D ERIVATIONS
OF
S ELECTED
E QUATIONS
Equation 2–3
The average value of a half-wave rectified sine wave is the area under the curve divided by
the period (2p). The equation for a sine wave is
v = Vpsin u
p
Vp
area
1
=
Vpsin u du =
(-cos u)|p0
2p
2p L0
2p
Vp
Vp
Vp
=
3-cos p - (-cos 0)4 =
3-(-1) - (-1)4 =
(2)
2p
2p
2p
Vp
=
p
VAVG =
VAVG
Equation 2–12
Refer to Figure B–1.
vC = Vp(rect)e
–
t
RLC
Vp(rect)
0
Vr( pp)
tdis
T
䊱
FIGURE B–1
When the filter capacitor discharges through RL, the voltage is
vC = Vp(rect)e - t>RLC
Since the discharge time of the capacitor is from one peak to approximately the next peak,
tdis ⬵ T when vC reaches its minimum value.
vC(min) = Vp(rect)e - T>RLC
Since RC W T, T/RLC becomes much less than 1 (which is usually the case); e - T>RLC
approaches 1 and can be expressed as
e - T>RLC ⬵ 1 -
T
RLC
Appendix
B
B-2
◆
A PPENDIX B
Therefore,
vC(min) = Vp(rect) a1 -
T
b
RLC
The peak-to-peak ripple voltage is
Vr(pp) = Vp(rect) - VC(min) = Vp(rect) - Vp(rect) +
Vr(pp) ⬵ a
Vp(rect)T
RLC
Vp(rect)T
=
RLC
1
bV
fRLC p(rect)
Equation 2–13
To obtain the dc value, one-half of the peak-to-peak ripple is subtracted from the peak value.
VDC = Vp(rect) VDC = a1 -
Vr(pp)
2
= Vp(rect) - a
1
bV
2fRLC p(rect)
1
bV
2fRLC p(rect)
Equation 6–1
The Shockley equation for the base-emitter pn junction is
IE = IR(eVQ>kT - 1)
where IE
IR
V
Q
k
T
=
=
=
=
=
=
the total forward current across the base-emitter junction
the reverse saturation current
the voltage across the depletion layer
the charge on an electron
a number known as Boltzmann’s constant
the absolute temperature
At ambient temperature, Q>kT ⬵ 40, so
IE = IR(eV40 - 1)
Differentiating yields
dIE
= 40IReV40
dV
Since IReV40 = IE + IR,
dIE
= 40(IE + IR)
dV
Assuming IR 6 6 IE,
dIE
⬵ 40IE
dV
The ac resistance r¿e of the base-emitter junction can be expressed as dV>dIE.
r¿e =
dV
1
25 mV
⬵
⬵
dIE
40IE
IE
Equation 6–14
The emitter-follower is represented by the r parameter ac equivalent circuit in Figure B–2(a).
D ERIVATIONS
C
R1
βac Ib
βac Ib
Rs || R1 || R2
B
Rs
r ′e
r ′e
E
Vs
R2
RE
Ie
(a)
䊱
RE
Ve = Vout
(b)
FIGURE B–2
Conventional current direction shown.
By thevenizing from the base back to the source, the circuit is simplified to the form
shown in Figure B–2(b). Vout = Ve, Iout = Ie, and Iin = Ib.
Ve
Ie
Ie ⬵ b acIb
Rout =
With Vs = 0 and with Ib produced by Vout, and neglecting the base-to-emitter voltage drop
(and therefore r¿e),
Ib ⬵
Ve
R1 || R2 || Rs
Assuming that R1 7 7 Rs and R2 7 7 Rs,
Ib ⬵
Ve
Rs
b acVe
Rs
Ve
Ve
Rs
=
=
=
Ie
b acVe>Rs
b ac
Iout = Ie =
Vout
Iout
Looking into the emitter, RE appears in parallel with Rs>b ac. Therefore,
Rout = a
Rs
b || RE
b ac
Midpoint Bias (Chapter 8)
The following proof is for the equation on page 400 that shows ID ⬵ 0.5IDSS when
VGS = VGS(off )/3.4.
Start with Equation 8–1:
ID ⬵ IDSS a1 -
VGS
VGS(off)
b
2
Let ID = 0.5IDSS.
0.5IDSS = IDSS a1 -
VGS
VGS(off)
b
2
OF
S ELECTED E QUATIONS
◆
B-3
B-4
◆
A PPENDIX B
Cancelling IDSS on each side,
0.5 = a1 -
VGS
VGS(off)
b
2
We want a factor (call it F) by which VGS(off) can be divided to give a value of VGS that will
produce a drain current that is 0.5IDSS.
0.5 = J 1 -
Solving for F,
10.5 = 1 -
a
a
VGS(off)
F
b
VGS(off)
VGS(off)
F
b
K
= 1 -
VGS(off)
10.5 - 1 = -
2
1
F
1
F
1
= 1 - 10.5
F
1
F =
⬵ 3.4
1 - 10.5
Therefore, ID ⬵ 0.5IDSS when VGS = VGS(off)/3.4.
Equation 9–2
ID ⬵ IDSS a1 = IDSS a1 -
IDRS 2
IDRS
IDRS
b = IDSS a1 b a1 b
VGS(off)
VGS(off)
VGS(off)
2IDRS
I2DR2S
2IDSSRS
IDSSR2S 2
+ 2
b = IDSS ID + 2
ID
VGS(off)
VGS(off)
VGS(off)
VGS(off)
Rearranging into a standard quadratic equation form,
a
IDSSR2S
V2GS(off)
bI2D - a1 +
2IDSSRS
b I + IDSS = 0
VGS(off) D
The coefficients and constant are
A =
R2SIDSS
V2GS(off)
B = - a1 +
2RSIDSS
b
VGS(off)
C = IDSS
In simplified notation, the equation is
AI2D + BID + C = 0
The solutions to this quadratic equation are
ID =
-B ; 2B2 - 4AC
2A
Equation 9–10
A general model of a switched-capacitor circuit, as shown in Figure B–3(a), consists of a
capacitor, two voltage sources, V1 and V2, and a two-pole switch. Let’s examine this circuit
D ERIVATIONS
OF
S ELECTED E QUATIONS
B-5
◆
0 T/2 T
I1
1
V1
Position 1
2
C
V2
Position 2
T/2
0
(a)
䊱
Position 1
Position 1
Position 2
T
(b)
FIGURE B–3
for a specified period of time, T. Assume that V1 and V2 are constant during the time period
T and V1 7 V2. Of particular interest is the average current I1 produced by the source V1
during the time period T.
During the first half of the time period T, the switch is in position 1, as indicated in
Figure B–3(b). The capacitor charges very rapidly to the source voltage V1. Therefore, an
average current I1 due to V1 is charging the capacitor during the interval from t = 0 to
t = T/2. During the second half of the time period, the switch is in position 2, as indicated.
Because V1 7 V2, the capacitor rapidly discharges to the voltage V2. The average current
produced by the source V1 over the time period T is
I1(avg) =
Q1(T/2) - Q1(0)
T
Q1(0) is the charge at t = 0 and Q1(T/2) is the charge at t = T/2. Therefore, Q1(T/2) - Q1(0)
is the net charge transferred while the switch is in position 1.
The capacitor voltage at T/2 is equal to V1, and the capacitor voltage at 0 or T is equal to
V2. By substituting CV for Q in the previous equation,
I1(avg) =
CV1(T/2) - CV2(0)
T
=
C1V1(T/2) - V2(0)2
T
Since V1 and V2 are assumed to be constant during T, the average current can be expressed as
I1(avg) =
C(V1 - V2)
T
Figure B–4 shows a conventional resistive circuit with two voltage sources. From Ohm’s
law, the current is
I1 =
I1
V1
V1 - V2
R
The current I1(avg) in the switched-capacitor circuit is equal to I1 in the resistive circuit.
I1(avg) =
C(V1 - V2)
V1 - V2
=
T
R
By solving for R and canceling the V1 - V2 terms,
T(V1 - V2)
C(V1 - V2)
T
R =
C
R =
As you can see, a switched-capacitor circuit can emulate a resistor with a value determined by the time period T and the capacitance C. Remember that the two-pole switch is
in each position for one-half of the time period T and that you can vary T by varying the
frequency at which the switches are operated.
䊱
F I G U R E B –4
R
V2
B-6
◆
A PPENDIX B
Since T = 1/f, the resistance in terms of frequency is
R =
1
fC
Equation 10–1
An inverting amplifier with feedback capacitance is shown in Figure B–5. For the input,
I1 =
Factoring V1 out,
I1 =
V1 - V2
XC
V1(1 - V2 > V1)
XC
The ratio V2>V1 is the voltage gain, -Av.
I1 =
䊳
V1(1 + Av)
V1
=
XC
XC > (1 + Av)
C
FIGURE B–5
I1
I2
Av
V1
The effective reactance as seen from the input terminals is
XCin(Miller) =
XC
1 + Av
or
1
1
=
2pfCin(Miller)
2pfC(1 + Av)
Cancelling and inverting,
Cin(Miller) = C(Av + 1)
Equation 10–2
For the output in Figure B–6,
I2 =
Since V1>V2 = -1>Av,
I2 =
V2(1 - V1>V2)
V2 - V1
=
XC
XC
V2(1 + 1>Av)
XC
=
V2
V2
=
XC>(1 + 1>Av)
XC>[(Av + 1)>Av]
The effective reactance as seen from the output is
XC
(Av + 1)>Av
1
=
2pfC[(Av + 1)>Av]
XCout(Miller) =
1
2pfCout(Miller)
V2
D ERIVATIONS
Cancelling and inverting yields
Cout(Miller) = C a
Av + 1
b
Av
Equations 10–29 and 10–30
The total gain, Av(tot), of an individual amplifier stage at the lower critical frequency equals
the midrange gain, Av(mid), times the attenuation of the high-pass RC circuit.
Av(tot) = Av(mid) a
fcl =
R
2R +
2
X2C
b = Av(mid) a
1
21 + X2C>R2
b
1
2pRC
Dividing both sides by any frequency f,
fcl
1
=
f
(2pfC)R
Since XC = 1 > 2pfC,
fcl
XC
=
f
R
Substitution in the gain formula gives
Av(tot) = Av(mid) a
1
21 + ( fcl>f )2
b
The gain ratio is
Av(tot)
Av(mid)
1
=
21 + ( fcl>f )2
For a multistage amplifier with n stages, each with the same fcl and gain ratio, the product
of the gain ratios is
a
1
21 + ( fcl>f )2
b
n
The critical frequency f cl
¿ of the multistage amplifier is the frequency at which
Av(tot) = 0.707Av(mid), so the gain ratio at f cl
¿ is
Av(tot)
Av(mid)
= 0.707 =
1
1
=
1.414
12
Therefore, for a multistage amplifier,
n
1
1
1
= c
d =
12
21 + ( fcl>f c¿ l)2
(21 + ( fcl>f c¿ l)2)n
So
21/2 = (21 + ( fcl>f cl
¿ )2)n
Squaring both sides,
2 = (1 + ( fcl>f cl
¿ )2)n
OF
S ELECTED E QUATIONS
◆
B-7
B-8
◆
A PPENDIX B
Taking the nth root of both sides,
21>n = 1 + ( fcl>f cl
¿ )2
fcl 2
b = 21>n - 1
f cl
¿
fcl
a b = 221>n - 1
f cl
¿
fcl
f cl
¿ =
1>n
22
- 1
a
A similar process will give Equation 10–30:
f cu
¿ = fcu 221>n - 1
Equations 10–31 and 10–32
The rise time is defined as the time required for the voltage to increase from 10 percent of
its final value to 90 percent of its final value, as indicated in Figure B–6.
Expressing the curve in its exponential form gives
v = Vfinal(1 - e - t>RC)
When v = 0.1Vfinal,
0.1Vfinal = Vfinal(1 - e - t>RC) = Vfinal - Vfinale - t>RC
Vfinale - t>RC = 0.9Vfinal
e - t>RC = 0.9
ln e - t>RC = ln (0.9)
t
= -0.1
RC
t = 0.1RC
䊳
FIGURE B–6
Vfinal
0.9 Vfinal
Vfinal 1 – e
–
t
RC
0.1 Vfinal
0
tr
When v = 0.9Vfinal,
0.9Vfinal = Vfinal(1 - e - t>RC) = Vfinal - Vfinale - t>RC
Vfinale - t>RC = 0.1Vfinal
ln e - t>RC = ln (0.1)
t
= -2.3
RC
t = 2.3RC
t
D ERIVATIONS
The difference is the rise time.
tr = 2.3RC - 0.1RC = 2.2RC
The critical frequency of an RC circuit is
1
2pRC
1
RC =
2pfc
fc =
Substituting,
2.2
0.35
=
2pfcu
fcu
0.35
=
tr
tr =
fcu
In a similar way, it can be shown that
fcl =
0.35
tf
Equation 12–21
The formula for open-loop gain in Equation 12–19 can be expressed in complex notation as
Aol =
Aol(mid)
1 + jf>fc(ol)
Substituting the above expression into the equation Acl = Aol>(1 + BAol) gives a formula
for the total closed-loop gain.
Acl =
Aol(mid)>(1 + jf>fc(ol))
1 + BAol(mid)>(1 + jf>fc(ol))
Multiplying the numerator and denominator by 1 + jf>fc(ol) yields
Acl =
Aol(mid)
1 + BAol(mid) + jf>fc(ol)
Dividing the numerator and denominator by 1 + BAol(mid) gives
Acl =
Aol(mid)>(1 + BAol(mid))
1 + j[ f>( fc(ol)(1 + BAol(mid)))]
The above expression is of the form of the first equation
Acl =
Acl(mid)
1 + jf>fc(cl)
where fc(cl) is the closed-loop critical frequency. Thus,
fc(cl) = fc(ol)(1 + BAol(mid))
Equation 14–1
In Figure B–7 the common-mode voltage, Vcm, on the noninverting input is amplified by
the small common-mode gain of op-amp A1. (Acm is typically less than 1.) The total output
voltage of op-amp A1 is
Vout1 = a1 +
R1
R1
bV - a bVin2 + Vcm
RG in1
RG
OF
S ELECTED E QUATIONS
◆
B-9
B-10
◆
A PPENDIX B
A similar analysis can be applied to op-amp A2 and results in the following output
expression:
Vout2 = a1 +
Vin1 + Vcm
+
Vout1 R3
A1
–
R2
R2
bV - a bVin1 + Vcm
RG in2
RG
R1
–
R2
RG
+
A2
Vin2 + Vcm
Vout = Acl (Vin2 – Vin1)
A3
–
䊱
R5
+
R4
Vout2
R6
FIGURE B–7
Op-amp A3 has Vout1 on one of its inputs and Vout2 on the other. Therefore, the differential input voltage to op-amp A3 is Vout2 - Vout1.
Vout2 - Vout1 = a1 +
R2
R1
R2
R1
+
bVin2 - a1 +
+
bV + Vcm - Vcm
RG
RG
RG
RG in1
For R1 = R2 = R,
Vout2 - Vout1 = a1 +
2R
2R
bV - a1 +
bV + Vcm - Vcm
RG in2
RG in1
Notice that, since the common-mode voltages (Vcm) are equal, they cancel each other.
Factoring out the differential gain gives the following expression for the differential input
to op-amp A3:
Vout2 - Vout1 = a1 +
2R
b(Vin2 - Vin1)
RG
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
Vout = 1(Vout2 - Vout1) = a1 +
2R
b(Vin2 - Vin1)
RG
The closed-loop gain is
Vout
Vin2 - Vin1
2R
Acl = 1 +
RG
Acl =
Equation 16–1
R(-jX)>(R - jX)
Vout
R(-jX)
=
=
Vin
(R - jX) + R(-jX)>(R - jX)
(R - jX)2 - jRX
D ERIVATIONS
Multiplying the numerator and denominator by j,
Vout
RX
RX
=
=
Vin
j(R - jX)2 + RX
RX + j(R2 - j2RX - X2)
RX
RX
=
=
2
2
RX + jR + 2RX - jX
3RX + j(R2 - X2)
For a 0° phase angle there can be no j term. Recall from complex numbers in ac theory
that a nonzero angle is associated with a complex number having a j term. Therefore, at fr
the j term is 0.
R2 - X2 = 0
Thus,
Vout
RX
=
Vin
3RX
Cancelling yields
Vout
1
=
Vin
3
Equation 16–2
From the derivation of Equation 16–1,
R2 - X2 = 0
R2 = X2
R = X
Since X =
1
,
2pfrC
1
2pfrC
1
fr =
2pRC
R =
Equations 16–3 and 16–4
The feedback circuit in the phase-shift oscillator consists of three RC stages, as shown in
Figure B–8. An expression for the attenuation is derived using the mesh analysis method
for the loop assignment shown. All Rs are equal in value, and all Cs are equal in value.
(R - j1>2pfC)I1 - RI2 + 0I3 = Vin
-RI1 + (2R - j1>2pfC)I2 - RI3 = 0
0I1 - RI2 + (2R - j1>2pfC)I3 = 0
C
Vin
䊱
FIGURE B–8
C
I1
R
C
I2
R
I3
R
Vout
OF
S ELECTED E QUATIONS
◆
B-11
B-12
◆
A PPENDIX B
In order to get Vout, we must solve for I3 using determinants:
3
(R - j1>2pfC)
-R
0
-R
(2R - j1>2pfC)
-R
Vin
0
0
3
(R - j1>2pfC)
-R
0
-R
(2R - j1>2pfC)
-R
0
3
-R
(2R - j1>2pfC)
I3 =
I3 =
3
R2Vin
(R - j1>2pfC)(2R - j1>2pfC)2 - R2(2R - j1>2pfC) - R2(R - 1>2pfC)
Vout
RI3
=
Vin
Vin
=
R3
(R - j1>2pfC)(2R - j1>2pfC)2 - R3(2 - j1>2pfRC) - R3(1 - 1>2pfRC)
=
R3
R3(1 - j1>2pfRC)(2 - j1>2pfRC)2 - R3[(2 - j1>2pfRC) - (1 - j1>2pfRC)]
=
R3
R3(1 - j1>2pfRC) (2 - j1>2pfRC)2 - R3(3 - j1>2pfRC)
Vout
1
=
Vin
(1 - j1>2pfRC) (2 - j1>2pfRC)2 - (3 - j1>2pfRC)
Expanding and combining the real terms and the j terms separately.
Vout
=
Vin
1
5
6
1
a1 b - ja
b
2 2 2 2
2pfRC
4p f R C
(2pf )3R3C 3
For oscillation in the phase-shift amplifier, the phase shift through the RC circuit must equal
180°. For this condition to exist, the j term must be 0 at the frequency of oscillation fr.
6
1
= 0
2pfrRC
(2pfr)3R3C3
6(2p)2fr2R2C2 - 1
(2p)3fr3R3C3
= 0
6(2p)2fr2R2C2 - 1 = 0
1
6(2p)2R2C2
1
fr =
2p16RC
fr2 =
Since the j term is 0,
Vout
=
Vin
1
5
1 2 2 2 2
4p fr R C
1
=
1 -
5
a
=
1
1
= 1 - 30
29
2
1
b R2C2
16RC
The negative sign results from the 180° inversion. Thus, the value of attenuation for the
feedback circuit is
B =
1
29