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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 =

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