Unwind di Analogue Integrated Circuit Design - 4 di 5

IBMio L 8Ma Ms1 ein inin

In the study of this two-stage OTA (5T-OTA + CS stage) there are the following sizing constrains:

• Symmetry: M1=M2, M3=M4, M6=M7, M8=M9;
• Bias: W5=N*W10, W8=W9=M*W10, W3=W4=N/2*W11;

Note that the constant N is only referred to the differential pair bias current IB while M is only referred to the second stage bias current.

Due to the symmetry of the circuit, the analysis can be done just considering only one side of the circuit. For example, considering a purely differential input and the left half side of the circuit, the small signal equivalent circuit is:

0Gdf laCoEx Cgs Ro7viInitial voi8m772IoRx 203 1 dissiCgd Cdb tfgd.sk17oz logRolo cdbztcgdqtld.bgCgsC ex tCetCi Co

The other half small signal circuit is perfectly identical but it's driven by the same signal with opposite sign:

Nor Voi

Let's study the frequency behaviour of this circuit applying KCL to nodes O1 and X:

ImaSCI tslgdzrt.it VxI S OgolaRo sei 8miVoi5C 0Tx5cgda t8D

NÈtComsidering: VoiNovoi 2VoceVoi NodeAnd avoiding all the maths calculation, the final result is:Im GmaRoRx GolaI 5Vod Ima5Adam i5 5Vid I Cl lat ttcgdzltfmr.ro litraCx tegoleRxdi rxr.CI CitdaIn the definition of Adm(s) is evidence the presence of Adm0:cgdztlilgdzitdmoiImrxfmr.RO 7ozRx Io Ros logRoThe CS stage of the 5T-OTA and the CS second stage are in cascade, so the mid-band gain Adm0 is simply equalto the product the 2 single gains. Moreover there is a right-hand-plane (RHP) zero:ImaWz iCgd7Usually the value of this zero is greater then the transit frequency so it can be not taken care of it:Wz WtObviously the same behaviour is present at the other side of the circuit and so for the device M6.The denominator is a second grade polynomial so it presents two roots that corresponds to the poles of thecircuit: StarsI 05 It b0StarsIta s St 0ecgdzltfmr.rob Cx't litra tegoleRxaida rxrocicitlicgdztlilgdz.tlè licit licealeRaro Cgd 1Let's determine now the roots

(poles):2b bPie 4 eb 1e It ba2 2 22b b C4PE b ie i be2 2 2The two real left-half-plane (LHP) poles are real only if the argument of the square root is real:be4Ce ib

When the previous condition is satisfied:Wp LupiP PaIn particular, wp1 acts like a dominant pole in the circuit because usually the operational amplifiers are used with feedback and a dominant pole behaviour increases stability of the system.

For example let’s consider: Bethvi voAls 0pPfH VoS 5a ati t.IS1 po WpaWpto o51 5 I SI t Clip Wpaw disco5OupWpaPao Pao

Usually, if beta is increased, the wp2 pole decreases it’s value and the wp1 pole increases it’s value until a certain point (that depends by their initial value and the mid-band gain A0) they became complex and conjugate.

If the closed loop TF features two complex conjugate poles, then the step response of the system may show ringing: visivo Sinodo E

Where vi(t) is the input signal, vo(t) is the output signal affected by ringing and vod(t) is the desired output voltage.

In order to avoid ringing conditions, the initial poles (the open loop poles) must be far from each other, in particular wp1 must be very larger then wp2:

WpaWp

In other words: b 4 e1 I beUn 1a 2bWpa 4 e1 1 be2 True only in the4 e Ioi b case of real poles he4 bee iii x ix bai t

Tipically this last condition is never satisfied, in fact in general they are of the same order of magnitude, in fact:20 2ozRx Io 12209203ei.ciei

Using the number of the exercises:

0,2 CadeCgd Caos fociIn 2 IoClIm 2oz GmRo 209

Then: Ct e te12 TRi tag ILI 2.671010 Recite e ReIO10CI 1 Re1,2 Using these values:

4ft 0,47

Clearly these value is not much more smaller then the unity so in a feedback behaviour this circuit should present ringing, so instability. In order to reduce the phenomenon of ringing it can be used the pole-splitting approach through compensation capacitance. In fact note that:

4124C licit tailgate12 Cgd Icgdzltfmr.roCx't lib tra tegoleRx

This parameter inversely depends by the gate-drain capacitance of

the device M7 so if that capacitance is increased, then the value of the previous parameter becomes smaller and can also become very small respect to the unity: him he 0pzGot toLet’s see how ti increase the value of Cgd7: VbbNA NAte µIBM IBMMa 146No GobGolace leIB 8Ms149 ein nin

The value of Cgd7 is a design parameter and cannot be changed. In order to change it’s value, it can be simply putted a parallel capacatior Cc whose value is added with the value of Cgd7: Cgd Ccgola 4124 licit tailgatee Cgd12 Icgdzltfmr.ro liCx'tb tra tegolaRx

If Cc tends to infinite, then: Le irx.IE 0ÈrxkIFIIi.rittir.cer rrb So the added capacitance Cc should be large in enough in order to: 1LetiLets consider now another time the poles: E b_4C eWp b 2C1 i iI tete lo22 I4C i_4C_4 i_2Cei 262 b2baba CitedCCup Rxr.ccCI rxleitfmr.rob Rxr.cc Role1I IRxleltfmzrotroGRxleltfmt.roRole rxleltfmr.ro It’s clear that wp1 uses the Miller effects because the added capacitance Cc is multiplied by the

Il fattore di Miller e a causa di questo il polo si trova a frequenze più basse. Questa compensazione è chiamata compensazione di Miller. Consideriamo ora cosa succede a wp2: I4Cba Ib b_4 E eIWpz.bz bI4 4_4Ce Ii e I 2Ci 262baba beI 2oCeE8m7RoRxb Rxccltfmzr.at ImaWp CItci CI Cirxr.cc Rothltfmzro fmt.rorxltfmr.ro RoIl secondo polo dipende dalla capacità al nodo X e dalla capacità di carico. Se la capacità di carico è sufficientemente grande, allora:Ima GmaWpa Ilieici ciQuesta ultima formula non è molto accurata ma dà un'idea approssimativa del valore di wp2.È quindi chiaro che aggiungendo la capacità Cc il polo dominante wp1 si sposta verso frequenze più basse, mentre il polo non dominante si sposta verso frequenze più alte fino a raggiungere un valore asintotico: gm7/CL' (se CL' è sufficientemente grande rispetto a Cx'). Con questa tecnica i poli sono quindi separati.Questa tecnica presenta comunque un problema. Infatti lo zero RHP è:Wz iGmagolaSe viene aggiunta la capacità di compensazione, allora wz diventa: eWz iIGma Wpa

Semmealetlegali CgdeIf Cc is large enough, then wz could become smaller then the second pole and this behaviour generates a lot ofproblems.

LESSON #16

LESSON #17

1.Let's firstly analyse the amplifier and let's indicate it's parasitic input capacitance Ci:

Ci

Applying a Norton transformation at the input:

5 AMPLIFIERes BETA-NETWORK

vs isEfCsci eii Tiseats

The voltage gain transfer function (TF) is:

51Adamo LUIs s_Ao 5 5I Iti 5 Wp Lope

Where: ImaEmiro GmGm Ro9miRoGma 20220Adamo Nè iZoe 208 sWzRo Irz.glcei le0,2ImaIma

The zero changes it's sign.

Choosing Rz such that e the zero changes it's sign and becomes a LHP zero. The voltage gain TF becomes:

SIAdamo WzSA 5 51 IWp Wpa

The poles are the following:

I GmaleNÉÌWp Rogier.ec cucitelo ci

The other parameters are:

CdbCoitcgdaxldbrtcgdztcgszici.cdbztcdbgtcgdgtlltlt.isi

The bias currents of the devices are:

100mATree IDE1dg Ins InIsco 1941400µAIssInId IdaIoIIs 200µA da

200µA200µA40014.1IosaIsaLet’s determine now the small signal parameters:Gmt 2,49MIo8mi 2ha µGma 2118mi Ing 5,78Mµ8 ITorRa ri5,71kIIRon MERo 621112oz 208 3e 61 lo Cdb llCoi 31,7ftff Cdb3 t 8 golpeThese are all the parameters that must be know about the OTA.Let’s analyse now the beta-network using the y-parameters model:il it ilisEfCs rt 1iNzv va4 122navi412824 4,2 setsi CsxCf iV Va OviOvi set42 set4,2 Èiovi viO OviThe beta-network can be seen so as:i iascent Cf Uzesta EfiVPutting together the equivalent beta-network and the amplifier, the equivalent circuit becomes:scuto settici ce liTiestais ToLet’s find an expression for the input voltage to determine the beta factor:Vi sleds Cs Ceist I vs voCi lileftest leftestCileftest5 Costsis paThe input voltage vi corresponds to the summation of two contributions. The first one depends by the inputvoltage vs scaled by a factor alfa while the second factor depends by the output voltage and the

beta-network tooby the beta-network gain beta: li legalip 32,3ftCe EggsCftlstli0,194ELet’s evalute now the loading of the feedback network on the output, remembering that the input must be shuttedoff (the input current generator becomes an open circuit): itSefro NtleEstee setticivi 0isZEB reit portoVi scarto viportoICftlstli5 P Cf Cileftest seesle.PTslfrti scfrto Vtscfrto PIit t.tttortovi 1le leCiaVtP PZEBit t tIt eppseThe capacitance CFB is the capacitance that the feedback network, being a non-ideal network, adds to the