Lecture 17: Frequency Response Of Amplifiers

Lecture 17:Frequency Response of AmplifiersGu-Yeon WeiDivision of Engineering and Applied SciencesHarvard Universityguyeon@eecs.harvard.eduWei1

Overview Reading– S&S: Chapter 7 Skim sections since mostly described using BJT circuits. Lecture notes focus onMOS circuits. WeiSupplemental Reading– Razavi, Design of Analog CMOS Integrated Circuits: Chapter 6Background– So far, our treatment of small-signal analysis of amplifiers has been for lowfrequencies where internal capacitances do not affect operation. However,we did see that internal capacitances do exist and we derived the fT oftransistors. Moreover, we spent some time looking at amplifiers modeledwith a single pole. Now, we will see how these capacitances affect thefrequency response of amplifiers.To fully understand and model the frequency response of amplifiers, weutilize Bode plots again. We will use a technique called open-circuit timeconstants (OCTs) to approximate frequency response calculations in thepresence of several capacitors and and Miller’s theorem to deal withbridging capacitors.ES154 - Lecture 172

Amplifier Transfer Function A (dB) A (dB)A0ωH Weiω-3dBωLωHωVoltage-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– AM-3dBωHCapacitively coupled amplifiers exhibit band-pass characteristics – attenuation at lowfrequency due to impedance from coupling capacitances increasing for low frequenciesWe will focus on the high-frequency portion of the response (ωH)– Gain drops due to effects of internal capacitances of the deviceBandwidth 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 AMωHwhere AM is the midband gain– We will see later that it is possible to trade off gain for bandwidthES154 - Lecture 173

Gain Function A(s) We can represent the frequency dependence of gain with the followingexpression:A(s ) AM FL (s )FH (s )– Where FL(s) and FH(s) are the functions that account for the frequencydependence of gain on frequency at the lower and upper frequency ranges– We can solve for AM by assuming that large coupling capacitors are shortcircuits and internal device capacitances are open circuits (what we havedone so far for low-frequency small-signal analysis)WeiES154 - Lecture 174

High-Frequency Response We can express function FH(s) with the general form:FH (s ) (1 s ωZ 1 )(1 s ωZ 2 ) (1 s ωZnH )(1 s ω P1 )(1 s ω P 2 ) (1 s ω PnH )Where ωP and ωZ represent the frequencies of high-frequency poles and zerosThe zeros are usually at infinity or sufficiently high frequency such that the numerator Æ 1and assuming there is one dominant pole (other poles at much higher frequencies), we canapproximate the function as FH (s ) ω H ω P1– This simplifies the determination of the BW or ωHIf 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 forderivation).ωH 1–Wei1(1 s ωP1 )1ω P21 1ω P2 2 2 2ω Z21 ω Z2 2 Note: if ωP1 is a dominant pole, then reduces to ωH ωP1ES154 - Lecture 175

Open-Circuit Time Constant Method It may be difficult to find the poles and zeros of the system (which is usually thecase). We can find approximate values of ωH using the following method.– We can multiply out factors and represent FH(s) in an alternative form:1 a1s a2 s 2 anH s nHFH (s ) 1 b1s b2 s 2 bnH s nH– Where a and b are coefficients related to the zero and pole frequencies111– We can show thatb1 ω P1 ω P 2ω PnHand b1 can be obtained by considering the various capacitances in the highfrequency equivalent circuit one at a time while reducing all other capacitorsto zero (or open circuits); and calculating and summing the RC timeconstant due to the circuit associated with each capacitor.– This is called the open-circuit time constant method (OCT)WeiES154 - Lecture 176

Calculating OCTsThe 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 timeconstants)nHb1 Ci Rioi 1 This method for determining b1 is exact. The approximation comes from usingthis result to determine ωH.1ω H nH Ci Rioi 1 Wei– This equation yields good results even if there is no single dominant polebut when all poles are realWe will see an example of this method when we analyze the high-frequencyresponse of different amplifier topologiesES154 - Lecture 177

Miller’s Theorem Before we begin analyzing the high-frequency response of amplifiers, there is an importantphenomenon 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 othernodes 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 andgnd.1I1YI2V1––1 I1V2V1I2 2Y1Y2V2The relationship between V2 and V1 is given by K V2/V1To find Y1 and Y2I1 Y (V1 V2 ) YV1 (1 V2 V1 )I1 YV1 (1 K )I 2 Y (V2 V1 ) YV2 (1 V1 V2 )I 2 YV2 (1 1 K )Caveat:I1 Y1V1I 2 Y2V2The Miller equivalent circuit is validonly as long as the conditions thatexisted in the network when K wasdetermined are not changed.Y1 Y (1 K )Wei2Y2 Y (1 1 K )ES154 - Lecture 178

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/vinLet’s investigate them all WeiES154 - Lecture 179

Using OCT on CS Amplifier Find the RC time constants associated with Cgd and Cgs in the following circuitC gdRsAM g m vgsvi vgsC gsR L rovoReplace Cgd with an open-ckt and find the resistance seen by CgsRsRgs gmvgsvtst vo g m (RL ro )viRL roItstvovtst Rsitstτ gs Rgs C gs Rs C gsReplace Cgs with an open-ckt and find the resistance seen by Cgditst v gs RsitstRsvtstvgsgmvgsRL roitst g m v gs v gs vtstRL roRgd vtst itst (RL ro ) g m Rs (RL ro ) Rsvoτ gd Rgd C gd [(RL ro ) g m Rs (RL ro ) Rs ]C gdWeiES154 - Lecture 1710

Using OCTs Cont’d Summing to two time constants yields ωH1ωH τ gs τ gdωH 1Rs C gs [(RL ro ) g m Rs (RL ro ) Rs ]C gd– From the above equation, it is not difficult to imagine that Cgd has a moresignificant effect on reducing BW– The resulting frequency dependence of gain is A(s ) WeiAMs ωH 1Let’s compare this result with what we get using Miller’s theoremES154 - Lecture 1711

Using Miller’s Theorem on CS Amplifier Redraw the high-frequency small-signal model using Miller’s theoremRsgmvgsCgd(1 gmRL')vivgsRL'CgsvoCgd[1 1/(gmRL')] Cgd–CTAssuming a dominant pole introduced by Cgd in parallel with CgsωH Wei11 C gs C gd (1 g m RL ') Rs CT Rs[]– Miller multiplication of Cgd results in a large input capacitanceNotice that this approximation for ωH is close to the approximation found using OCTassuming that RsCgd(1 gmRL’) dominatesLet’s verify our assumptions by deriving the exact high-frequency transfer function of the CSamplifierES154 - Lecture 1712

High-Frequency Response of CS Amplifier Replace the input source and series resistance with a Norton equivalentsCgd(vgs-vo)vi (s ) vgs sC gs v gs sC gd (v gs vo )RsRssC gd (vgs vo ) g m v gs vi/Rs gmvgsRsvo (s )RL '1 sg m C gdCgdvgsCgsR L'vovo (s ) A0vi (s )1 s Rs C gs Rs C gd (1 g m RL ') C gd RL ' s 2C gs C gd 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 ofmagnitude), we can rewrite the expression for the denominator as WeiES154 - Lecture 1713

HF Response of CS Amplifier Rewrite the denominator as:D(s ) (1 s ω P1 )(1 s ω P 2 ) 1 s (1 ω P1 1 ω P 2 ) s 2 ω P1ω P 2D(s ) 1 s ω P1 s 2 ω P1ω P 2– And from the solution on the previous slide we can write ω P1 ωP 2 Wei1Rs C gs Rs C gd (1 g m RL ') Rs C gd RL ' RsC gs C gd (1 g m RL ') C gd RL ' RsC gs C gd RL ' gmC gsSo the second pole is usually at a much higher frequency and we can assume adominant poleUsing either Miller’s theorem or OCTs enables a way to quickly findapproximations of the amplifier’s high-frequency responseES154 - Lecture 1714

Frequency Response of CG Amplifier One way to avoid the frequency limitations of Miller multiplication of Cgd is to utilizea CG amplifier configurationCgdRsvs -(gm gmb)vxCgsvxCdb-(gm gmb)vxRDCsbCD Cgd CdbRsvsvxRDCS Cgs CsbUsing OCT method, we find two time constantsCS– At the input (source node) τ S 1 Rs g m g mb– At the output (drain node) τ D C D RD The output usually drives additional load capacitance such that the output pole isdominantThe frequency response of CG amplifiers is when combined with a CS stage tobuild a cascode circuit WeiES154 - Lecture 1715

Cascode Stage Cascoding enables high bandwidth by suppressing Miller multiplication of Cgd. Let’sinvestigate how with the following high-frequency model of a cascode stage.RLvOUTCgd2-gm2 gmb2vxCdb2 CLCdb2 CLVbCgd1RsvINCgs2RsCdb1 Csb2vinCgs1voutRLCgd1gmvgs1Cgs1vxCdb1 Csb2 Cgs2– Use OCT method to find the time constants associated with each capacitor.The time constant associated with Cgd1 is τ gd 1 Wei 1 g m1 Rs ( g m 2 g mb 2 )Rsg m1 Rs C gd 1C gd 1 1 g m 2 g mb 2g gm2mb 2 ES154 - Lecture 17NOTICE:Miller multiplication is 216

Frequency Response of Source Followers Start with a high-frequency small-signal model of the source follower ctly solving for vout/vin yields:v gs sC gs g m v gs vout sC L v gs []sC Lvoutg m sC gsvin Rs vin sC gs (vin v gs )sC gd v gs voutω P1 gmg m Rs C gd C L C gsg m sC gsvout(s ) vinRs (C gs C L C gs C gd C gd C L )s 2 (g m Rs C gd C L C gs )s g m –WeiThe zero is due to Cgs that directly couples the signal from the input to the outputIf poles are far apart, then the s term represents the dominant poleES154 - Lecture 1717

More on Source Follower Other important aspects of a source follower are its input and outputimpedances (since they are often used as buffers)Let’s calculate the input impedance using the high-freq small-signalmodels g 11 1 m Z in sC gs sC gs g mb sC LNow calculate the output impedance (ignoring gmb for simplicity)Rs sC gs 1Z out g m sC gs– At low frequency, Zout 1/gm– At high frequency, Zout Rs– Shape of the response depends on the relative size of Rs and 1/gm Zout Cgsvgsgmvgsvout1/gmbCLRsCgsvgsgmvgsvout Zout 1/gmZoutRsRs1/gmωWeiZinωES154 - Lecture 17Zout can look inductiveor capacitive dependingon Rs and 1/gm18

Differential Pair We have seen that a symmetric differential amplifier can be analyzed with adifferential half circuit. This still holds true for high-frequency small-signalanalysis.RD vd/2-vd/2RsRDvoutRsCgdgmvgsRsvd/2CgsCdbRDvoutI– The response is identical to that of a common-source stageWeiES154 - Lecture 1719