MathCAD Listing, Steady-State Current-Sensing Output

Appendix D

MathCAD Listing, Steady-State Current-Sensing Output

K: = 10^ u: = 10"^

m:= 10-' 365

366

APPENDIXD. MATHCAD. STEADY-STATE CURRENT-SENSING OUTPUT

n:=10-'' p:= 10"'^ Fs:=50-K Vg: = 14 Vo:=5 L: = 37.5-u C: = 380-u Rc: = 20«m Ri: = 0.1 Rl: = 2.2-K Rf: = 30.5'K Cfz: = 5.8-n Cfp: = 220«p

-

^

Load resistance: R:=l Control voltage: Vc: = 0.5

8,: = Y8^.d.T

APPENDIX D. MA THCAD, STEAD Y-STA TE CURRENT-SENSING OUTPUT

T 1 : = Cfp<

Rl.Rf Rl+Rf

Ib:=IL4 2

T2: = Rf-Cfp T3:=R1-Cfp T4:=Rf.Cfz Sn: = V „ - V „ L Let ton: = d«T tof:=(l-J)-r Current slope alone: wp: coz:

(ox:

coy:

(oh:

1

1

T4"^T2

1

1

1

TI

T4

1 "T3

1 ~T2

367

368

APPENDIX D. MA THCAD, STEAD Y-STA TE CURRENT-SENSING OUTPUT

al: = Vg-Vo

L'lb

Bl: = Al: = i - . ( l - B l ) cop Dl: = - A l bl: =

Ri«Ib T3

(Al + Bl -ton + Dl .e-"P-'°") - — • (l - e-'P''"") • Vc ^ ' wp ^

p—wp * ton

all:

al2: = - 1 a2: =

-Vo L.(Ib + 8i)

B2: =

a2 (op

A2: = J_ . ( 1 - B 2 ) top D2: =: - A 2 b2::

Ri«(Ib + 8i) T3

a21: = - 1 a22: ^_

g-<»p * tof

bl al2 b2 a22 vbx: = all al2 a21 a22

• (A2 + B2.tof + D2-e-'-P''°f) - — • (1 - e-*'P-'°f) -Vc ^ ^' (Op ^

APPENDIX D. MA THCAD, STEAD Y-STA TE CURRENT-SENSING OUTPUT

vby: =

all bl a21 b2 all al2 a21 a22

vbl(t): = vbx.e-"'P-' + Vc. — . (1 - e--P-') '{Al+B\'t

+

(op ^ Dl'e-''P'')

'

^ ^ T3

vb2(t): = vbye--P<'-'°"' + Vc • — • [l - e'^P ••'-"""] - ^'^^^ '^ ^^ wp

'•

T3

• [A2 + B2{t - ton) + Dle-'^P'^''"'"^] Voltage at the Rf-Cfz node: vb(t): = vbl(t) • ((t) - *(t - ton)) + vb2(t) • (0(t - ton) - (t - T)) t: = 0, 0.002-T-.T Idc:

1

Rf-T

(Vc-vb(t))dt

The compensator output: vm(t): = vb(t) -

1 Cfz

;(x.^-,.).

The source function during ton: vl(t): = Ri-(Ib + Sn.t) during tof: v2(t): = Ri • lb + 8i

Vo

(t-d-T)

vi(t):= vl(t)-($(t) - OO - d-T)) + v2(t)'(0(t - d«T) - 0(t - T))

369

370

APPENDIX D. MA THCAD, STEAD Y-STA TE CURRENT-SENSING OUTPUT

Slope of vm(t) at t ^ ton according to the preceding calculation: T1

T3

T2

|Sm(ton)| = 1.714 X 10^ Slope of vni(ton) according to (2) of paper: Sn.Ri = 2.4x 10^ (01: =

1 Rl«(Cfp + Cfz)

Sn': = wi'Sn' d-T

.p-d-T\

\wz

wp/

>Ri

bii = 2.195 X 10^ 5 " ' . = 9,145 Sn«Ri |Sm(ton)| = 7.143 Sn'Ri Finding external ramp waveform: Se: = 0.15«10^ vx:= 1 Given Se • ton = vm(ton) se: = Find(Se) se = 6.647 x 10^ ve(t): = {se • t) • (