Electronic Power Conversion
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1 Electronic Power Conversion Challenge the future 1
2 8. Applications: AC motor drives Uninterruptible Power Supplies (UPS) Categories of voltage-source inverters (VSI,VSC): PWM inverters Square-wave inverters Single-phase inverters with voltage cancellation Not considered: CSI (Current-Source Inverters, CSI) Uni-directional power Source: Wikimedia Commons CC-BY-SA Bidirectional power (regenerative) 2
3 Switch-mode inverters basics Requirement for instantaneous power flow: Four-quadrant operation 3
4 Phase leg (phase arm or NL: fase tak) All dc-ac topologies in Chapt. 8 based on phase arm Point 0 is taken as reference for voltage, mostly not physically available 4
5 Consider one leg of an H-bridge and vary v control slowly 5
6 Synthesis of a Sinusoidal Output by PWM 1 v control > vtri T A+ is on è v A0 = Vd 2 1 v control < vtri T A- is on è v A0 = V 2 d 6
7 Spectrum f s - switching frequency, carrier frequency f 1 - modulating frequency, fundamental frequency Amplitude modulation ratio: Frequency modulation ratio: Vˆ m a = Vˆ control f tri s m f = f 1 7
8 Details of a Switching Time Period Control voltage can be assumed constant during a switching time-period Moving average of v A0 : From (1) and (2): where: v V V = with v < Vˆ Vˆ 2 control d A0 control tri tri v = V ˆ sinω t control control Vˆ V V V = sinωt = m sinωt control d d A0,1 1 a 1 Vˆ 2 2 tri 1 (1) (2) 8
9 Modulation frequency Small m f (m f < 21) Apply synchronous PWM to avoid subharmonics Use methods to eliminate specific low order harmonics Large m f (m f > 21) Good reproduction of reference wave Side band harmonics are small Switching harmonics can be eliminated with small filter 9
10 Overmodulation m > 1, Vˆ > Vˆ a control tri Voltage spectrum: 10
11 Output voltage Fundamental as a Function of m a ; Shows the linear and the over-modulation region; square-wave operation in the limit Overmodulation: Vd ( ˆ ) 4 Vd < V A0 < 2 1 π 2 11
12 Square wave operation 4 V V V = = π 2 2 ( ) A ˆ d d 0 1,max ( V ˆ A0 ) h = ( Vˆ A ) h
13 Single-phase inverters (half bridge) Current sharing between capacitors; i 0 cannot have a dc-component è no transformer saturation problem; Switch utilisation: V T = V d I T = I o 13
14 Full-bridge converter Preferred to half-bridge at high power levels v () t = v () t v () t = 2 v () t 0 A0 B0 A0 V ˆ = m V 0,1 4 V d < Vˆ 01, < V π a d d (m a <1) (m a >1) PWM with Bipolar voltage switching 14
15 DC-side current (* = averaged currents) f s è, so L and C è 0 From power balance: V i d * d () t = v () t i () t 0 0 = 2 V sin ω t 2 I sin( ω t - φ) = VI cosφ - VI cos( 2ω t-φ) VI VI i ( t ) = cosφ - cos( 2ω t-φ) * d Vd Vd Input current: DC component LF component of double fundamental frequency Strong HF components (steep slopes) 1 15
16 Full-bridge converter PWM with unipolar voltage switching Double control voltage Same output voltage and magnitude as bipolar switching Lower ripple current (about 50%) Less harmonics 16
17 Square wave operation output voltage control π h 0 π 2 ( V ˆ ) = π v cos( h θ) dθ 4 = V d sin( hβ ) π h 17
18 Voltage and current ripple Apply superposition: v () t = v () t + v () t 0 01 ripple i () t = i () t + i () t 0 01 ripple Fundamental (phasors): V = E + jω LI Ripple current: 1 t iripple() t = v ripple( ) d k L ζ ζ + 0 (b): largest ripple during zero crossing of voltage Square-wave PWM bipolar 18
19 Switch utilisation ratio (SUR) SUR = total apparent converter power S q V T I T q = total number of switches Example: Full bridge square-wave with sinusoidal current and output current I 0 P SUR = 0 switches P T 4 Output: V 01,max = Vd,max ; π 2 Switch T: V T =V ; I d,max T = 2 I 4 V d,max I 0,max P V 0 01,max I0,max 2 1 SUR = = = π = qp q V 2 I 4 V 2 I 2π T d,max 0,max d,max 0,max I 0 0,max 19
20 Effect of blanking time Blanking time (dead time) t Δ to avoid shoot-through ΔV = AN tδ + V d ( i A> 0) T s tδ - V d ( i A < 0) T s 20
21 Effect of blanking time (2) ΔV AB = ΔV AN ΔVBN ΔV = AB 2 tδ + V d ( i0 > 0) T s 2 tδ - V d ( i0 < 0) T s 21
22 Voltage versus Current Source 1) VSI with unipolar or bipolar modulation Behaves as voltage source 2a) Current mode modulation: VSI with hysteresis modulation Behaves as current source Variable switching frequency Constant current ripple amplitude 22
23 Voltage versus Current Source (2) 2b) VSI with Fixed frequency current control (triangular modulation with feedback) 23
24 Rectifier mode operation Phasor representation for fundamental: E VAn = EA + VL VL= jω L s I An A where s I A = jω Ls The phase and amplitude of v An (t) can be set through the input signal v control of the modulator V di s v L= L s dt di s v L= L s dt 24
25 i 1 Figure 2 shows a single- phase inverter connected to a single phase induc6on motor with counter emf e 0. The output voltage v 0 of the inverter is obtained by bipolar voltage switching modula6on scheme. To obtain a low distor6on linear modula6on is applied (no overmodula6on; m a <1). Given is further: V d = 350V Figure 2 (DC link voltage) ω 1 (fundamental frequency of v o and e 0 ) ω 1,nom =2π50 rad/s (nominal value of ω 1 ) V 01,nom =230 V (nominal rms value of fundamental of v 0 ) e 0 (counter emf) L = 30 mh (inductance of machine) f s =7.5 khz (switching frequency) At nominal speed and nominal voltage the input power of the loaded drive is 2kW at cos φ 1 =0.8. (a) Draw the relevant circuit 6me diagrams to show the opera6on of the inverter. Indicate which switch is on at what 6me period. Sketch equivalent circuit models to calculate the fundamental component of the current i o and the ripple component of the current i o. Calculate the maximum value of the peak- to- peak current ripple in i o that is caused by the switching. (b) Calculate the rms value of the fundamental of i o when the machine runs at rated speed and rated power. Sketch a phasor diagram with the phasors of e 0, v 0 and i 0. (c) Calculate the modula6on ra6o m a such that the machine runs at nominal speed and nominal voltage. (d) What will change in the answers to problems b) and c) if unipolar switching is used instead of bipolar switching? (e) Calculate the low- frequency (<1 khz) peak- to- peak voltage ripple Vd assuming that the current i1 is constant and C=1mF. 25
26 Given is a single- phase full bridge inverter opera6ng in a square- wave mode. The dc- voltage is 244V and the frequency of the output voltage that supplies a motor load is 47Hz. The inductance is L = 100 mh. Calculate the peak value of the ripple in the output current. 26
27 Image credits All uncredited diagrams are from the book Power Electronics: Converters, Applications, and Design by N. Mohan, T.M. Undeland and W.P. Robbins. All other uncredited images are from research done at the EWI faculty. 27
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