Home >Documents >RC Phase Shift Oscillator Full Derivation

RC Phase Shift Oscillator Full Derivation

Date post:19-Oct-2015
Category:
View:128 times
Download:7 times
Share this document with a friend
Description:
RC Phase Shift Oscillator Full Derivation
Transcript:
  • Appendix

    BDERIVATIONS OF SELECTEDEQUATIONS

    Equation 23

    The average value of a half-wave rectified sine wave is the area under the curve divided bythe period The equation for a sine wave is

    Equation 212

    Refer to Figure B1.

    VAVG =Vpp

    =Vp2p

    3-cos p - (-cos 0)4 =Vp2p

    3-(-1) - (-1)4 =Vp2p

    (2)

    VAVG =area

    2p=

    12pL

    p

    0Vpsin u du =

    Vp2p

    (-cos u)|p0

    v = Vpsin u

    (2p).

    tdis

    T

    0

    Vp(rect) Vr(pp)

    vC = Vp(rect)et

    RLC

    FIGURE B1

    When the filter capacitor discharges through the voltage is

    Since the discharge time of the capacitor is from one peak to approximately the next peak,when reaches its minimum value.

    Since becomes much less than 1 (which is usually the case); approaches 1 and can be expressed as

    e-T>RLC 1 -T

    RLC

    e-T>RLCRC W T, T/RLC

    vC(min) = Vp(rect)e-T>RLCvCtdis T

    vC = Vp(rect)e- t>RLCRL,

    Z05_FLOY9868_09_SE_APP2.qxd 1/13/11 3:46 PM Page B-1

  • Therefore,

    The peak-to-peak ripple voltage is

    Equation 213

    To obtain the dc value, one-half of the peak-to-peak ripple is subtracted from the peak value.

    Equation 61

    The Shockley equation for the base-emitter pn junction is

    where total forward current across the base-emitter junctionreverse saturation currentvoltage across the depletion layercharge on an electron

    number known as Boltzmanns constantabsolute temperature

    At ambient temperature, so

    Differentiating yields

    Since

    Assuming

    The ac resistance of the base-emitter junction can be expressed as

    Equation 614

    The emitter-follower is represented by the r parameter ac equivalent circuit in Figure B2(a).

    re =dVdIE

    1

    40IE

    25 mVIE

    dV>dIE.re

    dIEdV

    40IE

    IR 66 IE,

    dIEdV

    = 40(IE + IR)

    IReV40 = IE + IR,

    dIEdV

    = 40IReV40

    IE = IR(eV40 - 1)Q>kT 40,

    T = thek = a

    Q = theV = theIR = theIE = the

    IE = IR(eVQ>kT - 1)

    VDC = a1 -1

    2fRLCbVp(rect)

    VDC = Vp(rect) -Vr(pp)

    2= Vp(rect) - a

    12fRLC

    bVp(rect)

    Vr(pp) a1

    fRLCbVp(rect)

    Vr(pp) = Vp(rect) - VC(min) = Vp(rect) - Vp(rect) +Vp(rect)T

    RLC=

    Vp(rect)TRLC

    vC(min) = Vp(rect)a1 -T

    RLCb

    B-2 APPENDIX B

    Z05_FLOY9868_09_SE_APP2.qxd 1/13/11 3:46 PM Page B-2

  • DERIVATIONS OF SELECTED EQUATIONS B-3

    r e

    B

    RE

    Rs

    R2

    R1

    C

    E

    acIb

    Vs

    (a)

    r e

    RE

    Rs || R1 || R2acIb

    (b)

    Ve = VoutIe

    FIGURE B2

    Conventional current direction shown.

    By thevenizing from the base back to the source, the circuit is simplified to the formshown in Figure B2(b). .

    With and with produced by and neglecting the base-to-emitter voltage drop(and therefore ),

    Assuming that and

    Looking into the emitter, appears in parallel with Therefore,

    Midpoint Bias (Chapter 8)

    The following proof is for the equation on page 400 that shows when

    Start with Equation 81:

    Let

    0.5IDSS = IDSSa1 -VGS

    VGS(off)b

    2

    ID = 0.5IDSS.

    ID IDSSa1 -VGS

    VGS(off)b

    2

    VGS = VGS(off )/3.4.ID 0.5IDSS

    Rout = aRsbacb || RE

    Rs>bac.RE

    VoutIout

    =VeIe

    =Ve

    bacVe>Rs=

    Rsbac

    Iout = Ie =bacVe

    Rs

    Ib VeRs

    R2 77 Rs,R1 77 Rs

    Ib Ve

    R1 || R2 || Rs

    reVout,IbVs = 0

    Ie bacIb

    Rout =VeIe

    Vout = Ve, Iout = Ie, and Iin = Ib

    Z05_FLOY9868_09_SE_APP2.qxd 1/13/11 3:46 PM Page B-3

  • B-4 APPENDIX B

    Cancelling on each side,

    We want a factor (call it F) by which can be divided to give a value of that willproduce a drain current that is

    Solving for F,

    Therefore, when

    Equation 92

    Rearranging into a standard quadratic equation form,

    The coefficients and constant are

    In simplified notation, the equation is

    The solutions to this quadratic equation are

    Equation 910

    A general model of a switched-capacitor circuit, as shown in Figure B3(a), consists of acapacitor, two voltage sources, and and a two-pole switch. Lets examine this circuitV2,V1

    ID =-B ; 2B2 - 4AC

    2A

    AI2D + BID + C = 0

    C = IDSS

    B = - a1 +2RSIDSSVGS(off)

    b

    A =R2SIDSSV2GS(off)

    aIDSSR2SV2GS(off)

    bI2D - a1 +2IDSSRSVGS(off)

    b ID + IDSS = 0

    = IDSSa1 -2IDRS

    VGS(off)+

    I2DR2SV2GS(off)

    b = IDSS -2IDSSRSVGS(off)

    ID +IDSSR2SV2GS(off)

    I2D

    ID IDSSa1 -IDRS

    VGS(off)b

    2= IDSSa1 -

    IDRSVGS(off)

    b a1 -IDRS

    VGS(off)b

    VGS = VGS(off)/3.4.ID 0.5IDSS

    F =1

    1 - 10.5 3.4

    1F

    = 1 - 10.5

    10.5 - 1 = - 1F

    10.5 = 1 -a

    VGS(off)Fb

    VGS(off)= 1 -

    1F

    0.5 = J1 - aVGS(off)

    Fb

    VGS(off)K

    2

    0.5IDSS.VGSVGS(off)

    0.5 = a1 -VGS

    VGS(off)b

    2

    IDSS

    Z05_FLOY9868_09_SE_APP2.qxd 1/13/11 3:46 PM Page B-4

  • V1 V2

    I1

    C

    1

    0 T/2

    2

    T

    (a)

    Position 1 Position 1 Position 1

    Position 2 Position 2

    0 T

    (b)

    T/2

    FIGURE B3

    for a specified period of time, Assume that and are constant during the time periodand Of particular interest is the average current produced by the source

    during the time period .During the first half of the time period , the switch is in position 1, as indicated in

    Figure B3(b). The capacitor charges very rapidly to the source voltage . Therefore, anaverage current due to is charging the capacitor during the interval from to

    During the second half of the time period, the switch is in position 2, as indicated.Because the capacitor rapidly discharges to the voltage . The average currentproduced by the source over the time period is

    is the charge at and is the charge at Therefore, is the net charge transferred while the switch is in position 1.

    The capacitor voltage at is equal to , and the capacitor voltage at 0 or is equal to. By substituting for in the previous equation,

    Since and are assumed to be constant during , the average current can be expressed as

    Figure B4 shows a conventional resistive circuit with two voltage sources. From Ohmslaw, the current is

    The current I1(avg) in the switched-capacitor circuit is equal to I1 in the resistive circuit.

    By solving for R and canceling the terms,

    As you can see, a switched-capacitor circuit can emulate a resistor with a value deter-mined by the time period T and the capacitance C. Remember that the two-pole switch isin each position for one-half of the time period T and that you can vary T by varying thefrequency at which the switches are operated.

    R =TC

    R =T(V1 - V2)C(V1 - V2)

    V1 - V2

    I1(avg) =C(V1 - V2)

    T=

    V1 - V2R

    I1 =V1 - V2

    R

    I1(avg) =C(V1 - V2)

    T

    TV2V1

    I1(avg) =CV1(T/2) - CV2(0)

    T=

    C1V1(T/2) - V2(0)2T

    QCVV2TV1T/2

    Q1(T/2) - Q1(0)t = T/2.Q1(T/2)t = 0Q1(0)

    I1(avg) =Q1(T/2) - Q1(0)

    T

    TV1V2V1 7 V2,

    t = T/2.t = 0V1I1

    V1T

    TV1I1V1 7 V2.T

    V2V1T.

    V1 V2

    RI1

    FIGURE B4

    DERIVATIONS OF SELECTED EQUATIONS B-5

    Z05_FLOY9868_09_SE_APP2.qxd 1/13/11 3:46 PM Page B-5

  • B-6 APPENDIX B

    Since T = 1/f, the resistance in terms of frequency is

    Equation 101

    An inverting amplifier with feedback capacitance is shown in Figure B5. For the input,

    Factoring out,

    The ratio is the voltage gain,

    I1 =V1(1 + Av)

    XC=

    V1XC > (1 + Av)

    -Av.V2>V1

    I1 =V1(1 - V2 > V1)

    XC

    V1

    I1 =V1 - V2

    XC

    R =1fC

    AvV1 V2

    I2I1

    C FIGURE B5

    The effective reactance as seen from the input terminals is

    or

    Cancelling and inverting,

    Equation 102

    For the output in Figure B6,

    Since

    The effective reactance as seen from the output is

    12pfCout(Miller)

    =1

    2pfC[(Av + 1)>Av]

    XCout(Miller) =XC

    (Av + 1)>Av

    I2 =V2(1 + 1>Av)

    XC=

    V2XC>(1 + 1>Av)

    =V2

    XC>[(Av + 1)>Av]

    V1>V2 = -1>Av,

    I2 =V2 - V1

    XC=

    V2(1 - V1>V2)XC

    Cin(Miller) = C(Av + 1)

    12pfCin(Miller)

    =1

    2pfC(1 + Av)

    XCin(Miller) =XC

    1 + Av

    Z05_FLOY9868_09_SE_APP2.qxd 1/13/11 3:46 PM Page B-6

  • DERIVATIONS OF SELECTED EQUATIONS B-7

    Cancelling and inverting yields

    Equations 1029 and 1030

    The total gain, of an individual amplifier stage at the lower critical frequency equalsthe midrange gain, times the attenuation of the high-pass RC circuit.

    Dividing both sides by any frequency f,

    Since

    Substitution in the gain formula gives

    The gain ratio is

    For a multistage amplifier with n stages, each with the same and gain ratio, the productof the gain ratios is

    The critical frequency of the multistage amplifier is the frequency at whichso the gain ratio at is

    Therefore, for a multistage amplifier,

    So

    Squaring both sides,

    2 = (1 + ( fcl>f cl)2)n

    21/2 = (21 + ( fcl>f cl)2)n

    112

    = c1

    21 + ( fcl>f cl)2dn

    =1

    (21 + ( fcl>f cl)2)n

    Av(tot)Av(mid)

    = 0.707 =1

    1.414=

    112

    f clAv(tot) = 0.707Av(mid),f cl

    a1

    21 + ( fcl>f )2b

    n

    fcl

    Av(tot)Av(mid)

    =1

    21 + ( fcl>f )2

    Av(tot) = Av(mid)a1

    21 + ( fcl>f )2b

    fclf =

Embed Size (px)
Recommended