Objective: Analyze a circuit using a voltage divider bias circuit, and determine the change in the Q-point with a variation in β when the circuit contains an emitter resistor. For the circuit given in Figure 5.54(a), let R1 = 56 kΩ, R2 = 12.2 kΩ, RC = 2 kΩ, RE = 0.4 kΩ, VCC = 10 V, VB E (on) = 0.7

Question

Objective: Analyze a circuit using a voltage divider bias circuit, and determine the change in the Q-point with a variation in β when the circuit contains an emitter resistor. For the circuit given in Figure 5.54(a), let [latex]R_{1} = 56 kΩ, R_{2} = 12.2 kΩ, R_{C} = 2 kΩ, R_{E} = 0.4 kΩ, V_{CC} = 10 V, V_{B E} (on) = 0.7 V[/latex], and β = 100

Accepted Answer

Using the Thevenin equivalent circuit in Figure 5.54(b), we have
[latex]R_{T H} = R_{1} || R_{2} = 56 || 12.2 = 10.0 k \Omega[/latex]

and
[latex]V_{T H} = \left( \frac{R_{2}}{R_{1} + R_{2}} \right) \cdot V_{CC} = \left( \frac{12.2}{56 + 12.2} \right) (10) = 1.79 V[/latex]

Writing the Kirchhoff voltage law equation around the B–E loop, we obtain
[latex]I_{B Q} = \frac{V_{T H} − V_{B E} (on)}{R_{T H} + (1 + β)R_{E}} = \frac{1.79 − 0.7}{10 + (101)(0.4)} ⇒ 21.6 µA[/latex]
The collector current is
[latex]I_{C Q} = β I_{B Q} = (100)(21.6 µA) ⇒ 2.16 mA[/latex]
and the emitter current is
[latex]I_{E Q} = (1 + β) I_{B Q} = (101)(21.6 µA) ⇒ 2.18 mA[/latex]
The quiescent C–E voltage is then
[latex]V_{C E Q} = V_{CC} − I_{C Q} R_{C} − I_{E Q} R_{E} = 10 − (2.16)(2) − (2.18)(0.4) = 4.81 V[/latex]
These results show that the transistor is biased in the active region.
If the current gain of the transistor were to decrease to β = 50 or increase to β = 150, we obtain the following results:

β	50	100	150
	[latex] I_{B Q} = 35.9 µA[/latex]	[latex] I_{B Q} = 21.6 µA[/latex]	[latex] I_{B Q} = 15.5 µA[/latex]
Q-point values	[latex] I_{C Q} = 1.80 mA[/latex]	[latex] I_{C Q} = 2.16 mA[/latex]	[latex] I_{C Q} = 2.32 mA[/latex]
	[latex]V_{C E Q} = 5.67 V[/latex]	[latex]V_{C E Q} = 4.81 V[/latex]	[latex]V_{C E Q} = 4.40 V[/latex]

The load line and Q-points are plotted in Figure 5.55. The variation in Q-points for this circuit configuration is to be compared with the variation in Q-point values shown previously in Figure 5.52(b).
For a 3 : 1 ratio in β, the collector current and collector–emitter voltage change by only a 1.29 : 1 ratio.
Comment: The voltage divider circuit of [latex]R_{1}[/latex] and [latex]R_{2}[/latex] can bias the transistor in its active region using resistor values in the low kilohm range. In contrast, single resistor biasing requires a resistor in the megohm range. In addition, the change in [latex]I_{C Q}[/latex] and [latex]V_{C E Q}[/latex] with a change in β has been substantially reduced compared to the change shown in Figure 5.52(b). Including an emitter resistor [latex]R_{E}[/latex] has tended to stabilize the Q-point. This means that including the emitter resistor helps to stabilize the Q-point with respect to variations in β. Including the resistor [latex]R_{E}[/latex] introduces negative feedback, as we will see in Chapter 12. Negative feedback tends to stabilize circuits.

Question 5.15: Objective: Analyze a circuit using a voltage divider bias ci...

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