CHAPTER 13

Solutions for Exercises

E13.1 The emitter current is given by the Shockley equation:

For operation with , and we can write

Solving for , we have

E13.2

0.9 / 9
0.99 / 99
0.999 / 999

E13.3

E13.4 The base current is given by Equation 13.8:

which can be plotted to obtain the input characteristic shown in Figure 13.6a. For the output characteristic, we have provided that

Forfalls rapidly to zero at The output characteristics are shown in Figure 13.6b.

E13.5 The load lines for are shown:

As shown on the output load line, we find

E13.6 The load lines for the new values are shown:

As shown on the output load line, we have

E13.7 Refer to the characteristics shown in Figure 13.7 in the book. Select a point in the active region of the output characteristics. For example, we could choose the point defined by at which we find Then we have (For many transistors the value found for depends slightly on the point selected.)

E13.8 (a) Writing a KVL equation around the input loop we have the equation for the input load lines: The load lines are shown:

Then we write a KCL equation for the output circuit:

The resulting load line is:

From these load lines we find

(b) Inspecting the load lines, we see that the maximum of vin corresponds to IBmin which in turn corresponds to VCEmin. Because the maximum of vin corresponds to minimum VCE, the amplifier is inverting. This may be a little confusing because VCE takes on negative values, so the minimum value has the largest magnitude.

E13.9 (a) Cutoff because we have which is less than 0.5 V.

(b) Saturation because we have

(c) Active because we have

E13.10 (a) In this case (the BJT operates in the active region. Thus the equivalent circuit is shown in Figure 13.18d. We have

11.43 V

Because we have we are justified in assuming that the transistor operates in the active region.

(b) In this case (,the BJT operates in the saturation region. Thus the equivalent circuit is shown in Figure 13.18c. We have

Because we have we are justified in assuming that the transistor operates in the active region.

E13.11 For the operating point to be in the middle of the load line, we want and . Then we have

(a)

(b)

E13.12 Notice that a pnp BJT appears in this circuit.

(a) For it turns out that the BJT operates in the active region.

(b) For it turns out that the BJT operates in the saturation region.

Because we havewe are assured that the transistor operates in the active region.

E13.13

/ (mA) / (mA) / (V)
100 / 32.01 / 3.201 / 8.566
300 / 12.86 / 3.858 / 7.271

For the larger values of used in this Exercise, the ratio of the collector currents for the two values of b is 1.205, whereas for the smaller values of used in Example 13.7, the ratio of the collector currents for the two values of b is 1.0213. In general in the four-resistor bias network smaller values for lead to more nearly constant collector currents with changes in b.

E13.14

E13.15 First, we determine the bias point:

Now we can compute and the ac performance.

Answers for Selected Problems

P13.6*

P13.7*

P13.16* A

P13.18* At , the base-to-emitter voltage is approximately:

P13.19*

P13.24*

P13.28*

P13.29* (a) and (b)

(c)

(d) and (e) The sketches of vCE(t) are:

P13.36* In the active region, the base-collector junction is reverse biased and the base-emitter junction is forward biased.

In the saturation region, both junctions are forward biased.

In the cutoff region, both junctions are reverse biased. (Actually, cutoff applies for slight forward bias of the base-emitter junction as well, provided that the base current is negligible.)

P13.41* 1. Assume operation in saturation, cutoff, or active region.

2. Use the corresponding equivalent circuit to solve for currents and

voltages.

3. Check to see if the results are consistent with the assumption made in step 1. If so, the circuit is solved. If not, repeat with a different assumption.

P13.44* The results are given in the table:

Circuit / / Region of
operation /
(mA) /
(volts)
(a) / 100 / active / 1.93 / 10.9
(a) / 300 / saturation / 4.21 / 0.2
(b) / 100 / active / 1.47 / 5.00
(b) / 300 / saturation / 2.18 / 0.2
(c) / 100 / cutoff / 0 / 15
(c) / 300 / cutoff / 0 / 15
(d) / 100 / active / 6.5 / 8.5
(d) / 300 / saturation / 14.8 / 0.2

P13.47*

P13.49*

P13.56*

For 1 mA, we obtain 50 kW.

P13.63*

High impedance amplifier
(Problem 13.57) / Low impedance amplifier
(Problem 13.56)
ICQ / 0.0393 mA / 3.93 mA
/ 66.2 kΩ / 662 Ω
/ -75.5 / -75.5
/ -151 / -151
/ 54.8 kΩ / 548 Ω
/ -41.4 / -41.4
/ 3124 / 3124
/ 100 kΩ / 1 kΩ

P13.67*

11