Subscribe $4.99/month

Un-lock Verified Step-by-Step Experts Answers.

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

All the data tables that you may search for.

Need Help? We got you covered.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Need Help? We got you covered.

Chapter 5

Q. 5.9

Objective: Design a pnp bipolar transistor circuit to meet a set of specifications.

Specifications: The circuit configuration to be designed is shown in Figure 5.36(a). The quiescent emitter-collector voltage is to be V_{EC Q} = 2.5  V.

Choices: Discrete resistors with tolerances of ±10 percent are to be used, an emitter resistor with a nominal value of R_{E} = 2  k\Omega is to be used, and a transistor with β = 60 and V_{E B}(on) = 0.7  V is available.

5.36

Step-by-Step

Verified Solution

(ideal Q-point value): Writing the Kirchhoff’s voltage law equation around the C–E loop, we obtain
V^{+} = I_{E Q} R_{E} + V_{EC Q}
or
5 = I_{E Q}(2) + 2.5
which yields I_{E Q} = 1.25  mA. The collector current is
I_{C Q} = \left(\frac{β}{1  +  β} \right) \cdot I_{E Q} = \left(\frac{60}{61} \right) (1.25) = 1.23  mA
The base current is
I_{B Q} = \frac{I_{E Q}}{1  +  β} = \frac{1.25}{61} = 0.0205  mA
Writing the Kirchhoff’s voltage law equation around the E–B loop, we find
V^{+} = I_{E Q} R_{E} + V_{E B}(on) + I_{B Q} R_{B} + V_{B B}

or
5 = (1.25)(2) + 0.7 + (0.0205)R_{B} + (−2)
which yields R_{B} = 185  k \Omega.
(ideal load line): The load line equation is
V_{EC} = V^{+}  −  I_{E} R_{E} = V^{+}  −  I_{C} \left(\frac{1  +  β}{β} \right) R_{E}
or
V_{EC} = 5  −  I_{C} \left(\frac{61}{60} \right) (2) = 5  −  I_{C}(2.03)
The load line, using the nominal value of R_{E} , and the calculated Q-point are shown in Figure 5.37(a).
Trade-offs: As shown in Appendix C, a standard resistor value of 185 kΩ is not available. We will pick a value of 180 kΩ. We will consider R_{B} and R_{E} resistor tolerances of ±10 percent.
The quiescent collector current is given by
I_{C Q} = β \left[ \frac{V^{+}  −  V_{E B}(on)  −  V_{B B}}{R_{B}  +  (1  +  β)R_{E}} \right] = (60) \left[\frac{6.3}{R_{B}  +  (61)R_{E}} \right ]

and the load line is given by
V_{EC} = V^{+}  −  I_{C} \left(\frac{1  +  β}{β} \right) R_{E} = 5  −  \left(\frac{61}{60} \right) I_{C} R_{E}
The extreme values of R_{E} are:
2 kΩ − 10% = 1.8 kΩ           2 kΩ + 10% = 2.2 kΩ.

The extreme values of R_{B} are:
180 kΩ − 10% = 162 kΩ              180 kΩ + 10% = 198 kΩ.
The Q-point values for the extreme values of R_{B} and R_{E} are given in the following table.

R_{E}
R_{E} 1.8 kΩ 2.2 kΩ
162 kΩ I_{CQ} = 1.39  mA I_{CQ} = 1.28  mA
V_{ECQ} = 2.46  V V_{ECQ} = 2.14  V
198 kΩ I_{CQ} = 1.23  mA I_{CQ} = 1.14  mA
V_{ECQ} = 2.75  V V_{ECQ} = 2.45  V

Figure 5.37(b) shows the Q-points for the various possible extreme values of emitter and base resistances. The shaded area shows the region in which the Q-point will occur over the range of resistor values.
Comment: This example shows that an ideal Q-point can be determined based on a set of specifications, but, because of resistor tolerance, the actual Q-point will vary over a range of values. Other examples will consider the tolerances involved in transistor parameters.

5.37