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.