chapter 6 - Analog IC Design.org

Chapter 6 – Section 5 (5/2/04) Page 6.5-16
Frequency Response of the Folded Cascode Op Amp
The frequency response of the folded cascode op amp is determined primarily by the
output pole which is given as
p out =
-1
R out C out
where C out is all the capacitance connected from the output of the op amp to ground.
All other poles must be greater than GB = g m1 /C out . The approximate expressions for
each pole is
1.) Pole at node A: p A ≈ - g m6 /C A
2.) Pole at node B: p B ≈ - g m7 /C B
3.) Pole at drain of M6: p 6 ≈
-1
(R 2 +1/g m10 )C 6
4.) Pole at source of M8: p 8 ≈ -g m8 /C 8
5.) Pole at source of M9: p 9 ≈ -g m9 /C 9
6.) Pole at gate of M10: p 10 ≈ -g m10 /C 10
where the approximate expressions are found by the reciprocal product of the resistance
and parasitic capacitance seen to ground from a given node. One might feel that because
R B is approximately r ds that this pole might be too small. However, at frequencies where
this pole has influence, C out , causes R out to be much smaller making p B also non-dominant.
CMOS Analog Circuit Design © P.E. Allen - 2004
Chapter 6 – Section 5 (5/2/04) Page 6.5-17
Example 6.5-3 - Folded Cascode, CMOS Op Amp
Assume that all g mN = g mP = 100µS, r dsN = 2MΩ, r dsP = 1MΩ, and C L = 10pF. Find all
of the small-signal performance values for the folded-cascode op amp.
R II = 0.4GΩ, R A = 10kΩ, and R B = 4MΩ
v out
∴ k = 0.4x10 9(0.3x10-6)
100 = 1.2
⎛2+1.2⎞
v = ⎜ ⎟
in ⎝ 2+2.4 (100)(57.143) = 4,156V/V
⎠
R out = R II ||[g m7 r ds7 (r ds5 ||r ds2 )] = 400MΩ||[(100)(0.667MΩ)] = 57.143MΩ
1
1
|p out | = R out C out
= 57.143MΩ·10pF = 1,750 rads/sec. ⇒ 278Hz ⇒ GB = 1.21MHz
CMOS Analog Circuit Design © P.E. Allen - 2004