electronic-devices-9th-edition-by-floyd Floyd ed9 part5-derivations of selected

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


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


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


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



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



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


B-8 ◆ APPENDIX B

Taking the nth root of both sides,

A similar process will give Equation 10–30:


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,



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


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)


2R

RGbVin2 - a1 +

2R

RGbVin1 + Vcm - Vcm

R1 = R2 = R,


R2

RG+

R1

RGbVin2 - a1 +

R2

RG+

R1

RGbVin1 + Vcm - Vcm

Vout2 - Vout1.Vout2Vout1



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,


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


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,


electronic-devices-9th-edition-by-floyd Floyd ed9 part5-derivations of selected

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