Question 5.16: Objective: Design a bias-stable circuit to meet a set of spe...

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

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

\left(\frac{R_{2}}{R_{1}  +  R_{2}} \right) = \frac{1.447}{5} = 0.2894
Also,
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)
which yields
R_{1} = 21.3  k\Omega   and    R_{2} = 8.69  k\Omega
From Appendix C, we can choose standard resistor values of R_{1} = 20  k\Omega and  R_{2} = 8.2  k\Omega.
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
R_{T H} = R_{1}||R_{2} = 20||8.2 = 5.82  k\Omega
and
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
The base current is given by
I_{B Q} = \left[ \frac{V_{T H}  −  V_{B E} (on)}{R_{T H}  +  (1  +  β)R_{E}} \right]

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

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

Comment: The Q-point in this example is now considered to be stabilized against variations in β, and the voltage divider resistors R_{1} and R_{2} 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 V_{C E} = 2.80  V, 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.