6. Frequency Response - University of California, San...

6. Frequency Response

ECE 102, Winter 2011, F. Najmabadi

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

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: fLo Upper cut-off frequency: fHo 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)

Cc1vi = 0 → vo = 0Contributes 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

CLNo change in voDoes NOT contribute to fL

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

Cc1No change in voDoes NOT contribute to fH

CLvo = 0Contributes 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:

Example:

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.

Cc1 contributes to fL → short circuitCL 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

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 = 02. 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

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 and note that we exactly get the same poles).

Then fL = fp1 + fp2 + fp3


1. Consider Cc1 :

∞

)( 21

11

sigGcp RRC

f+

=π

2. Find resistance between Capacitor terminals

Terminals of Cc1


1. Consider CS :

Terminals of CS

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

2moLDmSS

p grRRgRCf

+=

π


moLD

m

grRRg

/||/1 +

moLD

m

grRRg

/)||(/1 +

moLD

m

grRRg

/)||(/1 +

moLD

m

grRRg

/)||(/1 +


1. Consider Cc2 : Terminals of Cc2

)||( 21

23

oDLcp rRRC

f+

=π


High-Frequency Response

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

Reverse Bias Forward 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 = 10-12 F

Capacitive Effects in MOS

1. Capacitance between Gate and channel(Parallel-plate capacitor)

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)

Capacitive Effects in MOS

mSB

sbsb VV

CC)/1( 0

0

+=

“Parallel-Plate” capacitances (depends on the channel shape)

oxovov CLWC ⋅⋅=oxgate CLWC ⋅⋅=Define:

Triode

ovgategs CCC +⋅=21

ovgategd CCC +⋅=21

Saturation


ovgd CC =

Cut-off

ovgdgs CCC ==

gategb CC =

Pinched Channel No Channel

“Junction” capacitances

mDB

dbdb VV

CC)/1( 0

0

+=

Small signal for MOS in high-frequencies

For source connected to body

Saturation


ovgd CC =

mDB

dbdb VV

CC)/1( 0

0

+=

High-frequency response of a CS amplifier

Cdb short:vo = 0

Cgd short:Input is connected to outputGain is reduced to 1

1) MOS “internal” capacitors are shown “outside” of the transistor to see their impact.2) All MOS capacitors contribute to fH (vo is reduced when f→∞ or caps short circuit)3) For f→∞ , all coupling (Cc1 and Cc2 ) and by-pass capacitors are short circuit

For f→∞ Cgs short:vi = 0 → vo = 0


)||||(

/11 x

1

LDomsigG

GM

pM

sig

o

RRrgRR

RA

sA

VV

+−=

+=

ω

See S&S pp712-714 for detailed calculations (S&S assumes Cdb → 0)

)||||1(

,)||||)((

1 ,)||(

121

LDomgdgsin

LDodbgdp

sigGinp

RRrgCCCRRrCCRRC

++=

+== ωω

In General:1) One internal capacitors shorts input to the ground (Cgs here)2) One internal capacitors shorts output to the ground (Cdb here)3) One internal capacitors shorts input to output (Cgd here)

Cgd appears in parallel to Cgs(with a much larger value)

Cgd appears in parallel to Cdb

High-frequency-relevant capacitors

High-frequency-relevant capacitors appear between o input & ground, o output & ground, ando input & output.

Capacitors that are connected between input & output provide feedback. In the case of CS amplifier, we saw that they appeared in the transfer function as capacitors in parallel to input & ground and output & ground capacitors.

We can use Miller’s Theorem to replace capacitors connected between input & output and simplify the analysis.

Miller’s Theorem

12 VAV ⋅=

ZVA

ZVVI 121

1)1( ⋅−

=−

=

)1( ,

)1/( 11

111 A

ZZZV

AZVI

−==

−=

AZZ

ZV

AAZVI

/11 ,

)1/( 22

222 −

==−⋅

=

AZVA

ZVA

ZVVI

⋅−

=−

=−

= 21122

)1()1(

Consider an amplifier with a gain A with an impedance Z attached between input and output

V1 and V2 “feel” the impedance of Z only through I1 and I2 We can replace Z with any circuit as long as a current I1 flows out of

V1 and a current I2 flows out of V2.

Miller’s Theorem

If an impedance Z is attached between input and output an amplifier with a gain A , it can be replaced with two impedances between input & ground and output & ground

12 VAV ⋅=

Other parts of the circuit

A

ZZ 112

−=

AZZ−

=11

12 VAV ⋅=

Example of Miller’s Theorem:Inverting amplifier

nnpo vAvvAv ⋅−=−⋅= 00 )(

1RR

vv f

i

o −=

Recall from ECE 100, if A0 is large

Solution using Miller’s theorem:

ff

f RA

RR ≈

+=

02 /1100

1 1 AR

AR

R fff ≈

+=

11

1

f

f

i

n

RRR

vv

+=

10101

00

11

100 )/()/(

)/(RR

ARRR

ARRARA

RRRA

vvA

vv f

f

f

f

f

f

f

i

n

i

o −≈

+

−=

+

−=

+

−=−=

AZZ

/112 −=

AZZ−

=11

Finding fH by inspection

1. Set vsig = 02. Use Miller’s Theorem to replace capacitor between input &

output with two capacitors at the input and output.3. Consider each capacitor separately (assume others are open

circuit!), e.g., Cn

4. Find the total resistance seen between the terminals of the capacitor, e.g., Rn (treat ground as a regular “node”).

5. Compute the

6. Upper cut-off frequency can be fund from:nn

pn CRf

π21

=

...1111

321

+++=pppH ffff

Caution:

Method in previous slide is called the time-constant approximation to fH(see S&S page 724).

Since , the above formula give

This is the correct formula to find fH

However, S&S gives a different formula in page 722 (contradicting formulas of pp724). Ignore this formula (S&S Eq. 9.68)

nnpn CR

fπ2

1=

...111 21

321n

+++=Σ=ppp

nnH fff

CRf

π

nnH CR

f 21

nΣ

=π (S&S Eq. 9.73)

...111123

22

21

+++=pppH ffff

Applying Miller’s Theorem to Capacitors

A

ZZ 112

−=

AZZ−

=11

12 VAV ⋅=

CACA

ZZ

CACA

ZZ

CjZ

)/11( /11

)1( 1

1

21

11

−=⇒−

=

−=⇒−

=

=ω

Example: High-frequency response of a CS amplifier

o Circuit includes CL which is often used to set the “dominate pole”.o The first step is to identify which capacitors are relevant to high-

frequency response (Cgs ,Cdb , Cgd , and CL ). The other capacitors, Cc1 and Cc2 are relevant to low-frequency response.

o At high frequency, Cc1 and Cc2 will be short.


Use Miller’s Theorem to replace capacitor between input & output (Cgd ) with two capacitors at the input and output.

gdmmgdgdigd CRgRgCACC ′≈′+=−= )1()1(,

gdLmgdgdogd CRgCACC ≈′+=−= )/11()/11(,

LmLDomg

d RgRRrgvvA ′−≡−== )||||(

* Assuming gmR’L >> 1igdgsin CCC ,+=

LogddbL CCCC ++=′ ,


1. Consider Cin :

∞

)||( 21

1sigGin

p RRCf

π=


Terminals of Cin


1. Consider C’L :

)||||( 21

2LDoL

p RRrCf

′=

π


Terminals of C’L


21

21

/1/1/1

)||||1(

)||||( 21/ ),||( 21/

)||||(

ppH

LdbgdL

LDomgdgsin

LDoLpsigGinp

LDomsigG

GM

fffCCCC

RRrgCCCRRrCfRRCf

RRrgRR

RA

+=

++=′

++=

′==

+−=

ππ

Miller’s Theorem vs Miller’s Approximation For Miller Theorem to work, ratio of V2/V1 (amplifier gain) should be

independent of feedback impedance Z.

This was correct for OpAmp example where the gain of the chip, A0 , remains constant when Rf is attached (output resistance of the chip is small).

However, the capacitor that connect the input and output changes the frequency response of the amplifier (i.e., its gain) and so we cannot “strictly” apply Miller’s Theorem.

In our analysis, we used mid-band gain of the transistor and ignored changes in the frequency response due to the feedback capacitor. This is called “Miller’s Approximation.”

Miller’s Approximation only gives “approximate” values of the poles and the higher cut-off frequency.

More importantly, Miller’s Approximation “misses” the zero introduced by the feedback resistor (which can cause “unstable” operation).

Example: High-frequency response of a CG amplifier

o Cdb is ignored in the above. Including body effect, one sees Cdbactually appears between drain and ground (parallel to CL in the above circuit) and is “absorbed” in CL.

o Note that Cgd is also between the drain and the ground and is in parallel to CL.

)( dbLgdL CCCC ++=′

om

LDo

rgRRr ||+


1. Consider Cgs :

]/)||(||[ 21

1omLDosiggs

p rgRRrRCf

+=

π


Terminals of Cgs

om

LDo

rgRRr ||+

≈

om

LDo

rgRRr ||+


1. Consider C’L :

)]1(||||[ 21

2sigmoLDL

p RgrRRCf

+′=

π


)1( sigmo Rgr +≈

)1( sigmo Rgr +

High-frequency response of a CG amplifier

21

2

1

/1/1/1

)(

)]1(||||[ 2/1

, )||( 2/1/)||(

)||||(

ppH

dbLgdL

sigmoLDLp

isiggsp

omLDoi

LDomsigi

iM

fffCCCC

RgrRRCfRRCf

rgRRrR

RRrgRR

RA

+=

++=′

+′=

=+=

++=

π

π

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