9. Frequency Response

Reading: Sedra & Smith: Sec. 1.6, Sec. 3.6 and Sec. 9 (MOS portions),

(S&S 5th Ed: Sec. 1.6, Sec. 3.7 (capacitive effects), Sec. 4.8, Sec. 4.9, ,Sec. 6. (Frequency response sections, i.e., 6.4, 6.6, …),

Sec. 7.6

ECE 102, Fall 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)

Cc1 open: vi = 0 → vo = 0 Contributes to fL

Example:

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

CL open: No change in vo Does NOT contribute to fL

How to find which capacitors contribute to the higher cut-off frequency

Cc1 short: No change in vo Does NOT contribute to fH

CL short: vo = 0 Contributes to fH

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:

Impact of various capacitors depend on the frequency of interest

f → ∞ All Caps are short. Used to find high-frequency C.

f → 0 All Caps are open. Used this to find low-frequency C.

Mid-band: High-f caps are open Low-f caps are short.

Computing fH : High-f caps are included. Low-f caps are short

Computing fL: High-f caps are open. Low-f caps included.

Impendence of capacitors (1/ωC)

Constructing appropriate circuits Example:

Cc1 contributes to fL CL contributes to fH

Mid-band: High-f caps are open Low-f caps are short.

Computing fH : High-f caps are included. Low-f caps are short

Computing fL: High-f caps are open. Low-f caps included.

Low-Frequency Response

Typical Low-frequency response of an amplifier

Each capacitors gives a pole. All poles contribute to fL (exact value of fL from computation or simulation)

A good approximation for design & hand calculations: fL ≈ fp1 + fp2 + fp3 + …

If one pole is at least a 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 within 20%)

321

x x x ppp

Msig

o

ss

ss

ssA

VV

ωωω +++=

Example: an amplifier with three poles

(Set s = jω to find Bode Plots)

Low-frequency response of a CS amplifier

Cc1 open: vi = 0 → vo = 0

Cc2 open: vo = 0

Cs open: Gain is reduced substantially (from CS amp. To CS amp. With RS)

)||||(

x x x 321

LDomsigG

GM

pppM

sig

o

RRrgRR

RA

ss

ss

ssA

VV

+−=

+++=

ωωω

See S&S pp689-692 for detailed calculations (S&S assumes ro → ∞ and RS → ∞ )

,)]/||/1(||[

1

)||(1 ,

)(1

2

23

11

moLDmSsp

LoDcp

sigGcp

grRRgRC

RrRCRRC

+≈

+=

+=

ω

ωω

All capacitors contribute to fL (vo is reduced when f → 0 or caps open circuit)

Finding poles by inspection

1. Set vsig = 0 2. Consider each capacitor separately, e.g., Cn (assume others

are short circuit!)

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

5. Lower-cut-off frequency can be found from fL ≈ fp1 + fp2 + fp3 + …

nnpn CR

fπ2

1=

* 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.)

Then fL ≈ fp1 + fp2 + fp3

Example: Low-frequency response of a CS amplifier

1. Consider Cc1 :

∞

)( 21

11

sigGcp RRC

f+

=π

2. Find resistance between Capacitor terminals

Terminals of Cc1

Example: Low-frequency response of a CS amplifier

1. Consider CS :

Terminals of CS

)]/||/1(||[ 21

2moLDmSS

p grRRgRCf

+=

π

2. Find resistance between Capacitor terminals

moLD

m

grRRg

/)||(/1 +

moLD

m

grRRg

/)||(/1 +

moLD

m

grRRg

/)||(/1 +

Example: Low-frequency response of a CS amplifier

1. Consider Cc2 : Terminals of Cc2

)||( 21

23

oDLcp rRRC

f+

=π

2. Find resistance between Capacitor terminals

High-Frequency Response

In addition to the impact of external capacitors, amplifier gain falls off due to the internal capacitive effects of transistors

Capacitive Effects in pn Junction

Majority Carriers Charge stored is a function of applied

voltage. We can define a “small-signal”

capacitance, Cj

In reverse-bias region, analysis show (see S&S pp154-156):

V0 : Junction built-in voltage Cj0 : Capacitance at zero reversed-bias

voltage. m : grading coefficient (1/2 to 1/3). For forward-bias region: Cj ≈ 2Cj0

QR VVR

Jj dV

dQC=

=

mR

jj VV

CC

)/1( 0

0

+=

Capacitive Effects in pn Junction

Minority Carriers Excess minority carriers are stored in p and n sides of the

junction. The charge depends on the minority carrier “life-time” (i.e.,

how long it would take for them to diffuse through the junction and recombine.

Gives Diffusion Capacitance, Cd

Cd is proportional to current (Cd = 0 for reverse-bias)

T

DTd V

IC ⋅=τ

Small Signal Model for a diode

Forward Bias Reverse Bias

02 jj CC ⋅≈

T

DTd V

IC ⋅=τ

0=dC

mR

jj VV

CC

)/1( 0

0

+=

rD

Cj + Cd

Junction capacitances are small and are given in femto-Farad (fF)

1 fF = 10−15 F

Capacitive Effects in MOS 1. Capacitance between Gate and channel (Parallel-plate capacitor) appears as 2 capacitors: between gate/source & between gate/drain

3. Junction capacitance between Source and Body (Reverse-bias junction)

4. Junction capacitance between Drain and Body (Reverse-bias junction)

2. Capacitance between Gate & Source and Gate & Drain due to the overlap of gate electrode (Parallel-plate capacitor)

MOS High-frequency small signal model

MOS high-frequency small signal model For source connected to body

(used by S&S) Accurate Model

(we use this model here)

Generally, transistor internal capacitances are shown outside the transistor so that we can use results from the mid-band calculations.