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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 =
XCR
XC = 1 > 2pfC,
fclf =
1(2pfC)R
fcl =1
2pRC
Av(tot) = Av(mid)aR
2R2 + X2Cb = Av(mid)a
121 + X2C>R2
b
Av(mid),Av(tot),
Cout(Miller) = CaAv + 1
Avb
Z05_FLOY9868_09_SE_APP2.qxd 1/13/11 3:46 PM Page B-7
B-8 APPENDIX B
Taking the nth root of both sides,
A similar process will give Equation 1030:
Equations 1031 and 1032
The rise time is defined as the time required for the voltage to increase from 10 percent ofits final value to 90 percent of its final value, as indicated in Figure B6.
Expressing the curve in its exponential form gives
When
t = 0.1RC
-t
RC= -0.1
ln e- t>RC = ln (0.9) e- t>RC = 0.9
Vfinale- t>RC = 0.9Vfinal 0.1Vfinal = Vfinal(1 - e- t>RC) = Vfinal - Vfinale- t>RC
v = 0.1Vfinal,
v = Vfinal(1 - e- t>RC)
f cu = fcu221>n - 1
f cl =fcl
221>n - 1
afclf clb = 221>n - 1
afclf clb
2= 21>n - 1
21>n = 1 + ( fcl>f cl)2
Vfinal
Vfinal 1 e
0.9 Vfinal
tRC
trt
0.1 Vfinal
0
FIGURE B6
When
t = 2.3RC
-t
RC= -2.3
ln e- t>RC = ln (0.1) Vfinale- t>RC = 0.1Vfinal
0.9Vfinal = Vfinal(1 - e- t>RC) = Vfinal - Vfinale- t>RCv = 0.9Vfinal,
Z05_FLOY9868_09_SE_APP2.qxd 1/13/11 3:46 PM Page B-8
DERIVATIONS OF SELECTED EQUATIONS B-9
The difference is the rise time.
The critical frequency of an RC circuit is
Substituting,
In a similar way, it can be shown that
Equation 1221
The formula for open-loop gain in Equation 1219 can be expressed in complex notation as
Substituting the above expression into the equation gives a formulafor the total closed-loop gain.
Multiplying the numerator and denominator by yields
Dividing the numerator and denominator by gives
The above expression is of the form of the first equation
where is the closed-loop critical frequency. Thus,
Equation 141
In Figure B7 the common-mode voltage, , on the noninverting input is amplified bythe small common-mode gain of op-amp A1. ( is typically less than 1.) The total outputvoltage of op-amp A1 is
Vout1 = a1 +R1RGbVin1 - a
R1RGbVin2 + Vcm
AcmVcm
fc(cl) = fc(ol)(1 + BAol(mid))fc(cl)
Acl =Acl(mid)
1 + jf>fc(cl)
Acl =Aol(mid)>(1 + BAol(mid))
1 + j[ f>( fc(ol)(1 + BAol(mid)))]
1 + BAol(mid)
Acl =Aol(mid)
1 + BAol(mid) + jf>fc(ol)
1 + jf>fc(ol)
Acl =Aol(mid)>(1 + jf>fc(ol))
1 + BAol(mid)>(1 + jf>fc(ol))
Acl = Aol>(1 + BAol)
Aol =Aol(mid)
1 + jf>fc(ol)
fcl =0.35
tf
fcu =0.35
tr
tr =2.2
2pfcu=
0.35fcu
RC =1
2pfc
fc =1
2pRC
tr = 2.3RC - 0.1RC = 2.2RC
Z05_FLOY9868_09_SE_APP2.qxd 1/13/11 3:46 PM Page B-9
B-10 APPENDIX B
A similar analysis can be applied to op-amp A2 and results in the following outputexpression:
Vout2 = a1 +R2RGbVin2 - a
R2RGbVin1 + Vcm
Vin1 + Vcm R3+
+
R1
R5
+
R2
A1
A2
A3
R4
R6
RG
Vout1
Vout2
Vout = Acl(Vin2 Vin1)
Vin2 + Vcm
FIGURE B7
Op-amp A3 has on one of its inputs and on the other. Therefore, the differ-ential input voltage to op-amp A3 is
For
Notice that, since the common-mode voltages are equal, they cancel each other.Factoring out the differential gain gives the following expression for the differential inputto op-amp A3:
Op-amp A3 has unity gain because R3 = R5 = R4 = R6 and Av = R5/R3. = R6/R4. Therefore,the final output of the instrumentation amplifier (the output of op-amp A3) is
The closed-loop gain is
Equation 161
VoutVin
=R(- jX)>(R - jX)
(R - jX) + R(- jX)>(R - jX) =R(- jX)
(R - jX)2 - jRX
Acl = 1 +2RRG
Acl =Vout
Vin2 - Vin1
Vout = 1(Vout2 - Vout1) = a1 +2RRGb (Vin2 - Vin1)
Vout2 - Vout1 = a1 +2RRGb (Vin2 - Vin1)
(Vcm)
Vout2 - Vout1 = a1 +2RRGbVin2 - a1 +
2RRGbVin1 + Vcm - Vcm
R1 = R2 = R,
Vout2 - Vout1 = a1 +R2RG
+R1RGbVin2 - a1 +
R2RG
+R1RGbVin1 + Vcm - Vcm
Vout2 - Vout1.Vout2Vout1
Z05_FLOY9868_09_SE_APP2.qxd 1/13/11 3:46 PM Page B-10
DERIVATIONS OF SELECTED EQUATIONS B-11
Multiplying the numerator and denominator by j,
For a phase angle there can be no j term. Recall from complex numbers in ac theorythat a nonzero angle is associated with a complex number having a j term. Therefore, at the j term is 0.
Thus,
Cancelling yields
Equation 162
From the derivation of Equation 161,
Equations 163 and 164
The feedback circuit in the phase-shift oscillator consists of three RC stages, as shown inFigure B8. An expression for the attenuation is derived using the mesh analysis methodfor the loop assignment shown. All Rs are equal in value, and all Cs are equal in value.
0I1 - RI2 + (2R - j1>2pfC)I3 = 0 -RI1 + (2R - j1>2pfC)I2 - RI3 = 0
(R - j1>2pfC)I1 - RI2 + 0I3 = Vin
fr =1
2pRC
R =1
2pfrC
Since X =1
2pfrC,
R = X R2 = X2
R2 - X2 = 0
VoutVin
=13
VoutVin
=RX3RX
R2 - X2 = 0
fr0
=RX
RX + jR2 + 2RX - jX2 =RX
3RX + j(R2 - X2)
VoutVin
=RX
j(R - jX)2 + RX =RX
RX + j(R2 - j2RX - X2)
RRRVin VoutI2 I3I1
C C C
FIGURE B8
Z05_FLOY9868_09_SE_APP2.qxd 1/13/11 3:47 PM Page B-11
B-12 APPENDIX B
In order to get we must solve for using determinants:
Expanding and combining the real terms and the j terms separately.
For oscillation in the phase-shift amplifier, the phase shift through the RC circuit must equalFor this condition to exist, the j term must be 0 at the frequency of oscillation
Since the j term is 0,
The negative sign results from the inversion. Thus, the value of attenuation for thefeedback circuit is
B =129
180
VoutVin
=1
1 -5
4p2f2r R2C2=
1
1 -5
a1
16RCb
2R2C2
=1
1 - 30= -
129
fr =1
2p16RC
f2r =1
6(2p)2R2C2
6(2p)2f2r R2C2 - 1 = 0
6(2p)2f2r R2C2 - 1(2p)3f3r R3C3
= 0
62pfrRC
-1
(2pfr)3R3C3= 0
fr.180.
VoutVin
=1
a1 -5
4p2f2R2C2 b - ja6
2pfRC -1
(2pf )3R3C3 b
VoutVin
=1
(1 - j1>2pfRC) (2 - j1>2pfRC)2 - (3 - j1>2pfRC)
=R3
R3(1 - j1>2pfRC) (2 - j1>2pfRC)2 - R3(3 - j1>2pfRC)
=R3
R3(1 - j1>2pfRC)(2 - j1>2pfRC)2 - R3[(2 - j1>2pfRC) - (1 - j1>2pfRC)]
=R3
(R - j1>2pfC)(2R - j1>2pfC)2 - R3(2 - j1>2pfRC) - R3(1 - 1>2pfRC)
VoutVin
=RI3Vin
I3 =R2Vin
(R - j1>2pfC)(2R - j1>2pfC)2 - R2(2R - j1>2pfC) - R2(R - 1>2pfC)
I3 =
3
(R - j1>2pfC) -R Vin-R (2R - j1>2pfC) 00 -R 0
33 (R - j1>2pfC) -R 0-R (2R - j1>2pfC) -R
0 -R (2R - j1>2pfC)3
I3Vout,
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