Chapter 3
Q. 3.18
Objective: Design the biasing of the cascode circuit to meet specific requirements.
For the circuit shown in Figure 3.53, the transistor parameters are: V_{T N1} = V_{T N2} = 1.2 V, K_{n1} = K_{n2} = 0.8 mA/V² , and λ_{1} = λ_{2} = 0. Let R_{1} + R_{2} + R_{3} = 300 kΩ and R_{S} = 10 kΩ. Design the circuit such that I_{DQ} = 0.4 mA and V_{DSQ1} = V_{DSQ2} = 2.5 V
Step-by-Step
Verified Solution
The dc voltage at the source of M_{1} is
V_{S1} = I_{DQ} R_{S} − 5 = (0.4)(10) − 5 = −1 V
Since M_{1} and M_{2} are identical transistors, and since the same current exists in the two transistors, the gate-to-source voltage is the same for both devices. We have
I_{D} = K_{n}(V_{GS} − V_{T N})^{2}
or
0.4 = 0.8(V_{GS} − 1.2)^{2}
which yields
V_{GS} = 1.907 V
Then,
V_{G1} =\left( \frac{R_{3}}{R_{1} + R_{2} + R_{3}} \right) (5) = V_{GS} + V_{S1}
or
\left( \frac{R_{3}}{300} \right) (5) = 1.907 − 1 = 0.907
which yields
R_{3} = 54.4 k\Omega
The voltage at the source of M_{2} is
V_{S2} = V_{DSQ1} + V_{S1} = 2.5 − 1 = 1.5 V
Then,
V_{G2} = \left( \frac{R_{2} + R_{3}}{R_{1} + R_{2} + R_{3}} \right) (5) = V_{GS} + V_{S2}
or
\left( \frac{R_{2} + R_{3}}{300} \right) (5) = 1.907 + 1.5 = 3.407 V
which yields
R_{2} + R_{3} = 204.4 k \Omega
and
R_{2} = 150 k\Omega
Therefore
R_{1} = 95.6 k \Omega
The voltage at the drain of M_{2} is
V_{D2} = V_{DSQ2} + V_{S2} = 2.5 + 1.5 = 4 V
The drain resistor is therefore
R_{D} = \frac{5 − V_{D2}}{I_{DQ}} = \frac{5 − 4}{0.4} = 2.5 k \Omega
Comment: Since V_{DS} = 2.5 V \gt V_{GS} − V_{T N} = 1.91 − 1.2 = 0.71 V, each transistor is biased in the saturation region.