Question 5.20: Objective: Design the circuit shown in Figure 5.63, called a...
Objective: Design the circuit shown in Figure 5.63, called a cascode circuit, to meet the following specifications: V_{C E1} = V_{C E2} = 2.5 V, V_{R E} = 0.7 V, I_{C1} ≅ I_{C2} ≅ 1 mA, and I_{R1} ≅ I_{R2} ≅ I_{R3} ≅ 0.10 mA
The initial design will neglect base currents. We can then define I_{Bias} = I_{R1} = I_{R2} = I_{R3} = 0.10 mA. Then
R_{1} + R_{2} + R_{3} = \frac{V^{+}}{I_{Bias}} = \frac{9}{0.10} = 90 k \Omega
The voltage at the base of Q_{1} is
V_{B1} = V_{R E} + V_{B E} (on) = 0.7 + 0.7 = 1.4 V
Then
R_{3} = \frac{V_{B1}}{I_{Bias}} = \frac{1.4}{0.10} = 14 k \Omega
The voltage at the base of Q_{2} is
V_{B2} = V_{R E} + V_{C E1} + V_{B E} (on) = 0.7 + 2.5 + 0.7 = 3.9 V
Then
R_{2} = \frac{V_{B2} − V_{B1}}{I_{Bias}} = \frac{3.9 − 1.4}{0.10} = 25 k \Omega
We then obtain
R_{1} = 90 − 25 − 14 = 51 k \Omega
The emitter resistor R_{E} can be found as
R_{E} = \frac{V_{R E}}{I_{C1}} = \frac{0.7}{1} = 0.7 k \Omega
The voltage at the collector of Q_{2} is
V_{C2} = V_{R E} + V_{C E1} + V_{C E2} = 0.7 + 2.5 + 2.5 = 5.7 V
Then
R_{C} = \frac{V^{+} − V_{C2}}{I_{C2}} = \frac{9 − 5.7}{1} = 3.3 k \Omega
Comment: By neglecting base currents, the design of this circuit is straightforward.
A computer analysis using PSpice, for example, will verify the design or show that small changes need to be made to meet the design specifications.
We will see the cascode circuit again in Section 6.9.3 of the next chapter.
One advantage of the cascode circuit will be determined in Chapter 7. The cas-code circuit has a larger bandwidth than just a simple common-emitter amplifier