potential barrier, and current is significantly reduced. This is called the reverse bias configuration. On the other hand, if the positive side of a battery is connected to the p-type material, the depletion layer is narrowed, the potential energy difference across the p-n junction is reduced, and electrons (holes) flow easily. This is called the forward bias configuration of the diode. In sum, the diode allows current to flow freely in one direction but prevents current flow in the opposite direction. In this sense, the semiconductor diode is a one-way valve.

We can estimate the mathematical relationship between the current and voltage for a diode using the electric potential concept. Consider N negatively charged majority carriers (electrons donated by impurity atoms) in the n-type material and a potential barrier V across the p-n junction. According to the Maxwell-Boltzmann distribution, the fraction of electrons that have enough energy to diffuse across the potential barrier is $N e to the power − e V / k sub B T$ . However, if a battery of voltage $V_{b}$ is applied in the forward-bias configuration, this fraction improves to $N e^{− e (V - V_{b}) / k_{B} T}$ . The electric current due to the majority carriers from the n-side to the p-side is therefore

$I = N e^{− e V / k_{B} T} e^{e V_{b} / k_{B} T} = I_{0} e^{e V_{b} / k_{B} T},$

where $I sub 0$ is the current with no applied voltage and T is the temperature. Current due to the minority carriers (thermal excitation of electrons from the valence band to the conduction band on the p-side and subsequent attraction to the n-side) is $− I sub 0$ , independent of the bias voltage. The net current is therefore

$I_{net} = I_{0} (e^{e V_{b} / k_{B} T} - 1) .$

A sample graph of the current versus bias voltage is given in Figure 9.25. In the forward bias configuration, small changes in the bias voltage lead to large changes in the current. In the reverse bias configuration, the current is $I sub net approximately equals − I sub 0$ . For extreme values of reverse bias, the atoms in the material are ionized which triggers an avalanche of current. This case occurs at the breakdown voltage.

Graph of I subscript net versus V. An arrow pointing right from the y axis is labeled forward bias. An arrow pointing left from the y axis is labeled reverse bias. The curve goes up and right in the first quadrant and then becomes almost vertical at higher values of x and y. It crosses the positive x axis into the fourth quadrant and then the negative y axis at minus I subscript 0. It travels left in a horizontal line till a point where it turns sharply down into what becomes an almost vertical line. The x value of the turning point is labeled breakdown voltage. — **Figure 9.25** Current versus voltage across a *p-n* junction (diode). In the forward bias configuration, electric current flows easily. However, in the reverse bias configuration, electric current flow very little.

Example 1 Diode Current

Attaching the positive end of a battery to the p-side and the negative end to the n-side of a semiconductor diode produces a current of $4.5 \times 10^{−1} A .$ The reverse saturation current is $2.2 \times 10^{−8} A .$ (The reverse saturation current is the current of a diode in a reverse bias configuration such as this.) The battery voltage is 0.12 V. What is the diode temperature?

Strategy The first arrangement is a forward bias configuration, and the second is the reverse bias configuration. In either case, Equation 9.2 gives the current.

Solution The current in the forward and reverse bias configurations is given by

$I_{net} = I_{0} (e^{e V_{b} / k_{B} T} - 1) .$

The current with no bias is related to the reverse saturation current by

$I_{0} \approx - I_{sat} = 2.2 \times 10^{−8} .$

Therefore

$\frac{I_{net}}{I_{0}} = \frac{4.5 \times 10^{−1} A}{2.2 \times 10^{−8} A} = 2.05 \times 10^{7} .$

Equation 9.2 can be written as

$\frac{I_{net}}{I_{0}} + 1 = e^{e V_{b} / k_{B} T} .$

This ratio is much greater than one, so the second term on the left-hand side of the equation vanishes. Taking the natural log of both sides gives

$\frac{e V_{b}}{k_{B} T} = 16.8 .$

The temperature is therefore

$T = \frac{e V_{b}}{k_{B}} (\frac{1}{19}) = \frac{e (0.12 V)}{8.617 \times 10^{−5} eV / K} (\frac{1}{19}) = 73 K .$

Significance The current moving through a diode in the forward and reverse bias configuration is sensitive to the temperature of the diode. If the potential energy supplied by the battery is large compared to the thermal energy of the diode’s surroundings, $k_{B} T,$ then the forward bias current is very large compared to the reverse saturation current.

Junction Transistor

If diodes are one-way valves, transistors are one-way valves that can be carefully opened and closed to control current. A special kind of transistor is a junction transistor. A junction transistor has three parts, including an n-type semiconductor, also called the emitter; a thin p-type semiconductor, which is the base; and another n-type semiconductor, called the collector (Figure 9.26). When a positive terminal is connected to the p-type layer (the base), a small current of electrons, called the base current $I_{B},$ flows to the terminal. This causes a large collector current $I_{c}$ to flow through the collector. The base current can be adjusted to control the large collector current. The current gain is therefore

$I_{c} = β I_{B} .$

Three blocks in contact are shown. From left to right they are labeled: emitter, n-type, base, p-type and collector, n-type. A voltage source is connected across the collector and emitter, with the collector being positive. Another voltage source is connected across the emitter and base, with the base being positive. A thick arrow starts from the emitter, goes through the other two blocks, comes out of the collector and travels along the first voltage loop. The arrow is labeled large electron flow, collector current. A thinner arrow from the base travels into the second voltage loop. This is labeled small electron flow, base current. — **Figure 9.26** A junction transistor has three parts: emitter, base, and collector. Voltage applied to the base acts as a valve to control electric current from the emitter to the collector.

A junction transistor can be used to amplify the voltage from a microphone to drive a loudspeaker. In this application, sound waves cause a diaphragm inside the microphone to move in and out rapidly (Figure 9.27). When the diaphragm is in the “in” position, a tiny positive voltage is applied to the base of the transistor. This opens the transistor “valve” and allows a large electrical current flow to the loudspeaker. When the diaphragm is in the “out” position, a tiny negative voltage is applied to the base of the transistor, which shuts off the transistor valve so that no current flows to the loudspeaker. This shuts the transistor “valve” off so no current flows to the loudspeaker. In this way, current to the speaker is controlled by the sound waves, and the sound is amplified. Any electric device that amplifies a signal is called an amplifier.

Figure a shows three blocks in contact. From left to right they are labeled: emitter, n-type, base, p-type and collector, n-type. A loudspeaker is connected across the collector and emitter. A microphone is connected across the emitter and base. It is labeled diaphragm moves out. The voltage across the microphone is labeled base voltage. It is negative at the base. The figure is labeled little or no electron flow. Figure b shows a similar arrangement to figure a. A thick arrow starts from the emitter, goes through the other two blocks, comes out of the collector and travels along the loudspeaker loop. The arrow is labeled large electron flow, collector current. A thinner arrow from the base travels into the microphone loop. This is labeled small electron flow, base current. The polarity of the base voltage across the microphone is opposite from that in figure a. The microphone is labeled diaphragm moves in. The loudspeaker is shown to give out sound waves. — **Figure 9.27** An audio amplifier based on a junction transistor. Voltage applied to the base by a microphone acts as a valve to control a larger electric current that passes through a loudspeaker.

In modern electronic devices, digital signals are used with diodes and transistors to perform tasks such as data manipulation. Electric circuits carry two types of electrical signals: analog and digital (Figure 9.28). An analog signal varies continuously, whereas a digital signal switches between two fixed voltage values, such as plus 1 volt and zero volts. In digital circuits like those found in computers, a transistor behaves like an on-off switch. The transistor is either on, meaning the valve is completely open, or it is off, meaning the valve is completely closed. Integrated circuits contain vast collections of transistors on a single piece of silicon. They are designed to handle digital signals that represent ones and zeroes, which is also known as binary code. The invention of the ic helped to launch the modern computer revolution.

Two graphs of V versus t are shown. The first, labeled analog signal has an irregular sinusoidal wave. The second, labeled digital signal has a square wave. — **Figure 9.28** Real-world data are often analog, meaning data can vary continuously. Intensity values of sound or visual images are usually analog. These data are converted into digital signals for electronic processing in recording devices or computers. The digital signal is generated from the analog signal by requiring certain voltage cut-off value.

Summary

A diode is produced by an n-p junction. A diode allows current to move in just one direction. In forward biased configuration of a diode, the current increases exponentially with the voltage.
A transistor is produced by an n-p-n junction. A transistor is an electric valve that controls the current in a circuit.
A transistor is a critical component in audio amplifiers, computers, and many other devices.

Conceptual Questions

When p- and n-type materials are joined, why is a uniform electric field generated near the junction?

When p- and n-type materials are joined, why does the depletion layer not grow indefinitely?

The electric field produced by the uncovered ions reduces further diffusion. In equilibrium, the diffusion and drift currents cancel so the net current is zero. Therefore, the resistance of the depletion region is large.

How do you know if a diode is in the forward biased configuration?

Why does the reverse bias configuration lead to a very small current?

The positive terminal is applied to the n-side, which uncovers more ions near the junction (widens the depletion layer), increases the junction voltage difference, and therefore reduces the diffusion of holes across the junction.

What happens in the extreme case that where the n- and p-type materials are heavily doped?

Explain how an audio amplifier works, using the transistor concept.

Sound moves a diaphragm in and out, which varies the input or base current of the transistor circuit. The transistor amplifies this signal (p-n-p semiconductor). The output or collector current drives a speaker.

Problems

Show that for V less than zero, $I sub net approximately equals − I sub 0.$

A p-n diode has a reverse saturation current $1.44 \times 10^{−8} A$ . It is forward biased so that it has a current of $6.78 \times 10^{−1} A$ moving through it. What bias voltage is being applied if the temperature is 300 K?

$V sub b equals 0.458 V$

The collector current of a transistor is 3.4 A for a base current of 4.2 mA. What is the current gain?

Applying the positive end of a battery to the p-side and the negative end to the n-side of a p-n junction, the measured current is $8.76 \times 10^{−1} A$ . Reversing this polarity give a reverse saturation current of $4.41 \times 10^{−8} A$ . What is the temperature if the bias voltage is 1.2 V?

$T equals 829 K$

The base current of a transistor is 4.4 A, and its current gain 1126. What is the collector current?