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SPICE problem

R

Robert Baer

Jan 1, 1970
0
I have not found any models for the LM285-ADJ, so have attempted to
create one, by following the equivalent circuit given in a data sheet.
However, i cannot figure why it does not work.
Please critique:
VREF+MOSFET reg (25V ref)
..OPTIONS ACCT LIST NODE OPTS NUMDGT=6 RELTOL=0.00001 NOPAGE
..TEMP 25
..DC V1 -50 50 1
V1 4 0 DC 180V ;Vreg
R1 3 1 7.87MEG ;for Vref 25V appx
R2 4 3 412K ;3uA at 1.24V
VR 0 1 0.0V ; ground current sense
* + FB -
X3 4 3 1 LM285ADJ
..PRINT DC I(VR) V(3) V(4)
..PLOT DC I(VR) V(3) V(4)
..SUBCKT LM285ADJ 1 2 3 ; 1=+ cathode, 2=FB, 3=- anode
ISINK 0 81 5UA
RSINK 81 0 1E15 ; dummy - get +V term 81 rel term 0
VREF 82 81 1.24V ; vary this +/- 1%
VP 83 0 15V
VN 84 0 -15V
* C B E SUB
Q1 82 85 0 0 QN2222
RA 3 0 1E-6OHM
RC 1 82 1E-6OHM
X1 81 2 83 84 85 TL031
..ENDS LM285ADJ
..MODEL QN2222 NPN(
+ IS= 2.560E-14 BF= 200. NE= 2.00 IKF= .560
+ BR= 5.00 NC= 2.00 ISE= 1.280E-11
+ RB= 10.0 RC= .500 ISC= 1.280E-11
+ CJE= 2.500E-11 TF= 5.333E-10 CJC= 8.000E-12 TR= 4.000E-08)
* TL031 operational amplifier "macromodel" subcircuit
* n.i. input inverting input
* | | positive power supply
* | | | negative power supply
* | | | | v--output
..subckt TL031 1 2 3 4 5
c1 11 12 3.498E-12
c2 6 7 15.00E-12
css 10 99 11.38E-12
dc 5 53 dx
de 54 5 dx
dlp 90 91 dx
dln 92 90 dx
dp 4 3 dx
egnd 99 0 poly(2) (3,0) (4,0) 0 .5 .5
fb 7 99 poly(5) vb vc ve vlp vln 0 936.5E3 -900E3 900E3 900E3
-900E3
ga 6 0 11 12 113.1E-6
gcm 0 6 10 99 2.257E-9
iss 3 10 dc 76.50E-6
hlim 90 0 vlim 1K
j1 11 2 10 jx
j2 12 1 10 jx
r2 6 9 100.0E3
rd1 4 11 8.841E3
rd2 4 12 8.841E3
ro1 8 5 135
ro2 7 99 135
rp 3 4 138.5E3
rss 10 99 2.614E6
vb 9 0 dc 0
vc 3 53 dc 1.700
ve 54 4 dc 1.800
vlim 7 8 dc 0
vlp 91 0 dc 8
vln 0 92 dc 8
..model dx D(Is=800.0E-18)
..model jx PJF(Is=1.000E-12 Beta=140e-6 Vto=-1)
..ends TL031
..END
 
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