Control of Drvetran Boost DC DC Converter Small sgnal model Converter transfer functons Current control loop bat current sensng c 2 v ge2 _ S 2 Q 2 D 2 v s S 1 C c voltage sensng I v v bat _ c 1 v ge1 _ Q 1 D 1 _ HV ref _ cv R s ref R s _ c v c c 2 PWM c Dead-tme Pulse-wdth modulator c 1 Hv H 1
arge sgnal, non lnear dynamc equatons d :1 v bat C v _ C d d dt v dt C bat v 1 d v d 1 2
Small sgnal AC equvalent crcut model V d D :1 v bat I d C v _ 3
PWM operaton and model c(t) c(t)
Open loop control to current transfer functon V d D :1 v bat I d C v _ Fnd d ( s) ˆ dˆ vˆ bat, ˆ 5
Open loop control to current transfer functon s V d D 2 sc D dˆ sc I 6
Open loop control to current transfer functon d ˆ ( s) dˆ do vˆ, ˆ bat 1 s 1 s z 2 2 o do I I D' D' 2 f z 1 1 2 C I V f o 1 2 D' C 7
Example Converter parameters: f s = 2 khz = 15 H C = 5 F DC operatng pont: V = 5 V I = 3 A V bat = 2 V D = 1 V bat /V =.6 I = I /D = 75 A do 2log f z f o I D' do 1A I 2 D' 1 1 I 2 C V 1 2 D' C 187.5 A 45.5 dba 19 Hz 23 Hz 8
Magntude and phase responses of d s = tf('s'); % defnes s as the aplace transform varable % % setup d do = I/(1-D); % low-frequency gan wz = (1-D)*I/C/V; % zero [rad/s] fz = wz/2/p % frequency of the zero [Hz] wo = (1-D)/sqrt(*C); % frequency of the par of poles [rad/s] fo = wo/2/p % frequency of the par of poles [Hz] % d = do*(1s/wz)/(1s^2/wo^2); % % set Bode plot optons BodeOptons = bodeoptons; BodeOptons.FreqUnts = 'Hz'; BodeOptons.Xlm = [1 1]; BodeOptons.Ylm = {[,1];[-18,18]}; BodeOptons.YmMode = {'manual';'manual'}; BodeOptons.rd = 'on'; % % plot magntude and phase responses bode(d,bodeoptons); MATAB scrpt 9
Magntude and phase responses of d 2log f do z f o I D' do 1A 1 1 I 2 C V 1 2 D' I 2 D' 187.5 A 45.5 dba C 19 Hz 23 Hz Magntude (db) Phase (deg) 1 8 6 4 2 18 135 9 45-45 -9-135 Bode Dagram -18 1 1 1 1 2 1 3 1 4 Frequency (Hz) 1
Current control loop oop gan T Compensator desgn Reference: Textbook Sectons 9.1 9.5 R s = gan of the current sensor 1/V M = gan of the pulse wdth modulator d = control tocurrent transfer functon c = current loop compensator 11
oop gan T, and closed loop transfer functon Closed loop response ˆ ˆ ref T 1 T oop gan T c 1 V M d R s Open loop control to current TF found earler d ˆ ( s) dˆ do vˆ, ˆ bat 1 s 1 z 2 s 2 2 o do I I D' D' f z f o 1 1 I 2 C V 1 D' 2 C 12
Feedback loop desgn objectves T(s) To meet the control objectves, desgn c (s) so that T s as large as possble n as wde frequency range as possble,.e. wth as hgh f c as possble mtaton: stablty and qualty of closed loop responses 13
Stablty Textbook Secton 9.4 Feedback may lead to nstablty: closed loop transfer functons may contan rght half plane poles Even f the closed loop system s stable, closed loop transent responses may exhbt rngng and long settlng tmes eneral stablty crteron: Nyqust Specal case: phase margn test* * Apples to the cases when T(s) has no rght half plane poles, and has exactly one cross over frequency f c 14
Phase margn test Stablty: m 15
Basc feedback loop desgn Closed loop response oop gan T ˆ ˆ ref c T 1 T 1 V M d R s Desgn compensator transfer functon c (s) to obtan: arge T over wde range of frequences, and hgh cross over f c Suffcently large phase margn m Cross over frequency should be well below swtchng frequency, e.g. f c < f s /5 16
Start from uncompensated loop gan, c = 1 T T c,uncomp 1 V M 1 V M d R s d R Example Converter parameters f s = 2 khz = 15 H C = 5 F V M = 1 R s = 1 DC operatng pont: V = 5 V I = 3 A V bat = 2 V D = 1 V bat /V =.6 I = I /D = 75 A s d Magntude (db) Phase (deg) 1 8 6 4 2 18 135 9 45-45 -9-135 2log T, uncomp( j) T (, uncomp j) Bode Dagram -18 1 1 1 1 2 1 3 1 4 Frequency (Hz) 17
T T c K p K ( s) c p 1 V M d d oop gan wth P compensator R s Magntude (db) Bode Dagram 6 4 2-2 -4-6 18 9 Phase (deg) -9-18 1 1 1 1 2 1 3 1 4 Frequency (Hz) 18
Closed loop response wth P compensator ( s) c K p ˆ ˆ ref T 1 T Magntude (db) Bode Dagram 2-2 -4-6 18 9 Phase (deg) -9-18 1 1 1 1 2 1 3 1 4 Frequency (Hz) 19
Proportonal Integral (PI) Compensator c ( s) K p K s K p 1 s zc f f f zc 2
c oop gan wth Proportonal Integral (PI) compensator ( s) f zc f c K p / 5 1 s zc 6 4 Bode Dagram Magntude (db) 2-2 -4-6 18 9 Phase (deg) -9-18 1 1 1 1 2 1 3 1 4 Frequency (Hz) 21
MATAB code: fnd f c and m % fnd phase margn and cross-over frequency [m,pm,wg,wp] = margn(t); Pm % phase margn Fc_actual = Wp/2/p % cross-over frequency Pm = 78.3553 Fc_actual = 2.374e3 22
Closed loop response wth PI compensator c ( s) ˆ ˆ f zc f c ref K p / 5 T 1 T 1 s zc Magntude (db) 2-2 -4 Bode Dagram -6 18 9 Phase (deg) -9-18 1 1 1 1 2 1 3 1 4 Frequency (Hz) 23