High-frequency response of a CG amplifier

High-frequency response of a CG amplifier

gd db L L C C C C′ = + + sb gs in C C C = + C_gs between source & ground

C_gd between drain & ground

“Input Pole”

“Output Pole” Low-pass

High-f response of a CG amplifier – Exact Solution (1)

gd db L L C C C C′ = + + sb gs i C C C = + 0 ) ( / 1 : Node 0 ) ( / 1 : Node = − + − + ′ + ′ = − + − − + − o i o i m L o L o o o o i i m in i sig sig i i r v v v g C s v R v v r v v v g sC v R v v v

High-f response of a CG amplifier – Exact Solution (2)

) / 1 || ( 1 1 / 1 / 1 ) 1 /( 1 1 1 1 1 1 0 ) ( : Node m sig i sig m m sig i sig m sig i sig m sig i sig m sig i i sig m i sig i i g R sC R g g v v R g R sC R g R sC R g v v v R g sC v v v + × + = + + × + = + + = = + + − L L L m i o i m o L L o o R C s R g v v v g v C s R v v ′ ′ + ′ = ⇒ = − ′ + ′ ( ) 0 1 : Node Voltage divider

(R_i=1/g_m and R_sig) “Input Pole”

“Output Pole” Mid-band Gain L L m sig i L m sig i i sig o R C s g R sC R g R R R v v ′ ′ + × + × ′ × + = 1 1 ) / 1 || ( 1 1 ) (

b₁ can be found by the open-circuit time-constants method. 1. Set v_sig = 0

2. Consider each capacitor separately, e.g., C_j (assume others are open circuit!) 3. Find the total resistance seen between the terminals of the capacitor, e.g., R_j

(treat ground as a regular “node”). 4.

A good approximation to f_H is:

1 2 1 b f_H

π

=

Open-Circuit Time-Constants Method

... 1 1 )... / 1 )( / 1 ( ... 1 ... 1 ... 1 ) ( 2 1 1 2 1 2 2 1 2 2 1 2 2 1 + + = + + + + + = + + + + + + = p p p p b s s s a s a s b s b s a s a s H

ω

ω

ω

ω

j j n j R C b₁ = Σ ₌₁

High-f response of a CG amplifier –

time-constant method (input pole)

1. Consider C_i : )] 1 /( ) ( || [ 1 =Cin Rsig ro +RL′ + gmro

τ

2. Find resistance between Capacitor terminals Terminals of C_in o m L o r g R r + ′ + 1 gd db L L C C C C′ = + + sb gs in C C C = + o m L o r g R r + ′ + = 1 o m L o r g R r + ′ + = 1

High-f response of a CG amplifier – time-constant

method (output pole)

1. Consider C’_L : )] 1 ( || [ 2 = CL′ RL′ ro + gmRsig

τ

2. Find resistance between Capacitor terminals ) 1 ( _{m sig} o g R r + ≈ ) 1 ( _{m sig} o g R r + gd db L L C C C C′ = + + sb gs in C C C = +

High-frequency response of a CG amplifier

o m L o i L o m sig i i M r g R r R R r g R R R A / ) ( ) || ( ′ + = ′ + + = ) ( 2 1 2 1 )] 1 ( || [ )] 1 /( ) ( || [ 2 1 1 2 1

τ

τ

π

π

τ

τ

+ = = + ′ ′ = + ′ + = b f R g r R C r g R r R C H sig m o L L o m L o sig in gd db L L C C C C′ = + + sb gs in C C C = + L L m sig in L m sig m m sig o R C s g R sC R g R g g v v ′ ′ + × + × ′ × + = 1 1 ) / 1 || ( 1 1 ) ( / 1 / 1

Comparison of time-constant method with the exact solution (r_o → ∞)

A_M 1 1 1/ : pole Input

τ

ω

_p = ₂ 1/ ₂ : pole Output

τ

ω

_p =

High-frequency response of a CS amplifier

db L C C + C_sb is shorted out. C_gd is between output and input!

Miller’s Theorem

1 2 A V V = ⋅ Z V A Z V V I 1 2 1 1 ) 1 ( − ⋅ = − = ) 1 ( , ) 1 /( ₁ 1 1 1 1 A Z Z Z V A Z V I − = = − = A Z Z Z V A ZA V I / 1 1 , ) 1 /( ₂ 2 2 2 2 = ₋ = = ₋ A Z V A Z V A Z V V I₂ = 2 − 1 = ( −1) 1 = ( −1) 2



_{Consider an amplifier with a gain}

A

_{with an impedance}

Z

_attached

between input and output



V

₁

_and

V

₂

_{“feel” the presence of}

Z

_{only through}

I

₁

_and

I

₂



_{We can replace}

Z

_{with any circuit as long as a current}

I

₁

_{flows out of}

V

₁

and a current

I

₂

flows out of

V

₂

.

Miller’s Theorem – statement



_{If an impedance}

Z

_{is attached between input and output an}

amplifier with a gain

A

, Z can be replaced with two

impedances between input & ground and output & ground

1

2 A V

V = ⋅

Other parts of the circuit

A Z Z 1 1 2 − = A Z Z − = 1 1 1 2 A V V = ⋅

1 0 1 0 1 0 0 1 1 1 0 0 ) / ( ) / ( ) / ( R R v v A R R R A R R A R A R R R A v v A v v f i o f f f f f f i n i o − ≈ + − = + − = + − = − =

Example of Miller’s Theorem: Inverting amplifier

n n p o A v v A v v = ₀ ⋅( − ) = − ₀⋅ : OpAmp 1 R R v v f i o = −

Recall from ECE 100, if A₀ is large

Solution using Miller’s theorem:

f f f R A R R ≈ + = 0 2 / 1 1 0 0 1 1 A R A R R_f f ≈ f + = 1 1 1 f f i n R R R v v + = A Z Z / 1 1 2 = ₋ A Z Z − = 1 1

Applying Miller’s Theorem to Capacitors

A Z Z 1 1 2 − = A Z Z − = 1 1 1 2 A V V = ⋅ ) / 1 1 ( / 1 1 ) 1 ( 1 1 2 1 1 1 C A C A Z Z C A C A Z Z C j Z − = ⇒ − = − = ⇒ − = = ω

Large capacitor at

the input for A >> 1

High-frequency response of a CS amplifier –

Using Miller’s Theorem

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

)] || ( 1 [ ) 1 ( ,i gd gd m o L gd C A C g r R C = − = + ′ * )] || ( / 1 1 [ ) / 1 1 ( , gd L o m gd gd o gd C R r g C A C C ≈ ′ + = − = ) || ( _{o L} m g d R r g v v A= = − ′ * Assuming g_mR’_L >> 1 i gd gs in C C C = + _, CL′ =Cdb +Cgd,o+CL

Note: C_gd appears in the input (C_gd,i) as a “much larger” capacitor.

High-f response of a CS amplifier –

Miller’s Theorem and time-constant method

Output Pole (C’_L ): ) || ( 2 =CL′ ro RL′

τ

1 = CinRsig

τ

Input Pole (C_in ): ) || ( 2 1 1 in sig L o L H R r C R C b f = = + ′ ′ π o r ∞ L gd db L L o m gd gs in C C C C R r g C C C + + = ′ ′ + + = [1 ( || )]

Miller & time-constant method:

1. Same b₁ and same f_H as exact solution

2. Although, we get the same f_H, there is a substantial error in individual input and output poles.

3. Miller approximation did not find the zero!

High-f response of a CS amplifier – Exact solution

Solving the circuit (node voltage method):

gd gs gd gs db L L o L sig in m gd L o m sig o C C C C C C b R r C R C b s b s b g sC R r g v v + + + = ′ ′ + = + + − × ′ − = ) )( ( ) || ( 1 ) / 1 ( ) || ( 2 1 2 2 1 L gd db L L o m gd gs in C C C C R r g C C C + + = ′ ′ + + = [1 ( || )] ) || ( 2 1 1 in sig L o L H R r C R C b f = = + ′ ′ π

Miller’s Theorem vs Miller’s Approximation



_{For Miller Theorem to work, ratio of}

V

₂

/V

₁

_{(amplifier gain) should be}

calculated in the presence of impedance

Z.



_{In our analysis, we used mid-band gain of the amplifier and ignored changes}

in the gain due to the feedback capacitor,

C

_gd

. This is called “Miller’s

Approximation.”

o In the OpAmp example the gain of the chip, A₀ , remains constant when R_f is

attached (output resistance of the chip is small).



_{Because the amplifier gain in the presence of}

C

_gd

_{is smaller than the}

mid-band gain (we are on the high-

f

portion of the gain Bode plot), Miller’s

approximation overestimates

C

_gd,i

and underestimates

C

_gd,o

o There is a substantial error in individual input and output poles. However, b₁

and f_H are estimated well.



_{More importantly, Miller’s Approximation “misses” the zero introduced by}

the feedback resistor (This is important for stability of feedback amplifiers

as it affects gain and phase margins).

1) By definition,

2) Because v_o = 0, zero current will flow in r_o, C_L+C_db and R’_L

3) By KCL, a current of g_mv_gs will flow in C_gd.

4) Ohm’s law for C_gd gives:

Finding the “zero” of the CS amplifier

gd z gs m gs C s v g i Z v −0 = = 0 ) ( : Zero v_o s = s_z = gs mv g i= 0 = i gd m z C g s =

Zero of CS amplifier can play an important role

in stability of feedback amplifiers

z f 2 p f 1 p f f_p1 f_z f_p2

Since the input pole is at Small R_sig can push f_p2 to very large values!

) 2 /( 1 ) 1/(2πτ₁ = π C_inR_sig 1 2 of Case f_z >> f_p > f_p Case of f_p2 > f_z > f_p1

Caution:



_{The time constant method approximation to}

f

_H

_{(see S&S page 724).}



_Since,

This is the correct formula to find

f

_H



_{However, S&S gives a different formula in page 722 (contradicting}

formulas of pp724).

Ignore this formula (S&S Eq. 9.68)



_{Discussions (and some conclusions re Miller’s theorem) in Examples 9.8}

to 9.10 are incorrect. The discrepancy between

f

_H

from Miller’s

approximation and exact solution is due to the use of Eq. 9.68 (Not

Miller’s fault!)

...

1

1

1

1

2 3 2 2 2 1

+

+

+

=

p p p H

ω

ω

ω

ω

... 1 1 1 ... 1 1 2 1 2 1 1 = + + ⇒ ≈ + + p p H p p b

ω

ω

ω

ω

ω

H j j n j R C b

ω

1 1 1 = Σ = ≈