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NTUEE Electronics – L.H. Lu 9-1

CHAPTER 9 FREQUENCY RESPONSE

Chapter Outline9.1 Low-Frequency Response of the CS and CE Amplifiers9.2 Internal Capacitive Effects and the High-Frequency Model9.3 High-Frequency Response of the CS and CE Amplifiers9.4 Tools for the Analysis of the High-Frequency Response of Amplifiers9.5 A Closer Look at the High-Frequency Response9.6 High-Frequency Response of the CG and Cascode Amplifiers


Frequency response of amplifiersMidband: The frequency range of interest for amplifiers Large capacitors act as short circuits and small capacitors act as open circuits Gain is constant and can be obtained by small-signal analysis

Low-frequency band: Large capacitors no longer act as short circuits (small capacitors still act as open circuits) The gain roll-off is mainly due to coupling and by-pass capacitors Gain drops at frequencies lower than fL

High-frequency band: Small capacitors no longer act as open circuits (large capacitors still act as short circuits) The gain roll-off is mainly due to parasitic capacitances of the MOSFETs and BJTs Gain drops at frequencies higher than fH

9.1 Low-Frequency Response of the CS and CE Amplifiers


The CS amplifierSmall-signal analysis for transfer function:

𝑉 =𝑅

𝑅 +1

𝑠𝐶+ 𝑅

𝑉

𝐼 =1

1𝑔

+1

𝑠𝐶||𝑅

𝑉

𝑉 = −𝑅

𝑅 + 𝑅 +1

𝑠𝐶

𝑅 𝐼

𝑉

𝑉𝑠 =

𝑉

𝐼

𝐼

𝑉

𝑉

𝑉= 𝐴

𝑠

𝑠 + 𝜔

𝑠 + 𝜔

𝑠 + 𝜔

𝑠

𝑠 + 𝜔

𝐴 = −𝑅

𝑅 + 𝑅𝑔 𝑅 ||𝑅

𝜔 =1

𝐶 𝑅 + 𝑅

𝜔 =𝑔 + 1/𝑅

𝐶

𝜔 =1

𝐶 𝑅 + 𝑅

𝜔 =1

𝐶 𝑅

The complete frequency response (gain and phase) are determined by the transfer function analysisDetermining the lower 3-dB frequency (fL) by dominant pole Coupling and by-pass capacitors result in a high-pass frequency response with 3 poles and 1 zero If poles are sufficiently separated (dominant-pole case):Bode plot can be used to evaluate the response for simplicityThe lower 3-dB frequency is the highest-frequency polefP2 is typically the highest-frequency pole due to small resistance of 1/gm


𝑅 + 𝑅 ≫ 𝑅 + 𝑅 ≫1

𝑔

𝑓 ≪ 𝑓 ≪ 𝑓 ≈ 𝑓

Determining the lower 3-dB frequency (fL) by poles and zeros If poles are located closely, the lower 3-dB frequency (at a frequency higher than poles and

zeros) can be evaluated by the transfer function

Determining the pole and zero frequencies by inspection (capacitors do not interact) A pole is determined by the time constant of the capacitor


𝑓 ≈ 𝑓 + 𝑓 + 𝑓 − 2𝑓

𝑉

𝑉𝑠 = 𝐴

𝑠

𝑠 + 𝜔

𝑠 + 𝜔

𝑠 + 𝜔

𝑠

𝑠 + 𝜔

𝑉

𝑉𝑗𝜔 = 𝐴

1

1 + 𝜔 /𝑗𝜔

1 + 𝜔 /𝑗𝜔

1 + 𝜔 /𝑗𝜔

1

1 + 𝜔 /𝑗𝜔

𝐴1

1 + 𝑓 /𝑗𝑓

1 + 𝑓 /𝑗𝑓

1 + 𝑓 /𝑗𝑓

1

1 + 𝑓 /𝑗𝑓=

𝐴

2

1 +𝑓

𝑓1 +

𝑓

𝑓1 +

𝑓

𝑓= 2 1 +

𝑓

𝑓→ 1 +

𝑓

𝑓+

𝑓

𝑓+

𝑓

𝑓≈ 2 1 +

𝑓

𝑓

𝜔 =1

𝜏=

1

𝐶 𝑅 + 𝑅

𝜔 =1

𝜏=

𝑔 + 1/𝑅

𝐶

𝜔 =1

𝜏=

1

𝐶 𝑅 + 𝑅

A transmission zero is the value of s at which the input does not reach the output

Coupling capacitor CC1: Vo = 0 as the capacitor is open (zero at s = 0)By-pass capacitor CS: Vo = 0 as ZS is open (zero at s = −1/RSCS)Coupling capacitor CC2: Vo = 0 as the capacitor is open (zero at s = 0)

Selecting values for coupling and by-pass capacitors These capacitors are typically required for discrete amplifier designs It is desirable to minimize the total capacitance for coupling and by-pass purposes CS is first determined to satisfy fP2 fL

CC1 and CC2 are chosen such that poles fP1 and fP3 are 5 to 10 times lower than fL


The method of short-circuit time constantUseful technique to estimate the lower 3-dB frequency for circuits where the capacitors interactTime-constant method is used to evaluate fL without deriving the transfer functionThe method is predicated on the assumption that one of the poles is dominantThe estimation is usually very good even a dominant pole does not exist Set independent source to zero (Vsig = 0) Consider each capacitor one at a time by set all other capacitors ideal (as short circuit) For each capacitor Ci, find the total equivalent resistance Ri seen by Ci

Calculate the 3-dB frequency fL using:

For the CS amplifier example:


𝑓 =1

2𝜋

1

𝐶 𝑅

𝑓 =1

2𝜋

1

𝐶 𝑅

𝐶 = 𝐶 → 𝑅 = 𝑅 + 𝑅

𝐶 = 𝐶 → 𝑅 = 𝑅 ||(1/𝑔 )

𝐶 = 𝐶 → 𝑅 = 𝑅 + 𝑅

=1

2𝜋

1

𝐶 𝑅 + 𝑅+

1

𝐶 𝑅 ||1

𝑔

+1

𝐶 𝑅 + 𝑅


The CE amplifierSmall-signal equivalent circuit:

The capacitors interact (3 capacitors not necessarily for 3 poles) Transfer function can be derived but is rather complex

Method of short-circuit time constants to evaluate fL: Considering only C1 = CC1:Set Vsig = 0 and treat CE and CC2 as shortThe equivalent resistance:

R1 = RC1 = Rsig+RB||r


Considering only C2 = CE:Set Vsig = 0 and treat CC1 and CC2 as shortThe equivalent resistance:

R2 = RCE = re+(Rsig||RB)/(1+) Considering only C3 = CC2:Set Vsig = 0 and treat CC1 and CE as shortThe equivalent resistance:

R3 = RC2 = RC+RL

Estimate the lower 3-dB frequency: fL [1/1+1/2+1/3]/2

= [1/CC1RC1+1/CERCE+1/CC2RC2]/2

Selecting values for the capacitors Typically required for discrete amplifier designs CE is first determined to satisfy needed fL

(choose 1/CERCE = 80% of L)

CC1 and CC2 are chosen for smaller time constant(1/CC1RC2 and 1/CC2RC2 10% of L)

9.2 Internal Capacitive Effects and the High-Frequency Model

The MOSFET deviceThere are basically two types of internal capacitance in the MOSFET Gate capacitance effect: the gate electrode forms a parallel-plate capacitor with gate oxide Junction capacitance effect: the source/body and drain/body are pn-junctions at reverse bias

The gate capacitive effect MOSFET in triode region:

MOSFET in saturation region:

and

MOSFET in cutoff region:and

Overlap capacitance:

The junction capacitance Includes components from the bottom

and from side walls The simplified expression are given by

and


𝐶 = 𝐶 =1

2𝑊𝐿𝐶 + 𝐶

𝐶 =2

3𝑊𝐿𝐶 + 𝐶 𝐶 = 𝐶

𝐶 = 𝑊𝐿 𝐶

𝐶 =𝐶

1 + 𝑉 /𝑉𝐶 =

𝐶

1 + 𝑉 /𝑉

𝐶 = 𝐶 = 𝐶 𝐶 = 𝑊𝐿𝐶

The high-frequency MOSFET model Based on low-frequency small-signal models including gm and ro

All parasitic capacitances (Cgs, Cgd, Csb and Cdb) for MOSET in saturation are includedSimplified high-frequency MOSFET model Source and body terminals are shorted Cdb is also neglected to simplify the analysis Cgs is typically the largest parasitic capacitance and plays

the most important role in amplifier frequency response


simplified model

The unity-gain frequency (fT) The frequency at which the current gain of the MOSFET becomes unity Is typically used as an indicator to evaluate the high-frequency capability Smaller parasitic capacitances Cgs and Cgd are desirable for higher unity-gain frequency

The unity-gain frequency can also be expressed as

The unity-gain frequency is strongly influenced by the channel lengthMinimizing the channel length effectively increases the unity-gain frequency of MOSFETHigher unity-gain frequency can be achieved for a given MOSFET by increasing the bias

current (ID) or the overdrive voltage (VOV)


𝐼 = 𝑠𝐶 𝑉 + 𝑠𝐶 𝑉

𝑠𝐶 𝑉 + 𝐼 = 𝑔 𝑉

→𝐼

𝐼=

𝑔 − 𝑠𝐶

𝑠 𝐶 + 𝐶≈

𝑔

𝑠 𝐶 + 𝐶

→ 𝑓 =1

2𝜋

𝑔

𝐶 + 𝐶

𝑓 =1

2𝜋

𝑔

𝐶 + 𝐶≈

1

2𝜋

𝜇 𝐶𝑊𝐿

𝑉

23

𝑊𝐿𝐶=

3𝜇 𝑉

4𝜋𝐿

The BJT DeviceHigh-frequency small-signal model: Includes all parasitic capacitances in the hybrid- model or T-model Parasitic capacitances:The base-charging or diffusion capacitance (Cde)The base-emitter junction capacitance (Cje)The total base-emitter capacitance (C = Cde+Cje)The collector-base junction capacitance (C) C is typically in the range of a few pF to a few tens of pF C is typically in the range of a fraction of a pF to a few pF


The cutoff (unity-gain) frequency: The frequency at which the current gain of the MOSFET becomes unity Is typically used as an indicator to evaluate the high-frequency capability


𝑉 =𝐼

1/𝑟 + 𝑠𝐶 + 𝑠𝐶

𝑔 𝑉 = 𝐼 + 𝑠𝐶 𝑉

→ ℎ =𝐼

𝐼=

𝑔 − 𝑠𝐶

1/𝑟 + 𝑠 𝐶 + 𝐶

→ 𝑓 =1

2𝜋

𝑔

𝐶 + 𝐶=

1

2𝜋

𝐼

(𝐶 + 𝐶 )𝑉

≈𝑔 𝑟

1 + 𝑠 𝐶 + 𝐶 𝑟=

𝛽

1 + 𝑠/𝜔

9.3 High-Frequency Response of the CS and CE Amplifiers

The common-source amplifierThe coupling and by-pass capacitors act as shirt circuits at high frequenciesThe MOSFET is represented by its high-frequency small-signal model


high-frequency equivalent circuit

simplified small-signal model

𝑉 = 𝑉 𝑅 / 𝑅 + 𝑅

𝑅 = 𝑅 ||𝑅

𝑅 = 𝑅 ||𝑅 𝑅 = 𝑟 ||𝑅 ||𝑅

Analysis of the high-frequency transfer function:

The transfer function is a second-order one with two poles and one zero The poles and zero result in gain roll-off at higher frequencies The high-frequency response can be evaluated by

The bandwidth is typically defined by the 3-dB frequency (fH)


𝑉 − 𝑉

𝑅=

𝑉

𝑅+ 𝑠𝐶 𝑉 + 𝑠C 𝑉 − 𝑉

𝑠C 𝑉 − 𝑉 = 𝑔 𝑉 +𝑉

𝑅

𝑉

𝑉𝑠 = −

𝑔 𝑅 1 − 𝑠 𝐶 /𝑔

1 + 𝑠 𝐶 + 𝐶 1 + 𝑔 𝑅 𝑅 + 𝐶 + 𝐶 𝑅 + 𝑠 𝐶 𝐶 + 𝐶 𝐶 + 𝐶 𝐶 𝑅 𝑅

𝑉

𝑉𝑠 = −

𝑔 𝑅 𝑅

𝑅 + 𝑅

1 − 𝑠 𝐶 /𝑔


= 𝐴1 − 𝑠/𝜔

1 + 𝑠/𝜔 1 + 𝑠/𝜔

𝑉

𝑉𝑗𝜔 = −

𝑔 𝑅 𝑅𝑅 + 𝑅

1 − 𝑗𝜔 𝐶 /𝑔

1 − 𝜔 𝐶 𝐶 + 𝐶 𝐶 + 𝐶 𝐶 𝑅 𝑅 + 𝑗𝜔 𝐶 + 𝐶 1 + 𝑔 𝑅 𝑅 + 𝐶 + 𝐶 𝑅

Determining the upper 3-dB frequency (fH): The 3-dB frequency is defined by the dominant pole (lowest-frequency) of the transfer function If dominant pole does not exist, the 3-dB frequency can be evaluated by the poles and zeros of

the high-frequency response:

(for dominant-pole case)


𝐴 𝑠 = 𝐴 𝐹 (𝑠) = 𝐴1 + 𝑠/𝜔 1 + 𝑠/𝜔

1 + 𝑠/𝜔 1 + 𝑠/𝜔

𝐹 (𝑗𝜔 ) =1

2=

1 + 𝜔 /𝜔 1 + 𝜔 /𝜔

1 + 𝜔 /𝜔 1 + 𝜔 /𝜔≈

1 + 𝜔1

𝜔+

1𝜔

1 + 𝜔1

𝜔+

1𝜔

𝑓 ≈1

2𝜋

1

1𝜔

+1

𝜔− 2

1𝜔

+1

𝜔

𝑓 ≈𝜔

2𝜋


The common-emitter amplifierThe coupling and by-pass capacitors act as shirt circuits at high frequenciesThe BJT is represented by its high-frequency small-signal model

high-frequency equivalent circuit

simplified small-signal model

𝑉 = 𝑉 (𝑅 ||𝑟 )/ 𝑅 ||𝑟 + 𝑅

𝑅 = 𝑅 ||𝑅𝑅 = 𝑅 ||𝑅

𝑅 = 𝑟 ||𝑅 ||𝑅

Analysis of the high-frequency transfer function:

The analysis is very similar to common-source amplifier The transfer function is a second-order one with two poles and one zero Poles and zero result in gain roll-off at higher frequencies High-frequency response can be evaluated by

The bandwidth is typically defined by the 3-dB frequency (fH)


𝑉 − 𝑉

𝑅= 𝑠𝐶 𝑉 + 𝑠C 𝑉 − 𝑉

𝑠C 𝑉 − 𝑉 = 𝑔 𝑉 +𝑉

𝑅

𝑉

𝑉𝑠 = −

𝑔 𝑅 1 − 𝑠 C /𝑔

1 + 𝑠 𝐶 + C 1 + 𝑔 𝑅 𝑅 + C 𝑅 + 𝑠 𝐶 C 𝑅 𝑅

𝑉

𝑉𝑠 = −

𝑔 𝑅 (𝑅 ||𝑟 )

(𝑅 ||𝑟 ) + 𝑅

1 − 𝑠 C /𝑔

1 + 𝑠 𝐶 + C 1 + 𝑔 𝑅 𝑅 + C 𝑅 + 𝑠 𝐶 C 𝑅 𝑅

= 𝐴1 − 𝑠/𝜔

1 + 𝑠/𝜔 1 + 𝑠/𝜔

𝑉

𝑉𝑗𝜔 = −

𝑔 𝑅 (𝑅 ||𝑟 )(𝑅 ||𝑟 ) + 𝑅

1 − 𝑗𝜔 C /𝑔

1 − 𝜔 𝐶 C 𝑅 𝑅 + 𝑗𝜔 𝐶 + C 1 + 𝑔 𝑅 𝑅 + C 𝑅

Miller’s Theorem:A technique to replace the bridging capacitanceThe bridging impedance Z can be equivalently divided into shunt Z1 at node 1 and shunt Z2 at node 2The equivalent input and output impedances are:

and

Simplified analysis of CS amplifier by Miller Theorem:Assume the gain from G to D is nearly constant for frequencies close to the midbandBridging capacitance Cgd is equivalently divided into C1 = (1-K)Cgd and C2 = (1-K-1)Cgd


𝐼 =𝑉 − 𝑉

𝑍=

𝑉 − 𝐾𝑉

𝑍=

𝑉

𝑍/(1 − 𝐾)=

𝑉

𝑍

𝐼 =𝑉 − 𝑉

𝑍=

𝑉 /𝐾 − 𝑉

𝑍=

−𝑉

𝑍/(1 − 𝐾 )=

−𝑉

𝑍

→ 𝑍 = 𝑍/(1 − 𝐾) 𝑍 = 𝑍/ 1 −1

𝐾

𝐾 ≡ 𝑉 /𝑉 = −𝑔 𝑅

𝐶 𝐶

Transfer function of the CS amplifier by using Miller’s theorem:

The transfer function has two poles:

The assumption that gain from G to D is constant no longer valid as frequency increases The midband gain remains the same The transfer function can not be used to predict high-frequency response very accurately It can be used to simplify the analysis for the 3-dB frequency estimation if a dominant pole exists The equivalent response of the CS amplifier is typically approximated by a STC circuit:


𝑉

𝑉𝑠 = −

𝑔 𝑅

1 + 𝑠𝑅 𝐶 + 𝐶 1 + 𝑠𝑅 𝐶 + 𝐶

𝑉

𝑉𝑠 = −

𝑔 𝑅 𝑅

𝑅 + 𝑅

1

1 + 𝑠𝑅 𝐶 + 𝐶 1 + 𝑔 𝑅 1 + 𝑠𝑅 𝐶 + 𝐶 1 + 1/𝑔 𝑅

𝜔 =1

𝑅 𝐶 + 𝐶 1 + 𝑔 𝑅

𝜔 =1

𝑅 𝐶 + 𝐶 1 +1

𝑔 𝑅

≈1

𝑅 𝐶 + 𝐶

𝐴 𝑠 ≈ −𝑔 𝑅 𝑅

𝑅 + 𝑅

1

1 + 𝑠𝑅 𝐶 + 𝐶 1 + 𝑔 𝑅=

𝐴

1 + 𝑠/𝜔


Simplified analysis of CE amplifier by Miller Theorem:Assume the gain from B to C is nearly constant for frequencies close to the midbandBridging capacitance C is equivalently divided into C1 = Ceq and C2 (negligible)

The midband gain is defined as

Ceq is considered Miller capacitance multiplying C by a factor of the midband gainCin is effectively the largest capacitance in the CE amplifier due to Miller Effect It is not necessary to derive the poles and zeros as the high-frequency response deviates The 3-dB frequency can be estimated by the dominant pole as

𝑓 =1

2𝜋

1

𝑅 𝐶 + 𝐶 1 + 𝑔 𝑅

𝐴 = −𝑔 𝑅 𝑅 ||𝑟

𝑅 + 𝑅 ||𝑟

9.4 Useful Tools for the Analysis of the High-Frequency Response of Amplifiers

Open-circuit time constant method to evaluate amplifier bandwidthGeneral transfer function of the amplifier

Open-circuit time constant: Remove all independent sources Consider the capacitors one at a time and the others are considered idea (open-circuit) Find the equivalent shunt resistance for each capacitor The coefficient can be obtained as

Dominant pole approximation:

For an amplifier with 2nd order transfer function and a dominant pole:

and


𝑏 = 𝐶 𝑅 + 𝐶 𝑅 + ⋯ + 𝐶 𝑅

𝑏 =1

𝜔+

1

𝜔+ ⋯ +

1

𝜔

𝜔 ≈ 𝜔 ≈1

𝑏=

1

𝐶 𝑅 + 𝐶 𝑅 + ⋯ + 𝐶 𝑅

𝐴 𝑠 = 𝐴1 +

𝑠𝜔

1 +𝑠

𝜔

1 +𝑠

𝜔1 +

𝑠𝜔

= 𝐴

1 + 𝑎 𝑠 + 𝑎 𝑠

1 + 𝑎 𝑠 + 𝑎 𝑠

𝑏 =1

𝜔+

1

𝜔≈

1

𝜔→ 𝜔 ≈

1

𝑏𝑏 =

1

𝜔 𝜔≈

𝑏

𝜔→ 𝜔 ≈

𝑏

𝑏

𝐴 𝑠 = 𝐴1 +

𝑠𝜔

1 +𝑠

𝜔. . 1 +

𝑠𝜔

1 +𝑠

𝜔1 +

𝑠𝜔

. . 1 +𝑠

𝜔

= 𝐴1 + 𝑎 𝑠 + 𝑎 𝑠 + ⋯ + 𝑎 𝑠

1 + +𝑎 𝑠 + 𝑎 𝑠 + ⋯ + 𝑎 𝑠

Analysis of CS amplifier by using open-circuit time constants

Open-circuit time constants:


𝑅 = 𝑅 𝑅 = 𝑅 1 + 𝑔 𝑅 +𝑅 𝑅 = 𝑅

𝜔 ≈1

𝑏=

1

𝐶 𝑅 + 𝐶 𝑔 𝑅 𝑅 + 𝑅 + 𝑅 + 𝐶 𝑅

𝑏 = 𝐶 𝑅 + 𝐶 𝑅 + 𝐶 𝑅 = 𝐶 𝑅 + 𝐶 𝑔 𝑅 𝑅 + 𝑅 + 𝑅 + 𝐶 𝑅

Analysis by using Miller’s Theorem and open-circuit time constantsBridging capacitance Cgd is replaced by C1 and C2

Transfer function and H:

Open-circuit time constant analysis based on Miller’s equivalent circuit:


𝐾 ≡ ≈-𝑔 𝑅

𝐶 = 𝐶 1 − 𝐾 = 𝐶 1 + 𝑔 𝑅

𝐶 = 𝐶 1 − 𝐾 = 𝐶 1 + 1/𝑔 𝑅 ≈ 𝐶

𝑉

𝑉= −𝑔 𝑅

𝑅

𝑅 + 𝑅

1

1 + 𝑠/𝜔 1 + 𝑠/𝜔

𝜔 =1

𝐶 𝑅=

1

𝐶 + 𝐶 1 + 𝑔 𝑅 𝑅≈ 𝜔

𝜔 =1

𝐶 𝑅=

1

𝐶 + 𝐶 𝑅

𝑅 = 𝑅

𝑅 = 𝑅

𝜔 ≈1

𝑏=

1

𝐶 + 𝐶 1 + 𝑔 𝑅 𝑅 + 𝐶 + 𝐶 𝑅

𝑏 = 𝐶 𝑅 + 𝐶 𝑅 = 𝐶 + 𝐶 1 + 𝑔 𝑅 𝑅 + 𝐶 + 𝐶 𝑅

Analysis of CE amplifier by using open-circuit time constants

Open-circuit time constants:


𝜔 ≈1

𝑏=

1

𝐶 𝑅 + 𝐶 𝑔 𝑅 𝑅 + 𝑅 + 𝑅 + 𝐶 𝑅

𝑏 = 𝐶 𝑅 + 𝐶 𝑅 + 𝐶 𝑅 = 𝐶 𝑅 + 𝐶 𝑔 𝑅 𝑅 + 𝑅 + 𝑅 + 𝐶 𝑅

𝑅 = 𝑅 𝑅 = 𝑅 1 + 𝑔 𝑅 +𝑅 𝑅 = 𝑅

9.5 A Closer Look at the High-Frequency Response

Analysis by transfer functionTransfer function of the amplifier:

Poles and zero for dominant-pole approximation:

The 3-dB frequency is given by H P1 (is typically limited by Rsig and Miller effect)


𝐴 = −𝑔 𝑅 𝑅

𝑅 + 𝑅

𝜔 =𝑔

𝐶

𝜔 ≈1

𝑏=

1

𝐶 + 𝐶 1 + 𝑔 𝑅 𝑅 + 𝐶 + 𝐶 𝑅

𝜔 ≈𝑏

𝑏=

𝐶 + 𝐶 1 + 𝑔 𝑅 𝑅 + 𝐶 + 𝐶 𝑅

𝐶 𝐶 + 𝐶 𝐶 + 𝐶 𝐶 𝑅 𝑅

𝑉 = 𝑉 𝑅 / 𝑅 + 𝑅

𝑅 = 𝑅 ||𝑅

𝑅 = 𝑟 ||𝑅 ||𝑅

𝑉

𝑉𝑠 = −

𝑔 𝑅 𝑅

𝑅 + 𝑅

1 − 𝑠 𝐶 /𝑔


= 𝐴1 − 𝑠/𝜔

1 + 𝑠/𝜔 1 + 𝑠/𝜔

𝑉

𝑉𝑠 = −

𝑔 𝑅 1 − 𝑠 𝐶 /𝑔


CS amplifier with a small source resistanceFor the case where source resistance is zeroThe transfer function of the amplifier:

Midband gain:

The transfer function has one pole and one zero

Midband gain and zero are virtually unchangedThe lowest-frequency pole no longer exists

Gain-bandwidth product: Gain rolls off beyond fH (20 dB/decade) The gain becomes 0 dB at ft:


𝐴 = −𝑔 𝑅

𝜔 =1

𝐶 + 𝐶 𝑅

𝜔 =𝑔

𝐶

𝑓 = 𝐴 𝑓 =𝑔

2𝜋 𝐶 + 𝐶

𝑉 = 𝑉

𝑅 =0

𝑅 = 𝑟 ||𝑅 ||𝑅

𝑉

𝑉𝑠 = − 𝑔 𝑅

1 − 𝑠 𝐶 /𝑔

1 + 𝑠 𝐶 + 𝐶 𝑅

9.6 High-Frequency Response of the CG and Cascode Amplifiers

High-frequency response of the CG amplifierHigh-frequency circuit model for CS amplifier

The frequency response by neglecting ro:

The upper 3-dB frequency is typically defined by H P2 P1

The poles of CG amplifier is usually much higher than the dominant pole of the CS amplifier


𝜔 =1

𝐶 𝑅 ||1

𝑔

𝜔 =1

𝐶 + 𝐶 𝑅

𝑉

𝑉𝑠 =

𝑔 𝑅

1 + 𝑔 𝑅

1

1 + 𝑠/𝜔 1 + 𝑠/𝜔

Time constant method to evaluate the 3-dB frequency of CG amplifier by neglecting ro:

The frequency response as ro included (ro has to be taken into account for IC amplifiers) The transfer function derivation is rather complex if ro is not negligible Time constant method can be used to evaluate the 3-dB frequency


𝑏 = 𝜏 + 𝜏 = 𝐶 𝑅 ||1

𝑔+ 𝐶 + 𝐶 𝑅

𝜔 ≈1

𝑏=

1

𝐶 𝑅 ||1

𝑔+ 𝐶 + 𝐶 𝑅

𝜏 = 𝐶 𝑅 = 𝐶 𝑅 ||𝑟 + 𝑅

1 + 𝑔 𝑟𝜏 = 𝐶 + 𝐶 𝑅 = 𝐶 + 𝐶 𝑅 || 𝑔 𝑟 𝑅 + 𝑟 + 𝑅

High-frequency response of the cascode amplifierOpen-circuit time constant method: Capacitance Cgs1 sees a resistance Rsig

Capacitance Cgd1 sees a resistance Rgd1

Capacitance (Cdb1+Cgs2) sees a resistance Rd1

Capacitance (Cgd2+CL) sees a resistance Ro||RL

Effective time constant

The upper 3-dB frequency:


𝑅 = 𝑅 1 + 𝑔 𝑅 + 𝑅

𝑅 = 𝑟 ||𝑅 = 𝑟 ||𝑟 + 𝑅

1 + 𝑔 𝑟

𝑅 ||𝑅 = 𝑔 𝑟 𝑟 + 𝑟 + 𝑟 ||𝑅

𝑏 = 𝐶 𝑅 + 𝐶 𝑅 + 𝐶 + 𝐶 𝑅 + 𝐶 + 𝐶 𝑅 ||𝑅

𝜔 ≈1

𝑏=

1

𝐶 𝑅 + 𝐶 𝑅 + 𝐶 + 𝐶 𝑅 + 𝐶 + 𝐶 𝑅 ||𝑅

The effective time constant can be expressed as:

The 1st term arises at input node; the 2nd term at the middle node; the 3rd term at output node In the case of a large Rsig:The first term dominates if the Miller multiplier is large (typically with large Rd1 and RL)A small RL (to the order of ro) is needed for extended bandwidthThe midband gain drops as the value of RL decreasesA trade-off exists between gain and bandwidth In the case of a small Rsig:The 1st term becomes negligible A large RL (to the order of A0ro) can be used to boost the amplifier gainThe 3rd term dominates In the case of zero Rsig:

Choose Ro||RL larger than RL, which is defined in CS amplifier, by a factor of Ao

The fH of the cascade will be lower than that of the CS amplifier by the same factor Ao

The unity-gain frequency is unchanged at


𝜏 = 𝑅 𝐶 + 𝐶 1 + 𝑔 𝑅 + 𝑅 𝐶 + 𝐶 + 𝐶 + 𝑅 ||𝑅 𝐶 + 𝐶

𝑓 ≈1

2𝜋 𝑅 ||𝑅 𝐶 + 𝐶

𝑓 ≈𝑔

2𝜋 𝐶 + 𝐶

Summary of CS and cascode amplifiers


High-frequency response of the cascode amplifierOpen-circuit time constant method: Capacitance C1 sees a resistance R1

Capacitance C1 sees a resistance R 1

Capacitance (Ccs1+C2) sees a resistance Rc1

Capacitance (C 2+CL) sees a resistance Ro||RL

Effective time constant

High-frequency response:


𝑅 = 𝑅 ||𝑟 1 + 𝑔 𝑅 + 𝑅

𝑅 = 𝑟 ||𝑅 = 𝑟 || 𝑟𝑟 + 𝑅

𝑟 + 𝑅 /(𝛽 + 1)

𝑅 ||𝑅 ≈ 𝛽 𝑟 ||𝑅

𝑏 = 𝐶 𝑅 + 𝐶 𝑅 + 𝐶 + 𝐶 𝑅 + 𝐶 + 𝐶 𝑅 ||𝑅

𝐴 = −𝑟

𝑟 + 𝑅 𝑔 𝛽𝑟 ||𝑅

𝑅 = 𝑅 ||𝑟

𝜔 ≈1

𝑏=

1

𝐶 𝑅 + 𝐶 𝑅 + 𝐶 + 𝐶 𝑅 + 𝐶 + 𝐶 𝑅 ||𝑅

9.7 High-Frequency Response of the Source and Emitter Followers

The source follower:

High-frequency transfer function:


𝑉

𝑉(𝑠) = 𝐴

1 + 𝑠/𝜔

1 + 𝑏 𝑠 + 𝑏 𝑠

𝐴 =𝑅

𝑅 + 1/𝑔=

𝑔 𝑅

1 + 𝑔 𝑅≈ 1

𝜔 =𝑔

𝐶

𝑏 = 𝐶 +𝐶

1 + 𝑔 𝑅𝑅 +

𝐶 + 𝐶

1 + 𝑔 𝑅𝑅

𝑏 =𝐶 𝐶 + 𝐶 𝐶 + 𝐶 𝐶

1 + 𝑔 𝑅𝑅 𝑅

Frequency response with dominant pole approximation:

Frequency response for non-dominant pole cases (real poles):

Frequency response for complex pole cases (Q > 0.5):


1 + 𝑏 𝑠 + 𝑏 𝑠 = 1 + 𝑠/𝜔 1 + 𝑠/𝜔

𝜔 ≈ 𝜔 ≈1

𝑏= 𝐶 +

𝐶

1 + 𝑔 𝑅𝑅 +

𝐶 + 𝐶

1 + 𝑔 𝑅𝑅

𝜔 ≈𝑏

𝑏= 𝐶 +

𝐶

1 + 𝑔 𝑅𝑅 +

𝐶 + 𝐶

1 + 𝑔 𝑅𝑅 /

𝐶 𝐶 + 𝐶 𝐶 + 𝐶 𝐶

1 + 𝑔 𝑅𝑅 𝑅

𝑓 ≈1

2𝜋

1

1𝜔

+1

𝜔− 2

1𝜔

1 + 𝑏 𝑠 + 𝑏 𝑠 = 1 +1

𝑄

𝑠

𝜔+

𝑠

𝜔

𝜔 =1

𝑏=

𝑔 𝑅 + 1

𝑅 𝑅 𝐶 𝐶 + 𝐶 𝐶 + 𝐶 𝐶

𝑄 =𝑏

𝑏=

𝑔 𝑅 + 1 𝑅 𝑅 𝐶 𝐶 + 𝐶 𝐶 + 𝐶 𝐶

𝐶 + 𝐶 𝑔 𝑅 + 1 𝑅 + 𝐶 + 𝐶 𝑅

The emitter follower:

High-frequency transfer function:


𝑉

𝑉(𝑠) = 𝐴

1 + 𝑠/𝜔

1 + 𝑏 𝑠 + 𝑏 𝑠

𝐴 =𝑅

𝑅 + 𝑟 + 𝑅 / 𝛽 + 1≈ 1

𝜔 =1

𝐶 𝑟

𝑏 =𝐶 + 𝐶 1 + 𝑅 /𝑟 𝑅 + 𝐶 + 𝐶 1 + 𝑅 /𝑟 𝑅

1 + 𝑅 /𝑟 + 𝑅 /𝑟

𝑏 =𝐶 𝐶 + 𝐶 𝐶 + 𝐶 𝐶 𝑅 𝑅

1 + 𝑅 /𝑟 + 𝑅 /𝑟

9.8 High-Frequency Response of Differential Amplifiers

Resistively Loaded MOS Differential Amplifier:Differential-mode operation: Use differential half-circuit for analysis Identical to the case for common-source amplifier Can be approximated by a dominant-pole system

Common-mode operation: Use common-mode half-circuit for analysis CSS is usually significant and can not be neglected CSS results in a zero at lower frequency Other capacitance high-frequency poles and zeros


𝐴 = −𝑅

2𝑅

∆𝑅

𝑅

𝐴 𝑠 = −𝑅

2𝑍

∆𝑅

𝑅= −

𝑅

2

∆𝑅

𝑅

1

𝑅+ 𝑠𝐶

= −𝑅

2𝑅

∆𝑅

𝑅1 + 𝑠𝐶 𝑅 = 𝐴 1 +

𝑠

𝜔

𝜔 =1

𝐶 𝑅

Common-mode rejection ratio (CMRR): The frequency dependence of CMRR can be evaluated CMRR decreases at higher frequencies due to the pole of Ad and the zero of Acm


Active-Loaded MOS Differential Amplifier:Differential-mode operation: Equivalent capacitances:

Transconductance (Gm):

The pole and zero of Gm(s) are at very high frequenciesMinor pole and zero near unity-gain frequency


𝐶 = 𝐶 +𝐶 +𝐶 +𝐶 +𝐶

𝐶 = 𝐶 +𝐶 +𝐶 +𝐶 +𝐶

𝑉 = −𝑔 𝑉 /2

𝑔 + 𝑠𝐶

𝐼 = −𝑔 𝑉 =𝑔 𝑔 𝑉 /2

𝑔 + 𝑠𝐶=

𝑔 𝑉 /2

1 + 𝑠𝐶 /𝑔

𝐼 = 𝐼 + 𝐼 =𝑔 𝑉 /2

1 + 𝑠𝐶 /𝑔+ 𝑔 𝑉 /2

𝐺 ≡𝐼

𝑉= 𝑔

1 + 𝑠𝐶 /(2𝑔 )

1 + 𝑠𝐶 /𝑔

𝑓 =𝑔

2𝜋𝐶≈

𝑓

2

𝑓 =2𝑔

2𝜋𝐶≈ 𝑓

The transfer function of the amplifier:

The dominant pole is typically

Common-mode operation: By taking CSS into account for the mid-band common-mode gain:

Contributes to a zero at lower frequencyCommon-mode rejection ratio (CMRR): CMRR decreases at higher frequencies due to the dominant pole of Ad(s) and the zero of Acm(s)


𝑓 =1

2𝜋𝐶 𝑅=

1

2𝜋𝐶 𝑟 ||𝑟

𝑉 = 𝐺 𝑉𝑅

1 + 𝑠𝐶 𝑅= 𝐺 𝑉

𝑟 ||𝑟

1 + 𝑠𝐶 𝑟 ||𝑟

𝑉

𝑉= 𝑔 𝑅

1 + 𝑠𝐶 /(2𝑔 )

1 + 𝑠𝐶 /𝑔

1

1 + 𝑠𝐶 𝑅= 𝑔 𝑟 ||𝑟

1 + 𝑠𝐶 /(2𝑔 )

1 + 𝑠𝐶 /𝑔

1

1 + 𝑠𝐶 𝑟 ||𝑟

𝐴 ≈ −1

2𝑔𝑅

1 + 𝑠𝐶 𝑅

= −1 + 𝑠𝐶 𝑅

2𝑔 𝑅

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