# CMOS Design Engineer Essay

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6. Frequency Response
Reading: Sedra & Smith: Chapter 1.6, Chapter 3.6 and Chapter 9 (MOS portions),

ECE 102, Winter 2011, F. Najmabadi

Typical Frequency response of an Amplifier
 Up to now we have ignored the capacitors. To include the capacitors, we need to solve the circuit in the frequency domain (or use Phasors). o Lower cut-off frequency: fL o Upper cut-off frequency: fH o Band-width: B = fH − fL

Classification of amplifiers based on the frequency response
AC amplifier (capacitively-coupled)

DC amplifier (directly-coupled) fL = 0

Tuned or Band-pass amplifier (High Q)

How to find which capacitors contribute to the lower cut-off frequency
 Consider each capacitor individually. Let f = 0 (capacitor is open circuit): o If vo (or AM) does not change, capacitor does NOT contribute to fL o If vo (or AM) → 0 or reduced substantially, capacitor contributes to fL
Example:

Cc1 vi = 0 → v o = 0
Contributes to fL

CL
No change in vo
Does NOT contribute to fL

How to find which capacitors contribute to the higher cut-off frequency
 Consider each capacitor individually. Let f → ∞ (capacitor is short circuit): o If vo (or AM) does not change, capacitor does NOT contribute to fH o If vo (or AM) → 0 or reduced substantially, capacitor contributes to fH
Example:

Cc1
No change in vo
Does NOT contribute to fH

CL vo = 0
Contributes to fH

How to find “mid-frequency” circuit
 All capacitors that contribute to low-frequency response should be short circuit.
 All capacitors that contribute to high-frequency response should be open circuit.

Example:

Cc1 contributes to fL → short circuit
CL contributes to fH → open circuit

Low-Frequency Response

Low-frequency response of a CS amplifier

 Each capacitors gives a pole.
 All poles contribute to fL (exact value of fL from simulation)
 If one pole is at least two octave (factor of 4) higher than others (e.g., fp2 in the above figure), fL is approximately equal to that pole (e.g., fL = fp2 in above)
 A good approximation for design & hand calculations: fL = fp1 + fp2 + fp3 + …

Low-frequency response of a CS amplifier
All capacitors contribute to fL (vo is reduced when f → 0 or caps open circuit)

Cc1 open: vi = 0 → v o = 0

Cc2 open: vo = 0

Cs open:
Gain is reduced substantially
(from CS amp. To CS amp. With RS)
See S&S pp689-692 for detailed calculations (S&S assumes ro → ∞ and RS → ∞ )

Vo s s s = AM x x x
Vsig
s + ω p1 s + ω p 2 s + ω p 3
AM = −

RG g m (ro || RD || RL )
RG + Rsig

ω p1 =

1
1
, ω p3 =
Cc1 ( RG + Rsig )
Cc 2 ( RD || ro + RL )

ω p2 ≈

1
,
Cs [ RS || (1 / g m + RD || RL / ro g m )]

Finding poles by inspection
1. Set vsig = 0
2. Consider each capacitor separately (assume others are short circuit!), e.g., Cn
3. Find the total resistance seen between the terminals of the capacitor, e.g., Rn (treat ground as a regular “node”).
4. The pole associated with that capacitor is f pn =

1
2πRn Cn

5. Lower-cut-off frequency can be found from fL = fp1 + fp2 + fp3 + …
* Although we are calculating frequency response in frequency domain, we will use time-domain notation instead of phasor form (i.e., vsig instead of Vsig ) to avoid confusion with the bias values.

Example: Low-frequency response of a CS amplifier

 Examination of circuit shows that ALL capacitors contribute to the low-frequency response.  In the following slides with compute poles introduced by each capacitor (compare with the detailed calculations and note that we exactly get the same poles).
 Then fL = fp1 + fp2 + fp3

Example: Low-frequency response of a CS amplifier

f p1 =

1. Consider Cc1 :

1
2π Cc1 ( RG + Rsig )

Terminals of Cc1

2. Find resistance between
Capacitor terminals

Example: Low-frequency response of a CS amplifier f p2 =

1
2π C S [ RS || (1 / g m + RD || RL / ro g m )]

1/ gm +
( RD || RL ) / ro g m

1.