一种新的三相AC_DC PWM整流器控制策略
合集下载
相关主题
- 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
- 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
- 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。
1 Ts = 2fsw Ts T0 , T n Sj = 1, S j , j = a, b, c. ¯j , −1, S Tn
Ts fsw
Tn+1
Tn+1 S= 2 [Em − RL I ∗ (t)− v0 d jωLI ∗ (t) − L I ∗ (t)]ejωt . dt
vao 2 −1 −1 sa vbo = 1 v0 −1 sb . 2 −1 6 vco sc −1 −1 2
(ησ (v0 , v0 )∆v ) = ησ (v0 , v0 )∆v − Baidu Nhomakorabea I ∗ (t) Tn Tn+1
2 s2 a + sq
∆vW t σ (v0 , v0 ) W. ||W ||2 ss q
IL IL
ss d
4 4 [ cos 30◦ ]2 = . 3 3 2 √ 3 ss d ss q
v0
[9]
W
v0 vf
n
ˆ(t) − I ˆ∗ (t) I I ∗ (t) ωi σi (x). i n vf ∆v = vf − v0 f(x) =
i=1
ωi
σi
i
W = [ω1 ω2 · · · ωn ]T , σ = [σ1 σ2 · · · σn ]T . ∆W = ησ (v0 , v0 )∆v. f(x) = W T σ (x). η η
d 1 I + RL I + Sv0 , dt 2
2 π ), 3 4 π ). 3
1) 4 ej (n− 3 , n = 1 , 2, · · · , 6, Sn = 3 0, n = 7, 8.
e = [ea eb ec ]T , i = [ia ib ic ]T , v = [vao vbo vco ]T .
η
WM ∼
η
[9,10]
B (M ) =
{W : ||W ||
W } Kp = 0.0004, Ki = 0.2 2.5%
M
ησ (v0 , v0 )∆v, ||W || < W M ˆ || = W M ∆W = ||W ∆vW T σ (v0 , v0 ) 0; (ησ (v0 , v0 )∆v ), ||W || = W M ∆vW T σ (v0 .v0 ) > 0.
S
d−q
s S = ss d + jsq .
2 d { Em − RL I ∗ (t) − L I ∗ (t) cos(ωt)+ v0 dt ∗ ωLI (t) sin(ωt)}, 2 d ss { Em − RL I ∗ (t) − L I ∗ (t) sin(ωt)+ q = v0 dt ωLI ∗ (t) cos(ωt)}. ss d = Snd Snq Sn Tn Tn+1 1 s Tn = S(n+1)q ss d − S(n+1)d sd Ts , Γn(n+1) 1 s Snq ss Tn+1 = d − Snd sq Ts . Γn(n+1) Γn(n+1) = Snd S(n+1)q − Snq S(n+1)d sd sq Tn Tn+1 sd I ∗ (t) I ∗ (t) vf v0 v0 = (v0 − vf )dt ˆ∗ (t) = NN(v0 , v , W ). I 0 sq ˆ∗ (t) I I ∗ (t) v0 v0 I ∗ (t) W
j K K K -
_
@ _
Z
_
−
−
−
1 ,3
2
3
1 ,3
2
3
[1]
[2,3] [4] [7]
− ·
−
−
−
E = −L
2 E = (ea + ueb + u2 ec ) = Em ejωt . 3 2 2 I = (ia + uib + u2 ic ), S = (sa + U sb + u2 sc ), 3 3 u = ej(2π/3) I I = I ∗ (t)ejωt . I ∗ (t) ω ea = Em cos(ωt), eb = Em cos(ωt − ec = Em cos(ωt − ea , eb ec e = −L Em d i + RL i + v. dt
sds = sqs = ·
ss d ss q
2 3 (ss d) , s 2 2 4 (sd ) + (ss q) 2 (ss 3 q) . 2 s 2 4 (ss d ) + (sq )
IL IL
Em 940 m L =
Vf
ω =
C0 =
Y Z j K Z K K Y Z
K -
Z
Z Y Z Y -