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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

3.53

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.