electronic-devices-9th-edition-by-floyd floyd ed9 part5-derivations of selected
TRANSCRIPT
Appendix
BDERIVATIONS OF SELECTED
EQUATIONS
Equation 2–3
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 2–12
Refer to Figure B–1.
VAVG =
Vp
p
=
Vp
2p 3-cos p - (-cos 0)4 =
Vp
2p 3-(-1) - (-1)4 =
Vp
2p (2)
VAVG =
area
2p=
1
2pL
p
0Vpsin u du =
Vp
2p (-cos u)|p0
v = Vpsin u
(2p).
tdis
T
0
Vp(rect) Vr(pp)
vC = Vp(rect)et
RLC–
� FIGURE B–1
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>RLC
vCtdis � T
vC = Vp(rect)e- t>RLC
RL,
Z05_FLOY9868_09_SE_APP2.qxd 1/13/11 3:46 PM Page B-1
Therefore,
The peak-to-peak ripple voltage is
Equation 2–13
To obtain the dc value, one-half of the peak-to-peak ripple is subtracted from the peak value.
Equation 6–1
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 Boltzmann’s constantabsolute temperature
At ambient temperature, so
Differentiating yields
Since
Assuming
The ac resistance of the base-emitter junction can be expressed as
Equation 6–14
The emitter-follower is represented by the r parameter ac equivalent circuit in Figure B–2(a).
r¿e =
dV
dIE�
1
40IE�
25 mV
IE
dV>dIE.r¿e
dIE
dV� 40IE
IR 6 6 IE,
dIE
dV= 40(IE + IR)
IReV40= IE + IR,
dIE
dV= 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
1
2fRLCbVp(rect)
Vr(pp) � a1
fRLCbVp(rect)
Vr(pp) = Vp(rect) - VC(min) = Vp(rect) - Vp(rect) +
Vp(rect)T
RLC=
Vp(rect)T
RLC
vC(min) = Vp(rect)a1 -
T
RLCb
B-2 ◆ APPENDIX B
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DERIVATIONS OF SELECTED EQUATIONS ◆ B-3
r ′e
B
RE
Rs
R2
R1
C
E
βacIb
Vs
(a)
r ′e
RE
Rs || R1 || R2
βacIb
(b)
Ve = VoutIe
� FIGURE B–2
Conventional current direction shown.
By thevenizing from the base back to the source, the circuit is simplified to the formshown in Figure B–2(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 8–1:
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 = aRs
bacb || RE
Rs>bac.RE
Vout
Iout=
Ve
Ie=
Ve
bacVe>Rs=
Rs
bac
Iout = Ie =
bacVe
Rs
Ib �Ve
Rs
R2 7 7 Rs,R1 7 7 Rs
Ib �Ve
R1 || R2 || Rs
r¿e
Vout,IbVs = 0
Ie � bacIb
Rout =
Ve
Ie
Vout = Ve, Iout = Ie, and Iin = Ib
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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 9–2
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 9–10
A general model of a switched-capacitor circuit, as shown in Figure B–3(a), consists of acapacitor, two voltage sources, and and a two-pole switch. Let’s examine this circuitV2,V1
ID =
-B ; 2B2- 4AC
2A
AI2D + BID + C = 0
C = IDSS
B = - a1 +
2RSIDSS
VGS(off)b
A =
R2SIDSS
V2GS(off)
aIDSSR2
S
V2GS(off)
bI2D - a1 +
2IDSSRS
VGS(off)b ID + IDSS = 0
= IDSSa1 -
2IDRS
VGS(off)+
I2DR2
S
V2GS(off)
b = IDSS -
2IDSSRS
VGS(off)ID +
IDSSR2S
V2GS(off)
I2D
ID � IDSSa1 -
IDRS
VGS(off)b
2
= IDSSa1 -
IDRS
VGS(off)b a1 -
IDRS
VGS(off)b
VGS = VGS(off)/3.4.ID � 0.5IDSS
F =
1
1 - 10.5� 3.4
1
F= 1 - 10.5
10.5 - 1 = - 1
F
10.5 = 1 -
aVGS(off)
Fb
VGS(off)= 1 -
1
F
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 B–3
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 B–3(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 B–4 shows a conventional resistive circuit with two voltage sources. From Ohm’slaw, 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 =
T
C
R =
T(V1 - V2)
C(V1 - V2)
V1 - V2
I1(avg) =
C(V1 - V2)
T=
V1 - V2
R
I1 =
V1 - V2
R
I1(avg) =
C(V1 - V2)
T
TV2V1
I1(avg) =
CV1(T/2) - CV2(0)
T=
C1V1(T/2) - V2(0)2
T
QCVV2
TV1T/2
Q1(T/2) - Q1(0)t = T/2.Q1(T/2)t = 0Q1(0)
I1(avg) =
Q1(T/2) - Q1(0)
T
TV1
V2V1 7 V2,t = T/2.
t = 0V1I1
V1
TT
V1I1V1 7 V2.TV2V1T.
V1 V2
RI1
� FIGURE B–4
DERIVATIONS OF SELECTED EQUATIONS ◆ B-5
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B-6 ◆ APPENDIX B
Since T = 1/f, the resistance in terms of frequency is
Equation 10–1
An inverting amplifier with feedback capacitance is shown in Figure B–5. For the input,
Factoring out,
The ratio is the voltage gain,
I1 =
V1(1 + Av)
XC=
V1
XC > (1 + Av)
-Av.V2>V1
I1 =
V1(1 - V2 > V1)
XC
V1
I1 =
V1 - V2
XC
R =
1
fC
AvV1 V2
I2I1
C� FIGURE B–5
The effective reactance as seen from the input terminals is
or
Cancelling and inverting,
Equation 10–2
For the output in Figure B–6,
Since
The effective reactance as seen from the output is
1
2pfCout(Miller)=
1
2pfC[(Av + 1)>Av]
XCout(Miller)=
XC
(Av + 1)>Av
I2 =
V2(1 + 1>Av)
XC=
V2
XC>(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)
1
2pfCin(Miller)=
1
2pfC(1 + Av)
XCin(Miller)=
XC
1 + Av
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DERIVATIONS OF SELECTED EQUATIONS ◆ B-7
Cancelling and inverting yields
Equations 10–29 and 10–30
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
1
12= c
1
21 + ( fcl>f ¿cl)2dn
=
1
(21 + ( fcl>f ¿cl)2)n
Av(tot)
Av(mid)= 0.707 =
1
1.414=
1
12
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
fcl
f=
XC
R
XC = 1 > 2pfC,
fcl
f=
1
(2pfC)R
fcl =
1
2pRC
Av(tot) = Av(mid)aR
2R2+ X2
C
b = Av(mid)a1
21 + X2C>R
2b
Av(mid),Av(tot),
Cout(Miller) = CaAv + 1
Avb
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B-8 ◆ APPENDIX B
Taking the nth root of both sides,
A similar process will give Equation 10–30:
Equations 10–31 and 10–32
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 B–6.
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
afcl
f ¿clb = 221>n
- 1
afcl
f ¿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 B–6
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>RC
v = 0.9Vfinal,
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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 12–21
The formula for open-loop gain in Equation 12–19 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 14–1
In Figure B–7 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 +
R1
RGbVin1 - a
R1
RGbVin2 + Vcm
Acm
Vcm
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=
0.35
fcu
RC =
1
2pfc
fc =
1
2pRC
tr = 2.3RC - 0.1RC = 2.2RC
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B-10 ◆ APPENDIX B
A similar analysis can be applied to op-amp A2 and results in the following outputexpression:
Vout2 = a1 +
R2
RGbVin2 - a
R2
RGbVin1 + Vcm
Vin1 + Vcm R3+
–
–
+
R1
R5
–
+
R2
A1
A2
A3
R4
R6
RG
Vout1
Vout2
Vout = Acl(Vin2 – Vin1)
Vin2 + Vcm
� FIGURE B–7
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 16–1
Vout
Vin=
R(- jX)>(R - jX)
(R - jX) + R(- jX)>(R - jX )=
R(- jX )
(R - jX)2- jRX
Acl = 1 +
2R
RG
Acl =
Vout
Vin2 - Vin1
Vout = 1(Vout2 - Vout1) = a1 +
2R
RGb (Vin2 - Vin1)
Vout2 - Vout1 = a1 +
2R
RGb (Vin2 - Vin1)
(Vcm)
Vout2 - Vout1 = a1 +
2R
RGbVin2 - a1 +
2R
RGbVin1 + Vcm - Vcm
R1 = R2 = R,
Vout2 - Vout1 = a1 +
R2
RG+
R1
RGbVin2 - a1 +
R2
RG+
R1
RGbVin1 + Vcm - Vcm
Vout2 - Vout1.Vout2Vout1
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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 16–2
From the derivation of Equation 16–1,
Equations 16–3 and 16–4
The feedback circuit in the phase-shift oscillator consists of three RC stages, as shown inFigure B–8. 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
Vout
Vin=
1
3
Vout
Vin=
RX
3RX
R2- X2
= 0
fr
0°
=
RX
RX + jR2+ 2RX - jX2 =
RX
3RX + j(R2- X2)
Vout
Vin=
RX
j(R - jX)2+ RX
=
RX
RX + j(R2- j2RX - X2)
RRRVin Vout
I2 I3I1
C C C
� FIGURE B–8
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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 =
1
29
180°
Vout
Vin=
1
1 -
5
4p2f2r R2C2
=
1
1 -
5
a1
16RCb
2
R2C2
=
1
1 - 30= -
1
29
fr =
1
2p16RC
f2r =
1
6(2p)2R2C2
6(2p)2f2r R2C2
- 1 = 0
6(2p)2f2
r R2C2- 1
(2p)3f3r R3C3 = 0
6
2pfrRC-
1
(2pfr)3R3C3 = 0
fr.180°.
Vout
Vin=
1
a1 -
5
4p2f2R2C2 b - ja6
2pfRC-
1
(2pf )3R3C3 b
Vout
Vin=
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)
Vout
Vin=
RI3
Vin
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) 0
0 -R 0
33 (R - j1>2pfC) -R 0
-R (2R - j1>2pfC) -R
0 -R (2R - j1>2pfC)
3
I3Vout,
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