Implementation of Cascade Amplifier in 180nm CMOS Technology

DOI : 10.17577/IJERTV2IS120988

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Implementation of Cascade Amplifier in 180nm CMOS Technology

Khushboo, Jubli

Electronics & Communication Engineering Northern India Engineering College

FC-26, Shastri Park, Delhi-110053, India

Abstract

This paper described about the complete analysis and design of common source amplifier designed for application in a capacitive-micro machined-ultrasonic-transducer (CMUT) based intravenous imaging system. These CMUTs have recently gained much interest due to their numerous advantages as ultrasound transducers that can be integrated with electronics on silicon and are compact. In this paper the workability of the feedback biasing cascade amplifier circuit configurations has been analysed and tested experimentally using PSPICE and MATLAB simulation tool. It has been found that the new propositions have improved performance such as gain and bandwidth. The results have been taken on 180nm technology. It gives the derivations of all the equations described in the paper like, closed loop gain of negative feedback amplifier, transfer function, drain current equation, stability factor, FOM etc. The Simulation result of the CS amplifier with feedback biasing in 180nm CMOS technology using PSPICE and compared this with the MATLAB plot of the transfer function of the same.

Keywords-Cascade amplifier, CMUT, CLOSED loop gain and bandwidth, stability factor, CS amplifier, 180nm CMOS technology.

  1. Introduction

    The Nanoscale technologies can be a viable option for the analog circuitry as well. Some of the features of Nanoscale technologies that are otherwise not desirable to an analog designer may actually be useful in some circuit techniques, as we will show in this brief for the case of feedback biasing. Even though a feedback-biasing scheme is simple and ensures that the input MOSFET remains in saturation, irrespective of the process and temperature variations, its biggest disadvantage has been the limited voltage swing. We show that this disadvantage vanishes as one move to Nanoscale technologies [1]. Simulation and experimental results [1], presented therein for validation of the ideas, were for Feedback biased Cascade amplifier designed for application in a capacitive-micro machined-ultrasonic-transducer (CMUT) based intravenous imaging system. A number of works on CMUT operation and applications are available. Application of ultrasonic imaging fields such as dermatology,ophthalmology and cardiovascular

    medicine require very high frequency resolution. CMUTs can be made for high frequency operation. CMUT is an appropriate technology for building very high frequency arrays. A linear array of high voltage pulsar and amplifier circuits has also been designed for use with an array of CMUTs to enable real time imaging applications [7]. Due to the requirement of having an array of several transducers with local signal conditioning on a catheter tip for performing imaging inside the human arteries, the compactness of the circuitry is of utmost importance, followed by the constraint on the power consumption of the system. It is for this reason that a cascade amplifier topology is designed for this application, as the cascade amplifier is one of the most efficient amplifier stages that can be realized in CMOS technology [6]. To realize the biasing within a very small area sub threshold MOSFETs as high-value resistors have been employed to bias amplifiers [1]. The article rediscovers the attractiveness of feedback biasing when applied to circuits designed in Nanoscale CMOS technologies. It is shown that very compact amplifiers can be obtained by utilizing a type of biasing that imposes minimal area overhead. It presents measurement results of common-source (CS) amplifiers using feedback biasing designed for application in a capacitive micro machined ultrasonic transducer 30MHz (CMUT). The proposed feedback biasing is also suitable for amplifying signals from high impedance sources that pose challenges on maintaining high input impedance for the voltage amplifiers while maintaining a very low input capacitance value. Here CMUTs can be integrated with electronics on silicon and are compact.

    The basic idea behind the cascade amplifier is it can be designed to increase the DC gain and the gain-bandwidth. We have presented feedback biased cascade amplifier, to increase the overall gain and bandwidth simultaneously. The paper discovers the attractiveness of feedback biasing when applied to circuits designed in NanoscaleCMOS technologies. Utilizing a type of biasing very compact amplifiers can be obtained that cover minimum area in the NanoscaleCMOS technologies where the biasing point has found to be more stable

    The proposed feedback biasing is also suitable for amplifying signals from high impedance sources that pose challenges on maintaining high input impedance for the voltage amplifiers while maintaining a very low input capacitance

    value. The amplifiers were fabricated in 180-nm CMOS technology and measured to be just 20µm x 10µm.

  2. FEEDBACK BIASING IN NANOSCALE CMOS TECHNOLOGIES

The four basic feedback topologies

  1. Voltage amplifiers: – It amplifies an input voltage signal and provides an output voltage signal. A suitable feedback topology for the voltage amplifier is the voltage-mixing

  2. Current amplifiers: – It amplifies an input current signal and provides an output current signal. A suitable feedback topology for the current amplifier is the current-mixing current sampling.

  3. Trans conductance amplifiers: – It amplifies an input voltage signal and provides an output current signal. A suitable feedback topology for the Trans conductance amplifier is the voltage-mixing current sampling. It is also called as series-series feedback. We have used Tran conductance amplifier for our amplifier.

  4. Trans resistance amplifiers: – It amplifies an input current signal and provides an output voltage signal. A

    • DC Analysis Equations

    VDD VDS=0 (1)

    VDD = IDRD + VDS

    VDD – IDRD VGS=0 (2) VDD = IDRD + VGS

    After comparing equation (1) & (2) IDRD + VDS = IDRD + VGS

    VDS = VGS (3)

    i.e. Output voltage is controlled by Input voltage.

    In above fig.1, the voltage drop across RG is zero & IG = 0. Output is taken across VDS,

    So VDS = Vout i.e. Vout = VGS (4) We know VDS = VGS so put it in equation no. (1)

    VDD = IDRD + VDS

    VDD = IDRD + VGS

    suitable feedback topology for the Trans resistance

    V = V

    + I R

    (5)

    amplifier is the current-mixing voltage sampling. It is also called as shunt-shunt feedback

    1. BASIC ANALYSIS OF FEEDBACK BIASING

      1. DC Analysis

        VDD

        DD GS D D

        Next we have to drive the equation:- VOUT_MIN VDS_MIN VT

        As we know VT = Threshold voltage.

        The value of VGS at which a sufficient no. of mobile

        RD ID

        0

        RG

        ID

        +

        Vout

        electrons accumulates in the channel region to form a conducting channel is called the Threshold Voltage. Its value varied between 0.5 to 1.0 V.

        When VGS = VT(induced channel), but at this point ID (drain current) is usually very small.

        As VGS> VT, increased conductance, & reduce resistance.

        VGS

        Fig.1NMOS with feedback biasing

        VDD

        In fact the conductance of channel is proportional to the excess gate voltage (VGS- VT) also known as the Effective voltage or the Overdrive voltage.

        It follows that the current ID will be proportional to VGS VT and of course to the voltageVDS that causes ID to flow.

        RG CD

        Vin

        RD

        Req

        Vout

        Fig.2 when used as a voltage amplifier with a low impedance source.

        Fig.3 Operation of enhancement NMOS Tr.

        VOUT_MIN = Put ID = 0

        VDD-VSS 2

        VT

        VDD = VDS+VSS

        VDD-VSS

        ID =

        RD

        Tr.

        Fig.4 ID versus VDS plot for enhancement type NMOS

        When VDS increased that reduces the voltage between gate and drain end to VT,

        I.e. VGD = VT Or

        VGS VDS = VT VDS = VGS VT

        The channel depth at the drain end decreased to almost zero,

        and the channel is said to be Pinched off at this point it

        Fig.6 DC load line

        For faithful amplification the quiescent point Q will be on the middle of the load line.

        VDD-VSS

        enters the saturation region of operation. The voltage VDS at which saturation occurs is denoted VDSsat,

        VDSsat = VGS VT

        So VOUT_MIN =

        2

        VDD-VSS

        VT (9)

        Or

        VOUT_MIN VDS_MIN VGS_MAX VT (6)

        Where

        = VGS_MAX

        2

        Bias Point calculation

        Equ.no. (9) is the minimum output voltage to give the faithful amplification.

        VDD

        RD

        V2

        Assume amplifier gain is high i.e. AV =

        V1

        is high.

        RG

        +

        VGS

        VDS

        -VSS

        Vout

        It means that the value of V1 should be kept low in order to not clip off the input voltage.

        When Q shifted towards the y-axis i.e. ID side, the positive peaks of the output signal would be clipped off, because the MOSFET would turn off for part of the cycle. It is called as the circuit not having sufficient headroom.

        When Q shifted towards the x-axis i.e. VDS side, the output signal is distorted. It is called as the circuit not having sufficient legroom.

        Fig. 5 NMOS with feedback biasing

        VDD VDS VSS = 0

        VDD = IDRD+VDS +VSS (7)

        Put VDS = VGS

        VDD VGS VSS = 0

        VDD = IDRD+VGS +VSS (8)

        From equ.no.(7) plot DC load line to obtain the given equation:-

      2. HIGH FREQUENCY ANALYSIS OF FEDDBACK BIASING

      At high frequency we include the internal MOSFET capacitance also.

      And at high frequency the feedback biasing figure is also improved

      Here, CGD=CG; CGS=CIN; CDS= CL

      VDD G D

      RD

      CG RG

      Vout

      Vt ro

      gmVGS

      RD R'G

      CL

      CIN

      Vout

      Vf

      CL

      Cin

      S S

      Fig. 9 (a) High frequency model of the circuit

      G D

      Fig.7(a) Schematic showing a feedback biased amplifier, with important impedances shown

      VDD

      Vt

      gmVGS

      S

      Rout R'G

      CL

      CIN

      Vout

      Vf

      S

      CG RG

      Cin

      RD

      + +

      Vf Vt

      – –

      Vout

      CL

      Fig. 9(b) High frequency model of the circuit

      Gm = Transconductance of the MOSFET. The control that the VGS has over the ID is measured by

      ID

      Fig. 7(b) Feedback loop is broken to calculate the loop gain

      Gm =

      VGS

      constant VDS.

      MOSFET CAPACITANCES

      Based on their physical origins, the parasitic device capacitance can be classified into two major groups:

      1. Oxide related capacitances: – The two overlap capacitances

        Fig. 2.9(b) is used to calculate the loop gain response of the feedback, as expressed in [1].

        Vf(s)

        that arise as a result of this structural arrangement as shown

        LG(s) =

        V (s)

        in figure (8) are called CGD are voltage independent.

        (overlap) and CGS

        (overlap). They t

      2. Junction capacitances: – The voltage dependent source-

      1 1sC

      substrate and drain-substrate junction capacitance are, C

      = -gm Rout||

      IN

      and C .

      SB

      sCL

      1 1

      DB RG|| +

      CGD

      D

      CGB

      CDB

      Rout

      sCG

      1

      sCIN

      G

      CGS

      MOSFET B

      (DC MODEL)

      CSB

      LG(s)= -gm sRGCG (10)

      1+sRoutCL 1+sRG CG+CIN

      Apply KCL

      S

      Fig. 8 Lumped representation of the parasitic MOSFET

      Vout 1 +sCL+ 1 = -gmVGS

      capacitances

      Here, CGD=CG; CGS=CIN; CDS= CL

      In order to obtain this loop gain response we have to make the

      Rout

      Z= R'G+

      Z

      1

      high frequency model of this feedback biasing method. High frequency analysis (equivalent circuit)

      Where

      Z=

      sCIN

      sR'GCIN + 1 sCIN

      & R`G = RGCG & Rout = roRD

      Vout 1

      +sCL+

      sCIN

      = -gmVGS

      Io R'G+ 1

      Rout

      sR'GCIN+1

      & VOUT =

      sCIN

      1sR'GCIN+1+sCLRout 1+R'GCIN +sCINRout

      Vout = -gmV

      1

      -gmVGSRout 1+sCINR'G

      Rout sR'GCIN+1

      GS

      IoR'G+ =

      sCIN

      sR'GCIN+11+sCLRout +sCINRout

      Vout sR'GCIN+11+sCLRout +sCINRout =-gmVGSRout sR'GCIN+1

      sR' C +1 -gmV R 1+sC R'

      -gmVGSRout sR'GCIN+1

      Io sC = sR' C +11+sC R +sC R

      G IN GS out IN G

      G IN GS out IN G

      Vout =

      IN G IN L out IN out

      Vout = Vin

      sR'GCIN+11+sCLRout +sCINRout

      -gmVGSRout sR'GCIN+1

      sR'GCIN+11+sCLRout +sCINRout

      Io= -gmVGSRout 1+sCINR'G ×

      sR'GCIN+11+sCLRout +sCINRout

      Io= -gmVGSRoutsCIN

      sCIN

      sR'GCIN+1

      RG

      RG=RGCG = sRGCG+1

      gmRout 1 sRGCIN

      sR'GCIN+11+sCLRout +sCINRout

      1

      &Vf = IO × sCIN

      Vout

      sRGCG 1

      Vin

      1 sRGCIN 1 sCLRout sCINRout

      Vf(s)=

      -gmVGSRoutsCIN ×

      sR'GCIN+11+sCLRout +sCINRout

      1 sCIN

      sRGCG 1

      gmRout sRGCIN sRGCG 1

      Where V = V

      Vout

      sRGCG 1

      GS t

      Vin

      sRGCIN sRGCG 1 1 sCLRout sCINRout

      -gmVGSRout 1+sCINR'G sC

      G IN L out IN out

      G IN L out IN out

      sRGCG 1 IN

      gmRout sRGCIN sRGCG 1

      Io= sR' C +11+sC R +sC R ×

      sR'GCIN+1

      Vout

      sRGCG 1

      Vin

      sRGCIN sRGCG 1 1 sCLRout sCINRout sRGCG 1

      sRGCG 1

      We know R`G = RG CG

      Vout gmRout sRGCIN sRGCG 1

      RG 1 sCG RG

      =

      Vin sRGCIN sRGCG 11 sCLRout sCINRout sRGCG 1

      RG 1 sCG sRGCG+1

      =

      =

      Vout -gmRout-sgmRoutRGCIN-sgmRoutRGCG 2

      Vf(s) -gmRout

      Vin s RGCINCLRout+RGCGCLRout+CINRoutRGCG +sRGCIN+RGCG+CLRout+CINRout+1

      Vt(s)

      sRGCIN 11+sC Rout sC Rout

      sR C +1

      L IN

      G G

      Vout = -gmRout-sgmRout RGCIN+RGCG

      Vin s2 RGCINCLRout+RGCGCLRout+CINRoutRGCG +sRGCIN+RGCG+CLRout+CINRout+1

      (3.19)

      Vf(s) -gmRout

      Vt(s) sRGCIN+sRGCG+11+sCLRout+sCINRout sRGCG+1

      sRGCG+1

      Vf(s) = Feedback voltage Vt(s) = Input voltage

      Vf(s) -gmRout sRGCG+1

      Vf(s)

      Vt(s)

      = -gmRout 1+sCGRG

      sRG CIN+CG +11+sCLRout

      Vt(s)

      sRGCIN+sRGCG+11+sCLRout+sCINRout sRGCG+1

      Now divide numerator and denominator by

      Rout× 1sC 1sC

      = gm L IN

      sCINRout 1+sRGCG

      Rout+ 1sC

      RG× 1

      L sCG + 1

      Vf(s)

      -gmRout 1+sRGCG

      sCINRout 1+sRGCG

      Rout+1/sCG sCIN

      =

      Vt(s)

      1+sRGCG+sCINRG 1+sRoutCL

      sCINRout 1+sRGCG

      Rout× 1 1

      sCINRout 1+sRGCG

      sCINRout 1+sRGCG

      = gm sCL sCIN

      +

      +

      Rout+ 1sC

      RG× 1

      L sCG + 1

      gm

      Rout+1/sCG sCIN

      Vf(s) =

      – sC

      sCINRout 1+sRGCG

      1

      1

      IN

      R

      Vt(s)

      1+sRGCG+s2CINRG1+sRoutCL

      = gm

      out

      sCIN

      1+sRoutCL sRGCIN+ 1+sRGCG

      Vf(s) =

      Vt(s)

      -gmRout 1+sRGCG

      1+sRG CG+CIN 1+sRoutCL

      sCIN 1+sRGCG

      1+sRoutCL 1+sRG CG+CIN

      1+sRoutCL 1+sRG CG+CIN

      = gm Rout 1+sRGCG

      Vf(s)

      Vt(s)

      = -gmRout

      1+sRoutCL

      × 1+sRGCG

      1+sRG CG+CIN

      LG(s) = Vf(s) = -gm Rout 1+sRGCG (13)

      (11) Vt(s) 1+sRoutCL 1+sRG CG+CIN

      Vf(s) = -gmRout-sgmRoutRGCG

      Vt(s) s2 RGRoutCGCL+RGRoutCLCIN +sRoutCL+RGCG+RGCIN +1

      or

      The loop gain response has a dc gain (gmRout)

      When RG>> Rout & CL&CIN are of the same order the dominant pole of the loop gain response is set by the RC product at the input node i.e. RG (CG + CIN).

      Vf(s)

      Vt(s)

      sRGCIN

      -gmRout

      LG(s)= -gm Rout 1+sRGCG = Zeros

      sR C +1 11+sCLRout sCINRout

      1+sRoutCL 1+sRG CG+CIN

      Poles

      G G

      =

      =

      Vf(s) -gmRout-sgmRoutRGCG

      Ist pole:

      Vt(s) s2RGRout CGCL+CLCIN+CINCG +sRoutCL+RGCG+RGCIN+RoutCIN 1

      1 +sRG (CG

      + CIN

      ) = 0

      (12)

      Rechecking the above equation no. (12) from

      sRG (CG + CIN) = 1

      s =

      s =

      -1

      bottom to top,

      RG CG+CIN

      (14)

      LG(s) =

      Vf(s)

      Vt(s)

      1

      IInd pole:

      (1 + sCL Rout) = 0

      sCL Rout = 1, s =

      -1

      (15)

      R C

      = gmRout|| 1sC

      sCIN

      out L

      L

      Rout× 1sC

      RG||

      1

      sCG

      1

      sC

      + 1 sCIN

      Zeros:

      gm Rout (1+ sRGCG) = 0

      1+sRGCG = 0

      sRGCG = 1

      = gm L IN

      Rout+ 1sC

      RG× 1 -1

      L sCG + 1

      s = (16)

      RGCG

      Rout+1/sCG sCIN

      jw-axis(imaginary axis)

      So B1 =

      -1

      RG CG+CIN

      RGRoutCL CG CIN 1

      RG CG CIN 0

      2nd pole -wt

      0

      1st pole

      Real axis Wt=unity gain frequency

      -1

      R

      R

      B1 = × -RG CG+CIN G CG+CIN

      Fig. 10Poles & Zero

      S-plane

      B1 = 1

      So all the coefficients in the first column are of the same

      Assuming that CIN>> CG, it lies at much higher frequencies than the dominant pole.So Ist pole is the dominant pole.Assuming the IInd pole is lies far beyond its unity gain frequency ft.One can estimate wt as the product of the dc gain & the frequency of the dominant pole i.e.

      sign (positive), the given equation has no roots with positive real part. Hence the system is stable. Since the input signal VIN to the amplifier shown in Figure (3.13) (a) is capacitive coupled (By CD) to the gate of the MOSFET. CD forms a HPF (High pass filter) with REQ.

      Wt =

      gmRout RG CG+CIN

      gmRout (17)

      RGCIN

      The loop gain is less than unity beyond the 3-dB frequency hence the feedback loop is disabled beyond this frequency and the amplifier operates effectively in open loop. Therefore the feedback loop while biasing the MOSFET

      Where CIN>>CG

      Ft = unity gain frequency

      Dominant pole = It is the pole which is near to zero (origin)

      does not interfere with the signals beyond W3dB.

      Resistance REQ appears due to the miller effect at the gate of the MOSFET i.e.

      RG

      so that it dominate the amplifier frequency response; it is called a Dominant pole.Check the stability of the system using Routh Hurwitz criterion:-

      Let the equation aosm + a1sm-1 +. + am = 0

      1. All the coefficients of the equation should have same sign.

      2. There should be no missing term.

      These two conditions are necessary to make the system stable.

      REQ =

      Vin

      1-AV

      RD

      RG CD

      Req

      VDD

      Vout

      (18)

      1+sRoutCL 1+sRG CG+CIN

      1+sRoutCL 1+sRG CG+CIN

      Apply this criterion to our loop gain response LG(s): LG(s) = -gm Rout 1+sRGCG

      HPF

      Fig. 11(a) When used as a voltage amplifier with a low impedance source.

      (1 + sCL Rout)(1 + sRGCG + sRGCIN) = 0

      S2RGRoutCL (CG + CIN) + sRG(CG +CIN) +1 =0 S2 RGRoutCL(CG + CIN) 1

      REQ1=RG/(1-AV) REQ2=RG(1-1/AV)

      RG

      VDD

      RD

      Vout

      S1 RG (CG +CIN) 0

      S0 1

      -1 a0 a2

      CD

      Vin

      REQ1

      By Miller effect

      REQ2

      B1 =

      a1 a1 a3

      Fig.11(b) When used as a voltage amplifier with a low impedance source.

    2. FIGURE OF MERIT DEFINITION

      Next to drive the equation:

      RL =

      RG

      1-AV ,

      Vout =

      Vin 1

      sC

      R+ 1

      Gm

      ×

      ×

      FOM =

      Gm sC

      IDS 2CGS

      FOM: Figure of merit= AV BW

      Vout=

      Vin 1+sCR ,

      WH =

      1

      CR ,

      1

      2 fH =

      CR

      Vout

      A = Voltage gain =

      1

      Vout

      V

      BW = Bandwidth = fH

      Vin

      fL

      fH= AV =

      2CR ,

      Vin ,

      AV = -gmRD

      fH = Higher cut off frequency fL = Lower cut off frequency

      FOM = -gmRD

      1

      2CINR'G

      At higher frequency analysis we take fH only

      R'D Gm

      CGD

      =

      C R' 2CGS

      IN G

      G D

      Vin

      ro RD

      Vout

      Gm

      ×

      ×

      FOM =

      Gm

      R'L

      CGS

      IDS 2CGS

      S gmVGS

      Fig. 12Equivalent circuit to find the FOM

    3. FEEDBACK BIASED CASCADED CS

      AMPLIFIER

      1

      V V

      (sC

      ) = g V

      In Figure (9), we show the schematic of a voltage

      out R'L

      GS GD m GS

      amplifier that has been implemented in a 90nm CMOS technology. To achieve sufficient gain, two stages were

      IGD = small (so neglect it)

      required due to the relatively low intrinsic gain in 90nm. The amplifier employs feedback-biasing to allow for the

      Vout

      = gmV

      GSRD

      use of small transistor area and, hence large mismatch. This requires each stage to have different bias points. Therefore a decoupling capacitor is added between the two stages.

      RD = rd||RD

      VGS-Vout

      This techniques result in a very small input capacitance [5].

      IGD =

      = sCGD(VGS-Vout)

      1/ sCGD

      MX3a

      VDD

      MX3b

      VDD

      =sCGD[VGS+gmRLVGS]

      MX1a

      M2a

      VBIAS VBIAS

      M2b

      MX1b

      Vout

      =sCGDVGS(1+gmRL) Ceq = CGD (1+gmRL)

      By miller effect replace CGD

      Vin

      Cin

      MX2a

      Stage-a

      M1a

      Ccoup

      MX2b

      Stage-b

      M1b

      Cload

      CIN = CGS + Ceq

      CIN= CGS + CGD (1+gmRL)

      RL occurs due to miller effect

      Fig.13Schematic of a feedback biased Cascaded CS amplifier

      Diode-connected devices MX1 and MX3, which are connected in series across input device M1, form the feedback biasing resistance. Using two devices in series ensures that the voltage across them is not sufficient to turn

      them both on during large voltage swings. Hence, a high- resistance path is guaranteed at all voltage levels.

      Device MX2 is optional in the design for the feedback biasing to work. It typically exhibits a much higher resistance, compared to the feedback-resistive path (series combination of MX1 and MX3), and forms a direct high- resistance path to ground for the gate node of M1. It also acts as a voltage dependent resistance that forms a weak feedback loop that keeps M1 from entering deep into the linear region during higher-than-rated input voltages. It was also observed that MX2 helped improve the linearity performance of the amplifier by compensating for the nonlinear behavior of the main feedback path. Under normal operation of the amplifier, the input resistance (at the gate of M1) is dominaed by the equivalent resistance decided by devices MX1 and MX3 [1].

    4. SIMULATION RESULTS AND TRANSFER FUNCTION PLOTS USING PSPICE & MATLAB

      There is a simulation result of transfer function given in equ.no. (11) and their relevant codes are given in Appendix-I.

      .AC Analysis (Gain vs Frequency plot)

      Fig.14Simulated gain plots for the amplifier using PSPICE.

      Gain = 15.417DB

      Bandwidth = 132.56MHz.

      9

      8

      7

      6

      Gain

      Gain

      5

      4

      3

      2

      1

      Green Dots: PSPICE simulation result

      Blue Line: MATLAB transfer function plot result

      Figure (15) shows the comparison and agreement between PSPICE simulations results and MATLAB plot of the respective transfer function.

    5. CONCLUSION

      This paper described about the complete analysis and design of common source amplifier which is used in medical ultrasound imaging system. It gives the derivations of all the equations like, closed loop gain of negative feedback amplifier, transfer function, stability factor, FOM etc. The presented the simulation result of the CS amplifier with feedback biasing in 180nm CMOS technology using PSPICE and compared this with the MATLAB plot of the transfer function of the same.

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  2. A.-J. Annema, B. Nauta, R. van Langevelde, and H. Tuinhout; Analog Circuits in Ultra-Deep-Submicron CMOS IEEE J. Solid-State Circuits, Jan. 2005,vol. 40, no. 1, pp. 132143.

  3. T. Singh, T. Sæther, and T. Ytterdal, Common Source Amplifier with Feedback Biasing in 90 nm CMOS Res. Microelectro. Electron. Ph.D., Jun. 2006, pp. 161164.

  4. A. Sedra and K. Smith; Microelectronic Circuits 5th edition, Oxford Univ. Press, 2004.

  5. C.J.M. Verhoeven and A. van Staveren; Systematic Biasing of Negative Feedback Amplifiers Design, Automation and Test in Europe Conference and Exhibition 1999. Proceedings, March 1999, pp. 318-322.

  6. Trond ytterdal; Nanoscale Analog CMOS Circuits for Medical Ultrasound Imaging Applications Solid-State and Integrated- Circuit Technology, ICSICT 2008. 9th International Conference, Oct. 2008, pp. 1697 1700.

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B. Khuri-Yakub; High-frequency CMUT Arrays for High- Resolution Medical Imaging IEEE Ultrason. Symp., Aug. 2004, vol. 1, pp. 399402.

0

2 4 6

8 10

10 10

10 10 10

Frequency in Hz

Fig.15 PSpice Simulation and MATLAB transfer function plots for the amplifiers.

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