Objective: Design a bias-stable circuit to meet a set of specifications. Specifications: The circuit configuration to be designed is shown in Figure 5.54(a). Let VCC = 5 V and RC = 1 k. Choose RE and determine the bias resistors R1 and R2 such that the circuit is considered bias stable and that VC

Question

Objective: Design a bias-stable circuit to meet a set of specifications. Specifications: The circuit configuration to be designed is shown in Figure 5.54(a). Let [latex]V_{CC} = 5 V[/latex] and [latex]R_{C} = 1 k\Omega[/latex]. Choose [latex]R_{E}[/latex] and determine the bias resistors [latex]R_{1}[/latex] and [latex]R_{2}[/latex] such that the circuit is considered bias stable and that [latex]V_{CE Q} = 3 V[/latex].

Choices: Assume the transistor has nominal values of β = 120 and [latex]V_{B E} (on) = 0.7 V[/latex]. We will choose standard resistor values and will assume that the transistor current gain varies over the range 60 ≤ β ≤ 180.

Design Pointer: Typically, the voltage across [latex]R_{E}[/latex] should be on the same order of magnitude as [latex]V_{B E} (on)[/latex]. Larger voltage drops may mean the supply voltage [latex]V_{CC}[/latex] has to be increased in order to obtain the required voltage across the collector-emitter and across [latex]R_{C}[/latex] .

Accepted Answer

With [latex]β = 120, I_{C Q} \approx I_{E Q}[/latex] . Then, choosing a standard value of 0.51 kΩ for [latex]R_{E}[/latex] , we find
[latex]I_{C Q} \cong \frac{V_{CC} − V_{C E Q} }{R_{C} + R_{E}} = \frac{5 − 3}{1 + 0.51} = 1.32 mA[/latex]
The voltage drop across [latex]R_{E}[/latex] is now (1.32)(0.51) = 0.673 V, which is approximately the desired value. The base current is found to be
[latex]I_{B Q} = \frac{I_{C Q}}{β} = \frac{1.32}{120} ⇒ 11.0 µA[/latex]

Using the Thevenin equivalent circuit in Figure 5.54(b), we find
[latex]I_{B Q} = \frac{V_{T H} − V_{B E} (on)}{R_{T H} + (1 + β)R_{E}} [/latex]
For a bias-stable circuit, [latex]R_{T H} = 0.1(1 + β)R_{E}[/latex] , or
[latex]R_{T H} = (0.1)(121)(0.51) = 6.17 k\Omega[/latex]
Then,
[latex]I_{B Q} = 11.0 µA ⇒ \frac{V_{T H} − 0.7}{6.17 + (121)(0.51)}[/latex]
which yields
[latex]V_{T H} = 0.747 + 0.70 = 1.447 V[/latex]
Now
[latex]V_{T H} = \left(\frac{R_{2}}{R_{1} + R_{2}} \right) V_{CC} = \left(\frac{R_{2}}{R_{1} + R_{2}} \right) (5) = 1.447 V [/latex]
or

[latex]\left(\frac{R_{2}}{R_{1} + R_{2}} \right) = \frac{1.447}{5} = 0.2894 [/latex]
Also,
[latex]R_{T H} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = 6.17 k\Omega = R_{1}\left(\frac{R_{2}}{R_{1} + R_{2}} \right) = R_{1}(0.2894) [/latex]
which yields
[latex]R_{1} = 21.3 k\Omega[/latex] and [latex]R_{2} = 8.69 k\Omega[/latex]
From Appendix C, we can choose standard resistor values of [latex]R_{1} = 20 k\Omega[/latex] and [latex]R_{2} = 8.2 k\Omega[/latex].
Trade-offs: We will neglect, in this example, the tolerance effects of the resistors (end-of-chapter problems such as Problems 5.18 and 5.40 do include tolerance effects). We will consider the effect on the transistor Q-point values of the commonemitter current gain variation.
Using the standard resistor values, we have
[latex]R_{T H} = R_{1}||R_{2} = 20||8.2 = 5.82 k\Omega[/latex]
and
[latex]V_{T H} = \left(\frac{R_{2}}{R_{1} + R_{2}} \right) (V_{CC}) = \left(\frac{8.2}{20 + 8.2} \right) (5) = 1.454 V[/latex]
The base current is given by
[latex]I_{B Q} = \left[ \frac{V_{T H} − V_{B E} (on)}{R_{T H} + (1 + β)R_{E}} \right] [/latex]

while the collector current is [latex]I_{C Q} = β I_{B Q}[/latex], and the collector–emitter voltage is given by
[latex]V_{C E Q} = V_{CC} − I_{C Q} \left[R_{C} + \left(\frac{1 + β}{β} \right)R_{E} \right] [/latex]

The Q-point values for three values of β are shown in the following table

β	60	120	180
	[latex]I_{BQ} = 20.4 µA[/latex]	[latex]I_{BQ} = 11.2 µA[/latex]	[latex]I_{BQ} = 7.68 µA[/latex]
Q-point values	[latex] I_{CQ} = 1.23 mA[/latex]	[latex] I_{CQ} = 1.34 mA[/latex]	[latex] I_{CQ} = 1.38 mA[/latex]
	[latex] V_{CEQ} = 3.13 V[/latex]	[latex] V_{CEQ} = 2.97 V[/latex]	[latex] V_{CEQ} = 2.91 V[/latex]

Comment: The Q-point in this example is now considered to be stabilized against variations in β, and the voltage divider resistors [latex]R_{1}[/latex] and [latex]R_{2}[/latex] have reasonable values in the kilohm range. We see that the collector current changes by only −8.2 percent when β changes by a factor of 2 (from 120 to 60), and changes by only +3.0 percent when β changes by +50 percent (from 120 to 180). Compare these changes to those of the single-base resistor design in Example 5.14.

Computer Simulation: Figure 5.56 shows the PSpice circuit schematic diagram with the standard resistor values and with a standard 2N2222 transistor from the PSpice library for the circuit designed in this example. A dc analysis was performed and the resulting transistor Q-point values are shown. The collector–emitter voltage is [latex]V_{C E} = 2.80 V[/latex], which is close to the design value of 3 V. One reason for the difference is that the standard-valued resistors are not exactly equal to the design values.
Another reason for the slight difference is that the effective β of the 2N2222 is 157 compared to the assumed value of 120.

**** BIPOLAR JUNCTION TRANSISTORS
NAME	Q_Q1
MODEL	Q2N2222
IB	9.25E−06
IC	1.45E−03
VBE	6.55E−01
VBC	−2.15E+00
VCE	2.80E+00
BETADC	1.57E+02

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