The roots of a two-stage OP AMP using Miller compensation with a nulling resistor are given as: Dominant pole: P₁ Nulling pole: P₂ Nulling zero: Z₁ == == = 8ml A,Cc 1 R₂C₁ Output pole: P₂ Mirror pole: P3 1 RC-(C18m6) 8m6 CL 8m3 Cgs3 +C gs4 with the following parameters 8ml = 100 μs, R₁ = 500k, 8ml = 1000µS, R₁ = 50k, C₁=0.1pF, C₁ =3pF, P3 = -1 -1592.35 MHz с (a) What is the low frequency gain and the gain bandwidth (GB) in MHz

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The roots of a two-stage OP AMP using Miller compensation with a nulling
resistor are given as:
8 m6
8 ml
Output pole: P2
A,C.
= -
Dominant pole: P1
8 m3
1
Mirror pole: P3
R.C,
Nulling pole: Pz
gs3
gs4
1
with the following parameters
%3|
Nulling zero: Z =
R.C. -(C./ 8 m6 )
8 mi = 100 uS, R, = 500k2, 8 mll =1000uS, R,=50k2, C, =0.1pF ,
C. = 3 pF , p3 =-1592.35MHz
(a) What is the low frequency gain and the gain bandwidth (GB) in MHz
(b) What value of C, gives a phase margin of 60° if R, =
(c) What value of R, causes the RHP zero to cancel the output pole using
the C, of the previous question
(d) What is the new GB for 60° phase margin?
0?
Transcribed Image Text:The roots of a two-stage OP AMP using Miller compensation with a nulling resistor are given as: 8 m6 8 ml Output pole: P2 A,C. = - Dominant pole: P1 8 m3 1 Mirror pole: P3 R.C, Nulling pole: Pz gs3 gs4 1 with the following parameters %3| Nulling zero: Z = R.C. -(C./ 8 m6 ) 8 mi = 100 uS, R, = 500k2, 8 mll =1000uS, R,=50k2, C, =0.1pF , C. = 3 pF , p3 =-1592.35MHz (a) What is the low frequency gain and the gain bandwidth (GB) in MHz (b) What value of C, gives a phase margin of 60° if R, = (c) What value of R, causes the RHP zero to cancel the output pole using the C, of the previous question (d) What is the new GB for 60° phase margin? 0?
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