Lecture 17:

Frequency Response of Amplifiers

Gu-Yeon Wei Division of Engineering and Applied Sciences Harvard University [email protected]

Wei 1 Overview

• Reading – S&S: Chapter 7 • Skim sections since mostly described using BJT circuits. Lecture notes focus on MOS circuits. • Supplemental Reading – Razavi, Design of Analog CMOS Integrated Circuits: Chapter 6 • Background – So far, our treatment of small-signal analysis of amplifiers has been for low frequencies where internal capacitances do not affect operation. However, we did see that internal capacitances do exist and we derived the fT of transistors. Moreover, we spent some time looking at amplifiers modeled with a single pole. Now, we will see how these capacitances affect the frequency response of amplifiers. To fully understand and model the frequency response of amplifiers, we utilize Bode plots again. We will use a technique called open-circuit time constants (OCTs) to approximate frequency response calculations in the presence of several capacitors and and Miller’s theorem to deal with bridging capacitors.

Wei ES154 - Lecture 17 2 Amplifier Transfer Function

|A| (dB) |A| (dB)

A A 0 -3dB M -3dB

ωH ω ωL ωH ω • Voltage-gain frequency response of amplifiers seen so far take one of two forms – Direct-Coupled (DC) amplifiers exhibit low-pass characteristics – flat gain from DC to

ωH – Capacitively coupled amplifiers exhibit band-pass characteristics – attenuation at low frequency due to impedance from coupling capacitances increasing for low frequencies

• We will focus on the high-frequency portion of the response (ωH) – Gain drops due to effects of internal capacitances of the device • Bandwidth is the frequency range over which gain is flat

–BW= ωH or ωH-ωL ≈ωH (ωH >> ωL) • Gain-Bandwidth Product (GB) – Amplifier figure of merit

–GB ≡ AMωH

where AM is the midband gain – We will see later that it is possible to trade off gain for bandwidth

Wei ES154 - Lecture 17 3 Gain Function A(s)

• We can represent the frequency dependence of gain with the following expression: A(s)= AM FL (s)FH (s)

– Where FL(s) and FH(s) are the functions that account for the frequency dependence of gain on frequency at the lower and upper frequency ranges

– We can solve for AM by assuming that large coupling capacitors are short circuits and internal device capacitances are open circuits (what we have done so far for low-frequency small-signal analysis)

Wei ES154 - Lecture 17 4 High-Frequency Response

• We can express function FH(s) with theω general form: ω (1+ s ωZ1 )(1+ s ωZ 2 )⋅⋅⋅()1+ s ωZnH FH ()s = ()1+ s P1 ()1+ s P2 ⋅⋅⋅()1+ s ωPnH

• Where ωP and ωZ represent the frequencies of high-frequency poles and zeros • The zeros are usually at infinity or sufficiently high frequency such that the numerator Æ 1 and assuming there is one dominant pole (other poles at much higher frequencies), we can approximate the function as… 1 F ()s = H ωH ≅ ωP1 ()1+ s ωP1

– This simplifies the determination of the BW or ωH ω • If a dominant pole does not exist, the upper 3-dB frequency ωH can be found from a plot of |FH(jω)|. Alternatively, we can approximate with following formula (see S&S p593 for derivation). ω ω 1 1 2 2 H = 1 2 + 2 + ⋅⋅⋅− ω 2 − 2 ⋅⋅⋅ P1 P2 Z1 ωZ 2

–Note: if ωP1 is a dominant pole, then reduces to ωH=ωP1

Wei ES154 - Lecture 17 5 Open-Circuit Time Constant Method

• It may be difficult to find the poles and zeros of the system (which is usually the

case). We can find approximate values of ωH using the following method.

– We can multiply out factors and represent FH(s) in an alternative form: 2 nH 1+ a1s + a2s + ⋅⋅⋅+ anH s FH ()s = 2 nH 1+ b1s + b2s + ⋅⋅⋅+ bnH s – Where a and b are coefficientsω related to the zero and pole frequencies – We can show that 1 ω1 1 b1 = + + ⋅⋅⋅+ P1 P2 ωPnH

and b1 can be obtained by considering the various capacitances in the high- frequency equivalent circuit one at a time while reducing all other capacitors to zero (or open circuits); and calculating and summing the RC time constant due to the circuit associated with each capacitor. – This is called the open-circuit time constant method (OCT)

Wei ES154 - Lecture 17 6 Calculating OCTs

The approach: • For each capacitor: – set input signal to zero – replace all other capacitors with open circuits

– find the effective resistance (Rio) seen by the capacitor Ci • Sum the individual time constants (RCs or also called the open-circuit time constants) nH b1 = ∑Ci Rio i=1

• This method for determining b1 is exact. The approximation comes from using this result to determine ωH. 1 ωH ≅ nH ∑Ci Rio i=1 – This equation yields good results even if there is no single dominant pole but when all poles are real • We will see an example of this method when we analyze the high-frequency response of different amplifier topologies

Wei ES154 - Lecture 17 7 Miller’s Theorem

• Before we begin analyzing the high-frequency response of amplifiers, there is an important phenomenon that we should first investigate called “Miller Effect” Consider the circuit network below on the right with two nodes, 1 and 2. An admittance Y (Y=1/Z) is connected between the two nodes and these nodes are also connected to other nodes in the network. Miller’s theorem provides a way for replacing the “bridging” admittance Y with two admittances Y1 and Y2 between node 1 and gnd, and node 2 and gnd. Y 12I1 I2 12I1 I2

V1 V2 V1 Y1 Y2 V2

– The relationship between V2 and V1 is given by K=V2/V1

– To find Y1 and Y2

I1 = Y ()V1 −V2 = YV1()1−V2 V1 I 2 = Y (V2 −V1 )= YV2 (1−V1 V2 ) Caveat: () I1 = YV1 1− K I 2 = YV2 ()1−1 K The Miller equivalent circuit is valid I = YV I = Y V only as long as the conditions that 1()1 1 2 2 2 existed in the network when K was Y1 = Y 1− K Y2 = Y ()1−1 K determined are not changed.

Wei ES154 - Lecture 17 8 High-Frequency Response of CS Amp

• Take the following circuit and investigate its high-frequency response – First, redraw using a high-frequency small-signal model for the nMOS

• There are two ways to find the upper 3-dB frequency ωH – Use open-circuit time constant method – Use Miller’s theorem

– Brute force calculations to find vout/vin • Let’s investigate them all

Wei ES154 - Lecture 17 9 Using OCT on CS Amplifier

• Find the RC time constants associated with Cgd and Cgs in the following circuit

Cgd Rs vo g v m gs AM = = −gm ()RL || ro v v C R ||r v i gs gs L o o vi

•Replace Cgd with an open-ckt and find the resistance seen by Cgs

Rs vtst Rgs = = Rs g v m gs itst R ||r v vtst Itst L o o τ gs = RgsCgs = RsCgs

•Replace Cgs with an open-ckt and find the resistance seen by Cgd i vgs + vtst R tst s itst = −vgs Rs itst = gmvgs + RL || ro vtst gmvgs R ||r v vgs L o o Rgd = vtst itst = (RL || ro )+ gm Rs (RL || ro )+ Rs

τ gd = Rgd Cgd = [(RL || ro )+ gm Rs (RL || ro )+ Rs ]Cgd

Wei ES154 - Lecture 17 10 Using OCTs Cont’d

• Summing to two time constants yields ω ω H 1 H ≅ τ gs +τ gd 1 ωH ≅ RsCgs + []()RL || ro + gm Rs RL || ()ro + Rs Cgd

– From the above equation, it is not difficult to imagine that Cgd has a more significant effect on reducing BW – The resulting frequency dependence of gain is… A A()s = M s ωH +1

• Let’s compare this result with what we get using Miller’s theorem

Wei ES154 - Lecture 17 11 Using Miller’s Theorem on CS Amplifier

• Redraw the high-frequency small-signal model using Miller’s theorem

Rs

Cgd(1+gmRL') gmvgs

vi vgs Cgs RL' vo

Cgd[1+1/(gmRL')] ~= Cgd

CT

– Assuming a dominant pole introduced by Cgd in parallel with Cgs 1 1 ωH ≅ = []Cgs + Cgd ()1+ gm RL ' Rs CT Rs

– Miller multiplication of Cgd results in a large input capacitance

• Notice that this approximation for ωH is close to the approximation found using OCT assuming that RsCgd(1+gmRL’) dominates

• Let’s verify our assumptions by deriving the exact high-frequency transfer function of the CS amplifier

Wei ES154 - Lecture 17 12 High-Frequency Response of CS Amplifier

• Replace the input source and series resistance with a Norton equivalent

sCgd(vgs-vo) Cgd

v ()s vgs i vi/Rs g v = + sCgsvgs + sCgd ()vgs − vo m gs Rs Rs Rs vgs Cgs RL' vo () vo ()s sCgd vgs − vo = gmvgs + RL ' s 1− v ()s g C ⇒ o = −A m gd () 0 2 vi s 1+ s[]RsCgs + RsCgd ()1+ gm RL ' + Cgd RL ' + s CgsCgd Rs RL '

– The exact solution gives a zero (at a high frequency) and two poles – Notice that the s term is the same as the solution using the OCT method

• Unfortunately, the denominator is too complicated to extract any useful info… So, assuming the two poles are widely separated (greater than an order of magnitude), we can rewrite the expression for the denominator as…

Wei ES154 - Lecture 17 13 HF Response of CS Amplifier

• Rewrite theω denominator as: ω 2 D()s = ()1+ sω P1 ()1+ s P2 =1+ s(1 P1 +1 P2 )+ s P1 P2 D()s ≅ 1+ s + s2 ω ω ω P1 P1 P2 ω – And from the solution on the previous slide we can write… ω ω ω 1 P1 = RsCgs + RsCgd ()1+ gm RL ' + RsCgd RL ' Rs

Cgs + Cgd ()1+ gm RL ' + Cgd RL ' Rs gm ωP2 = ≅ CgsCgd RL ' Cgs

• So the second pole is usually at a much higher frequency and we can assume a dominant pole • Using either Miller’s theorem or OCTs enables a way to quickly find approximations of the amplifier’s high-frequency response

Wei ES154 - Lecture 17 14 Frequency Response of CG Amplifier

• One way to avoid the frequency limitations of Miller multiplication of Cgd is to utilize a CG amplifier configuration

Cgd

-(gm+gmb)vx -(gm+gmb)vx Cgs Cdb RD RD C =C +C Rs Rs D gd db

vs vx Csb vs vx CS=Cgs+Csb

• Using OCT method, we find two time constants – At the input (source node) CS τ S = 1 Rs + gm + gmb

– At the output (drain node) τ D = CD RD

• The output usually drives additional load capacitance such that the output pole is dominant • The frequency response of CG amplifiers is when combined with a CS stage to build a cascode circuit

Wei ES154 - Lecture 17 15 Cascode Stage

• Cascoding enables high bandwidth by suppressing Miller multiplication of Cgd. Let’s investigate how with the following high-frequency model of a cascode stage.

RL

Cgd2 v OUT v C +C -g +g v out db2 L m2 mb2 x C +C db2 L RL Vb

C C gs2 gd1 Rs Cgd1

Rs C +C gmvgs1 db1 sb2 Cdb1+Csb2+Cgs2 vin v Cgs1 x C vIN gs1

– Use OCT method to find the time constants associated with each capacitor.

The time constant associated with Cgd1 is… 1+ g R + ()g + g R  g  m1 s m2 mb2 s  m1  NOTICE: τ gd1 = Cgd1 ≅ 1+ RsCgd1 gm2 + gmb2  gm2 + gmb2  Miller multiplication is ~2

Wei ES154 - Lecture 17 16 Frequency Response of Source Followers

• Start with a high-frequency small-signal model of the source follower circuit Rs Rs

v vin gs gmvgs vOUT Cgd Cgs

vIN CL vout

CL

– Directly solving for vout/vin yields: sC v sC + g v = v sC → v = L v gs gs m gs out L gs out gm gm + sCgs ωP1 ≅ gm RsCgd + CL + Cgs vin = Rs []vin sCgs + ()vin + vgs sCgd + vgs + vout

vout gm + sCgs ()s = 2 vin Rs ()()CgsCL + CgsCgd + Cgd CL s + gm RsCgd + CL + Cgs s + gm

• The zero is due to Cgs that directly couples the signal from the input to the output – If poles are far apart, then the s term represents the dominant pole

Wei ES154 - Lecture 17 17 More on Source Follower

• Other important aspects of a source follower are its input and output impedances (since they are often used as buffers) v gs gmvgs • Let’s calculate the input impedance using the high-freq small-signal Zin Cgs models v 1  g  1 out Z = + 1+ m  in   sCgs  sCgs  gmb + sCL 1/gmb CL

• Now calculate the output impedance (ignoring gmb for simplicity)

Rs sCgs +1 Rs Zout = gm + sCgs

– At low frequency, Zout ≈ 1/gm v gs gmvgs Cgs – At high frequency, Zout ≈ Rs

– Shape of the response depends on the relative size of Rs and 1/gm vout

|Zout| |Zout| Z 1/gm Rs out

Zout can look inductive

Rs 1/gm or capacitive depending ωωon Rs and 1/gm

Wei ES154 - Lecture 17 18 Differential Pair

• We have seen that a symmetric differential amplifier can be analyzed with a differential half circuit. This still holds true for high-frequency small-signal analysis.

RD RD v out C Rs Rs gd +vd/2 gmvgs Rs vd/2 Cgs RD vout -vd/2 Cdb I

– The response is identical to that of a common-source stage

Wei ES154 - Lecture 17 19 High-Frequency CMRR

• The CMRR of a differential pair degrades at high frequency due to a number of factors. The most important is the increase in CM gain with R frequency due to capacitance on the tail node. D

• Use the common-mode equivalent half circuit to understand how CM vout,CM

gain increases with frequency vin,CM

– Draw the small-signal equivalent model and see the effect of CTAIL C on the vout/vin transfer function 2ro TAIL I/2

vout = gmvgs RD and vgs = vin-vx and vx = gmvgs (2ro||2 sCTAIL )

vin vout gm RD vgs = ⇒ = 1+ gm ()2ro||2 sCTAIL vin 1+ gm ()2ro||2 sCTAIL v v vout gm RD 1+ sr()oCTAIL in,CM out,CM ()s = gmvgs vin 1+ 2gmro + sroCTAIL vgs RD

– Zero at ωZ = 1/roCTAIL (since ro is big, ωZ occurs at a low frequency) C 2ro TAIL – There are additional poles at higher frequencies due to CTAIL and other internal capacitances (that we have ignored) • The zero causes the CM gain to increase with frequency until the higher frequency poles kick in Æ CMRR degrades due to the zero

Wei ES154 - Lecture 17 20 HF CMRR plots

• The impact of the zero in the CM gain on CMRR can |A | (dB) be illustrated as shown cm

– Remember CMRR = Ad/Acm

ωZ ω (log scale)

|Ad| (dB) • There is a trade off between CMRR and voltage headroom

– Wider current source devices enable lower vds ωP ω (log scale) – Wider current source device means larger CTAIL

CMRR (dB) -20dB/dec

-40dB/dec

ωZ ωP ω (log scale)

Wei ES154 - Lecture 17 21 Next Time

• Reading – S&S Chapter 8 • Supplemental Reading – Razavi: Chapter 8

• What to look forward to… – Negative feedback for amplifiers was invented in 1927 by Harold Black to stabilize the gain and correct the distortion of amplifiers used in long- distance telephone networks. Negative feedback (as well as positive feedback) is widely used in analog circuits today. In fact, we used negative feedback when we constructed op amps with gain set using resistors. Throughout the next lecture, we will investigate the general theory of feedback and look at four basic feedback topologies. We will also learn how to understand and analyze the stability of amplifiers.

Wei ES154 - Lecture 17 22