Question 4.13: Objective: Determine the small-signal voltage gain of the CM...
Objective: Determine the small-signal voltage gain of the CMOS amplifier. For the circuit shown in Figure 4.45(a), assume transistor parameters of V_{T N} = +0.8 V, V_{T P} = −0.8 V, k´_{n} = 80 µA/V^{2} , k´_{p} = 40 µA/V^{2} , (W/L)_{n} = 15, (W/L)_{p} = 30, and λ_{n} = λ_{p} = 0.01 V^{−1} . Also, assume I_{Bias} = 0.2 mA.
The transconductance of the NMOS driver is
g_{mn} = 2 \sqrt{K_{n} I_{DQ}} = 2 \sqrt{\left( \frac{k´_{n}}{2} \right) \left( \frac{W}{L} \right)_{n} I_{Bias}}= 2 \sqrt{\left(\frac{0.08}{2} \right) (15)(0.2)} = 0.693 mA/V
Since λ_{n} = λ_{p} the output resistances are
r_{on} = r_{op} = \frac{1}{\lambda I_{DQ}} = \frac{1}{(0.01)(0.2)} = 500 k \OmegaThe small-signal voltage gain is then
A_{v} = – g_{m}(r_{on} || r_{op}) = -(0.693)(500 || 500) = – 173Comment: The voltage gain of the CMOS amplifier is on the same order of magnitude as the NMOS amplifier with depletion load. However, the CMOS amplifier does not suffer from the body effect.
Discussion: In the circuit configuration shown in Figure 4.45(a), we must again apply a dc voltage to the gate of M_{1} to achieve the “proper” Q-point. We will show in later chapters using more sophisticated circuits how the Q-point is more easily established with current-source biasing. However, this circuit demonstrates the basic principles of the CMOS common-source amplifier.