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chapter 6 - Analog IC Design.org

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Chapter 6 – Section 3 (5/2/04) Page 6.3-16<br />

Example 6.3-2 - Continued<br />

It is worthwhile to check that the RHP zero has been moved on top of p 2 . To do this,<br />

first calculate the value of R z . V SG8 must first be determined. It is equal to V SG10 , which is<br />

V SG10 =<br />

2I 10<br />

K’ P S 10<br />

+ |V TP | =<br />

2·15<br />

50·1 + 0.7 = 1.474V<br />

Next determine R z .<br />

1<br />

R z = K’ P S 8 (V SG10 -|V TP |) = 106<br />

50·5.63(1.474-.7) = 4.590kΩ<br />

The location of z 1 is calculated as<br />

−1<br />

z 1 =<br />

(4.590 x 103)(3x10 -12 ) −<br />

The output pole, p 2 , is<br />

3x10 -12 = -94.46x106 rads/sec<br />

942.5x10 -6<br />

p 2 = 942.5x10 -6<br />

10x10 -12 = -94.25x10 6 rads/sec<br />

Thus, we see that for all practical purposes, the output pole is canceled by the zero<br />

that has been moved from the RHP to the LHP.<br />

The results of this design are summarized below.<br />

W 8 = 6 µm W 9 = 2 µm W 10 = 1 µm W 11 = 15 µm<br />

CMOS <strong>Analog</strong> Circuit <strong>Design</strong> © P.E. Allen - 2004<br />

Chapter 6 – Section 3 (5/2/04) Page 6.3-17<br />

An Alternate Form of Nulling Resistor<br />

To cancel p 2 ,<br />

z 1 = p 2 → R z = C c+C L 1<br />

g m6A C =<br />

C g m6B<br />

Which gives<br />

⎛<br />

⎜<br />

g m6B = g m6A⎝ ⎜<br />

C c<br />

⎞<br />

⎟<br />

⎟<br />

⎠<br />

C c +C L<br />

In the previous example,<br />

g m6A = 942.5µS, C c = 3pF<br />

and C L = 10pF.<br />

Choose I 6B = 10µA to get<br />

or<br />

g m6B = g m6AC c<br />

C c + C L<br />

→<br />

W 6B<br />

L 6B<br />

⎛<br />

= ⎜<br />

⎝<br />

3 ⎞<br />

⎟<br />

⎠<br />

I 13 2 6A W 6A<br />

I 6B<br />

L 6A<br />

⎛<br />

= ⎜<br />

⎝<br />

2K P W 6B I 6B<br />

L 6B<br />

- M1 M2<br />

v in<br />

+<br />

+<br />

VBias<br />

-<br />

M3<br />

M4<br />

M5<br />

V DD<br />

M11<br />

M6B<br />

M8<br />

V SS<br />

M10<br />

C c<br />

M9<br />

M6<br />

v out<br />

C L<br />

M7<br />

Fig. 6.3-4A<br />

3 ⎞<br />

⎟<br />

13 2 ⎛<br />

⎜<br />

⎠ ⎝<br />

⎟<br />

⎠<br />

⎛<br />

= ⎜<br />

⎝<br />

C c<br />

C c +C L<br />

⎞<br />

⎟<br />

⎠<br />

2K P W 6A I D6<br />

L 6A<br />

95⎞<br />

10 (94) = 47.6 → W 6B = 48µm<br />

CMOS <strong>Analog</strong> Circuit <strong>Design</strong> © P.E. Allen - 2004

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