Applications: DC Transistor Circuits

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Applications: DC Transistor Circuits

Most of us deal with electronic products on a routine basis and ha ve some e xperience with personal computers. A basic component for the integrated circuits found in these electronics and computers is the active, three-terminal device known as the transistor. Understanding the transistor is essential before an engineer can start an electronic circuit design.

Figure 3.37 depicts various kinds of transistors commercially available. There are tw o basic types of transistors: bipolar junction tr ansistors (BJTs) and field-effect transistors (FETs). Here, we consider only the BJTs, which were the first of the two and are still used today . Our objective is to present enough detail about the BJT to enable us to apply the techniques developed in this chapter to analyze dc transistor circuits.

Practice Problem 3.11

Courtesy of Lucent Technologies/Bell Labs

Historical

William Schockley (1910–1989), John Bardeen (1908–1991), and Walter Brattain (1902–1987) co-invented the transistor.

Nothing has had a greater impact on the transition from the “Industrial Age” to the “Age of the Engineer” than the transistor. I am sure that Dr. Shockley, Dr. Bardeen, and Dr. Brattain had no idea they would have this incredible effect on our history. While working at Bell Laboratories, they successfully demonstrated the point-contact transistor, invented by Bardeen and Brattain in 1947, and the junction transistor, which Shockley conceived in 1948 and successfully produced in 1951.

It is interesting to note that the idea of the field-effect transistor, the most commonly used one today, was first conceived in 1925–1928 by J. E. Lilienfeld, a German immigrant to the United States. This is evident from his patents of what appears to be a field-effect transistor. Unfortunately, the technology to realize this device had to wait until 1954 when Shockley’s field-effect transistor became a reality. Just think what today would be like if we had this transistor 30 years earlier!

For their contributions to the creation of the transistor, Dr. Shockley, Dr. Bardeen, and Dr. Brattain received, in 1956, the Nobel Prize in physics. It should be noted that Dr. Bardeen is the only individual to win two Nobel prizes in physics; the second came later for work in superconductivity at the University of Illinois.

Figure 3.38 Two types of BJTs and their circuit symbols: (a) npn, (b) pnp.

Figure 3.37 Various types of transistors. (© McGraw-Hill Education/Mark Dierker, photographer)

There are tw o types of BJTs: npn and pnp, with their circuit sym bols as shown in Fig. 3.38. Each type has three terminals, designated as emitter (E), base (B), and collector (C). For the npn transistor, the currents and voltages of the transistor are specified as in Fig. 3.39. Applying KCL to Fig. 3.39(a) gives

$$I_E = I_B + I_C \tag{3.27}$$

where IE, IC, and IB are emitter, collector, and base currents, respectively. Similarly, applying KVL to Fig. 3.39(b) gives

$$V_{CE} + V_{EB} + V_{BC} = 0 \tag{3.28}$$

where VCE, VEB, and VBC are collector -emitter, emitter-base, and basecollector voltages. The BJT can operate in one of three modes: acti ve, cutoff, and saturation. When transistors operate in the active mode, typically VBE ≃ 0.7 V,

$$I_C = \alpha I_E \tag{3.29}$$

where α is called the common-base curr ent gain. In Eq. (3.29), α denotes the fraction of electrons injected by the emitter that are col lected by the collector. Also,

$$I_C = \beta I_B \tag{3.30}$$

where β is known as the common-emitter current gain. The α and β are characteristic properties of a given transistor and assume constant values for that transistor. Typically, α takes values in the range of 0.98 to 0.999, while β takes values in the range of 50 to 1000. From Eqs. (3.27) to (3.30), it is evident that

$$I_E = (1 + \beta)I_B \tag{3.31}$$

and

$$\beta = \frac{\alpha}{1 - \alpha} \tag{3.32}$$

These equations show that, in the active mode, the BJT can be modeled as a dependent current-controlled current source. Thus, in circuit analysis, the dc equi valent model in Fig. 3.40(b) may be used to replace the npn transistor in Fig. 3.40(a). Since β in Eq. (3.32) is large, a small base current controls lar ge currents in the output circuit. Consequently , the bipolar transistor can serve as an amplifier, producing both current gain and voltage gain. Such amplifiers can be used to furnish a considerable amount of power to transducers such as loudspeakers or control motors.

(a) An npn transistor, (b) its dc equivalent model.

It should be observ ed in the follo wing e xamples that one cannot directly analyze transistor circuits using nodal analysis because of the potential difference between the terminals of the transistor . Only when the transistor is replaced by its equivalent model can we apply nodal analysis.

Figure 3.39

The terminal variables of an npn transistor: (a) currents, (b) voltages.

In fact, transistor circuits provide motivation to study dependent sources.

Example 3.12 Find IB, IC, and vo in the transistor circuit of Fig. 3.41. Assume that the transistor operates in the active mode and that β = 50.

Figure 3.41 For Example 3.12.

Solution:

For the input loop, KVL gives

$$-4 + I_B (20 \times 10^3) + V_{BE} = 0$$

Since VBE = 0.7 V in the active mode,

$$I_B = \frac{4 - 0.7}{200 \times 10^3} = 16.5 \,\mu\text{A}$$

But

$$I_C = \beta I_B = 50 \times 16.5 \,\mu A = 0.825 \text{ mA}$$

For the output loop, KVL gives

$$-v_o - 100I_C + 6 = 0$$

or

$$v_o = 6 - 100I_C = 6 - 0.0825 = 5.917$$

V

Note that v^o = VCE in this case.

Answer: 4.691 V, 888.5 mV.

Practice Problem 3.12

For Practice Prob. 3.12.

For the BJT circuit in Fig. 3.43, β = 150 and VBE = 0.7 V. Find vo.

Solution:

• 1. Define. The circuit is clearly defined and the problem is clearly stated. There appear to be no additional questions that need to be asked.
• 2. Present. We are to determine the output voltage of the circuit shown in Fig. 3.43. The circuit contains an ideal transistor with β = 150 and VBE = 0.7 V.
• 3. Alternative. We can use mesh analysis to solv e for vo. We can re place the transistor with its equivalent circuit and use nodal analysis. We can try both approaches and use them to check each other . As a third check, we can use the equi valent circuit and solv e it using PSpice.

4. Attempt.

■ METHOD 1 Working with Fig. 3.44(a), we start with the first loop.

−2 + 100kI¹ + 200k(I¹ − I2) = 0 or 3I¹ − 2I² = 2 × 10⁻⁵ (3.13.1)

Figure 3.44 Solution of the problem in Example 3.13: (a) Method 1, (b) Method 2, (c) Method 3.

Example 3.13

1 kΩ

Now for loop 2.

$$200k(I_2 - I_1) + V_{BE} = 0 \qquad \text{or} \qquad -2I_1 + 2I_2 = -0.7 \times 10^{-5}$$

\n(3.13.2)

Since we have two equations and two unknowns, we can solve for I¹ and I2. Adding Eq. (3.13.1) to (3.13.2) we get;

I¹ = 1.3 × 10⁻⁵ A and I² = (−0.7 + 2.6)10⁻⁵ ∕2 = 9.5 µA Since I³ = −150I² = −1.425 mA, we can now solve for vo using loop 3: −v^o + 1 kI³ + 16 = 0 or v^o = −1.425 + 16 = 14.575 V

■ METHOD 2 Replacing the transistor with its equivalent circuit produces the circuit shown in Fig. 3.44(b). We can now use nodal analysis to solve for vo.

At node number 1: V¹ = 0.7 V

$$(0.7 - 2)/100k + 0.7/200k + I_B = 0 \qquad \text{or} \qquad I_B = 9.5 \,\mu\text{A}$$

At node number 2 we have:

150IB + (v^o − 16)∕1k = 0 or v^o = 16 − 150 × 103 × 9.5 × 10⁻⁶ = 14.575 V

• 5. Evaluate. The answers check, b ut to further check we can use PSpice (Method 3), which gi ves us the solution sho wn in Fig. 3.44(c).
• 6. Satisfactory? Clearly, we have obtained the desired answer with a very high confidence level. We can now present our work as a solution to the problem.

Practice Problem 3.13

The transistor circuit in Fig. 3.45 has β = 80 and VBE = 0.7 V. Find v^o and Io.

Figure 3.45 For Practice Prob. 3.13. Answer: 12 V, 600 µ A.

• 1. Nodal analysis is the application of Kirchhoff’s current law at the nonreference nodes. (It is applicable to both planar and nonplanar circuits.) We express the result in terms of the node voltages. Solv ing the simultaneous equations yields the node voltages.
• 2. A supernode consists of two nonreference nodes connected by a (dependent or independent) voltage source.
• 3. Mesh analysis is the application of Kirchhoff’s voltage law around meshes in a planar circuit. We express the result in terms of mesh currents. Solving the simultaneous equations yields the mesh currents.

• 4. A supermesh consists of two meshes that have a (dependent or independent) current source in common.

• 5. Nodal analysis is normally used when a circuit has fewer node equations than mesh equations. Mesh analysis is normally used when a circuit has fewer mesh equations than node equations.
• 6. Circuit analysis can be carried out using PSpice.
• 7. DC transistor circuits can be analyzed using the techniques covered in this chapter.