Question 10.12: Consider an active-loaded MOS differential amplifier of the ...
Consider an active-loaded MOS differential amplifier of the type shown in Fig. 10.37(a). Assume that for all transistors, W/L = 7.2 μm/0.36 μm, Cgs = 20 fF, Cgd = 5 fF, and Cdb = 5 fF. Also, let μnCox = 387 μA/V², μpCox = 86 μA/V², V_{An}^{′} = 5 V/μm, and |V_{Ap}^{′}|= 6 V/μm. The bias current I = 0.2 mA, and the bias current source has an output resistance RSS = 25 kΩ and an output capacitance CSS = 0.2 pF. In addition to the capacitances introduced by the transistors at the output node, there is a capacitance Cx of 25 fF. It is required to determine the low-frequency values of Ad , Acm, and CMRR. It is also required to find the poles and zero of Ad and the dominant pole of CMRR.
Since I = 0.2 mA, each of the four transistors is operating at a bias current of 100 μA. Thus, for Q1 and Q2,
100 = \frac{1}{2} × 387 × \frac{7.2}{0.36} × V_{OV}^{2}
which leads to
VOV = 0.16 V
Thus,
g_{m} = g_{m1} = g_{m2} = \frac{2 × 0.1}{0.16} = 1.25 mA/V
r_{o1} = r_{o2} = \frac{5 × 0.36}{0.1} = 18 kΩ
For Q3 and Q4 we have
100 = \frac{1}{2} × 86 × \frac{7.2}{0.36} V_{OV3,4}^{2}
Thus,
|VOV3,4| = 0.34 V
and
g_{m3} = g_{m4} = \frac{2 × 0.1}{0.34} = 0.6 mA/V
r_{o3} = r_{o4} = \frac{6 × 0.36}{0.1} = 21.6 kΩ
The low-frequency value of the differential gain can be determined from
Ad = gm (ro2 || ro4)
= 1.25 (18 || 21.6) = 12.3 V/V
The low-frequency value of the common-mode gain can be determined from Eq. (9.157) as
A_{cm} = – \frac{1}{2g_{m3}R_{SS}} (9.157)
= – \frac{1}{2 × 0.6 × 25} = −0.033 V/V
The low-frequency value of the CMRR can now be determined as
CMRR = \frac{|A_{d}|}{|A_{cm}|} = \frac{12.3}{0.033} = 369
or,
20 log 369 = 51.3 dB
To determine the poles and zero of Ad we first compute the values of the two pertinent capacitances Cm and CL. Using Eq. (10.133),
Cm = Cgd1 + Cdb1 + Cdb3 + Cgs3 + Cgs4 (10.133)
= 5 + 5 + 5 + 20 + 20 = 55 fF
Capacitance CL is found using Eq. (10.134) as
CL = Cgd2 + Cdb2 + Cgd4 + Cdb4 + Cx (10.134)
= 5 + 5 + 5 + 5 + 25 = 45 fF
Now, the poles and zero of Ad can be found from Eqs. (10.145), (10.139), and (10.140) as
f_{P1} = \frac{1}{2πC_{L}R_{o}} (10.145)
= \frac{1}{2π × C_{L}(r_{o2} || r_{o4})}
= \frac{1}{2π × 45 × 10^{−15} (18 || 21.6) 10^{3}}
= 360 MHz
f_{P2} = \frac{g_{m3}}{2πC_{m}} (10.139)
f_{Z} = \frac{2 g_{m3}}{2πC_{m}} (10.140)
f_{P2} = \frac{g_{m3}}{2πC_{m}} = \frac{0.6 × 10^{−3}}{2π × 55 × 10^{−15}} = 1.74 GHz
fZ = 2fP2 = 3.5 GHz
Thus the dominant pole is that produced by CL at the output node. As expected, the pole and zero of the mirror are at much higher frequencies.
The dominant pole of the CMRR is at the location of the common-mode-gain zero introduced by CSS and RSS , that is,
f_{Z} = \frac{1}{2π C_{SS}R_{SS}}
= \frac{1}{2π × 0.2 × 10^{−12} × 25 × 10^{3}}
= 31.8 MHz
Thus, the CMRR begins to decrease at 31.8 MHz, which is much lower than fP1.