Lecture 6 ECEN 4517/5517

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Jemima Montgomery
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1 Lecture 6 ECEN 4517/5517 Experiment 4: inverter system Battery 12 VDC HVDC: VDC DC-DC converter Isolated flyback DC-AC inverter H-bridge v ac AC load 120 Vrms 60 Hz d d Feedback controller V ref Digital controller Step-up dc-dc converter with isolation (flyback) DC-AC inverter (H-bridge) Feedback controller to regulate HVDC ECEN

2 Due dates Right now: Prelab assignment for Exp. 3 Part 3 (one from every student) Due within five minutes of beginning of lecture This week in lab (Feb ): Nothing due. Try to finish Exp. 3. Next week in lecture (Feb. 26): Prelab assignment for Exp. 4 Part 1 (one from every student) Next week in lab (Feb ): Definitely finish Exp. 3, and begin Exp. 4 The following week in lab (Mar. 4-6): Exp. 3 final report due ECEN

3 Goals in upcoming weeks Exp. 4: A three-week experiment Exp. 4 Part 1: Design and fabrication of flyback transformer Snubber circuit Demonstrate flyback converter power stage operating open loop V batt snubber v HVDC PWM Compensator V ref Exp. 4 Part 2: Design feedback loop Measure loop gain, compare with simulation and theory Demonstrate closed-loop control of converter output voltage ECEN

4 Exp. 4, Part 3 H-bridge inverter, off grid v HVDC IR3101 IR3101 i ac AC load 120 Vrms 60 Hz v ac Filtering of ac output not explicitly shown IR 3101 half-bridge modules with integrated drivers Grid-tied: control i ac Off-grid: control v ac ECEN Digital controller Exp. 4 Part 3: off-grid inverter Demonstrate modified sine-wave inverter (required) Demonstrate PWM inverter (extra credit)

5 Modified Sine-Wave Inverter v ac has a rectangular waveform Inverter transistors switch at 60 Hz, T = 8.33 msec v ac V HVDC V HVDC DT/2 T/2 RMS value of v ac is: V ac,rms = 1 v 2 T ac t dt 0 T = D V HVDC Choose V HVDC larger than desired V ac,rms Can regulate value of V ac,rms by variation of D Waveform is highly nonsinusoidal, with significant harmonics ECEN

6 PWM Inverter Average v ac has a sinusoidal waveform Inverter transistors switch at frequency substantially higher than 60 Hz v ac t Choose V HVDC larger than desired V ac,peak Can regulate waveshape and value of V ac,rms by variation of d Can achieve sinusoidal waveform, with negligible harmonics Higher switching frequency leads to more switching loss and need to filter high-frequency switching harmonics and commonmode currents ECEN

7 The buck-boost converter 1 2 Switch in position 1: charges inductor i L V Switch in position 2: energy stored in inductor is transferred to output Conversion ratio: V = D 1D Subinterval 1 Subinterval 2 i L i L V V ECEN

8 The flyback converter: A transformer-isolated buck-boost converter See also: supplementary notes on Flyback converter, Exp. 4 web page buck-boost converter: Q 1 D 1 L V construct inductor winding using two parallel wires: L Q 1 D 1 1:1 V ECEN

9 Derivation of flyback converter, cont. Isolate inductor windings: the flyback converter L M Q 1 D 1 1:1 V Flyback converter having a 1:n turns ratio and positive output: L M 1:n D 1 C V Q1 ECEN

10 A simple transformer model Multiple winding transformer Equivalent circuit model i 1 n 1 : n 2 i 2 i 1 i M i 1 ' n 1 : n 2 i 2 v 1 v 2 i 3 v 3 v 1 L M v 1 n = v 2 1 n = v 3 2 n = =n 1 i 1 'n 2 i 2 n 3 i 3... : n 3 v 2 i 3 v 3 : n 3 Ideal transformer ECEN

The magnetizing inductance L M Models magnetization of transformer core material Appears effectively in parallel with windings If all secondary windings are disconnected, then primary winding behaves

11 The magnetizing inductance L M Models magnetization of transformer core material Appears effectively in parallel with windings If all secondary windings are disconnected, then primary winding behaves as an inductor, equal to the magnetizing inductance At dc: magnetizing inductance tends to short-circuit. Transformers cannot pass dc voltages Transformer saturates when magnetizing current i M is too large Transformer core B-H characteristic B v 1 dt saturation slope L M H i M ECEN

12 Volt-second balance in L M The magnetizing inductance is a real inductor, obeying di v 1 =L M M dt integrate: v 1 i M i M (0) = 1 t v L 1 (τ)dτ M 0 Magnetizing current is determined by integral of the applied winding voltage. The magnetizing current and the winding currents are independent quantities. Volt-second balance applies: in steady-state, i M (T s ) = i M (0), and hence i 1 i M L M i 1 ' n 1 : n 2 : n 3 i 2 v 2 i 3 v 3 0= T 1 T s v 1 dt s 0 Ideal transformer ECEN

13 The flyback transformer i g L M Transformer model i 1:n v L D 1 C i C R v A two-winding inductor Symbol is same as transformer, but function differs significantly from ideal transformer Energy is stored in magnetizing inductance Q 1 Magnetizing inductance is relatively small Current does not simultaneously flow in primary and secondary windings Instantaneous winding voltages follow turns ratio Instantaneous (and rms) winding currents do not follow turns ratio Model as (small) magnetizing inductance in parallel with ideal transformer ECEN

14 Subinterval 1 Transformer model i g i 1:n i C v L = L M v L C R v i C = v R i g = i CCM: small ripple approximation leads to Q 1 on, D 1 off v L = i C = V R i g = I ECEN

15 Subinterval 2 i g = 0 Transformer model i v L v/n 1:n i/n C i C R v v L = v n i C = i n v R i g =0 CCM: small ripple approximation leads to Q 1 off, D 1 on v L = V n i C = I n V R i g =0 ECEN

16 CCM Flyback waveforms and solution v L Volt-second balance: i C V/n I/n V/R v L = D D' V n =0 Conversion ratio is M(D)= V = n D D' Charge balance: i g V/R I DT s 0 D'T s t i C = D V R D' I n V R =0 Dc component of magnetizing current is I = D'R nv Dc component of source current is I g = i g = DI D' 0 Conducting devices: T s Q 1 D 1 ECEN

17 Equivalent circuit model: CCM Flyback v L = D D' V n =0 I g I i C = D V R D' I n V R =0 DI D'V D'I n R D n V I g = i g = DI D' 0 1 : D D' : n I g I R V ECEN

18 Step-up DC-DC flyback converter Need to step up the 12 V battery voltage to HVDC ( V) How much power can you get using the parts in your kit? Key limitations: MOSFET on-resistance (90 m ) Input capacitor rms current rating: 25 V 2200 μf: 2.88 A 35 V 2200 μf: 3.45 A Snubber loss Need to choose turns ratio, as well as D, f s, to minimize peak currents Possible project for expo: build a better (and more complex) step-up dc-dc converter ECEN

19 Design of CCM flyback transformer i M Transformer model I M Δi M i 1 i M L M v M n 1 : n 2 D 1 i 2 C R V 0 i 1 I M 0 i 2 Q 1 n 1 n 2 I M 0 v M 0 DT s ECEN

20 Approach Use your PQ 32/20 core Choose turns ratio n 2 /n 1, L M, D, and f s (choose your own values, don t use values in supplementary notes) Select primary turns n 1 so that total loss P tot in flyback transformer is minimized: P tot = P fe P cu = core loss plus copper loss Determine air gap length Determine primary and secondary wire gauges Make sure that core does not saturate ECEN

21 Core loss CCM flyback example B-H loop for this application: B B sat The relevant waveforms: B B max ΔB B max ΔB H c Minor BH loop, CCM flyback example BH loop, large excitation 0 v M 0 n 1 A c DT s B vs. applied voltage, from Faraday s law: db dt = v M n 1 A c For the first subinterval: db dt = n 1 A c Solve for B: ECEN B = DT s 2n 1 A c

Calculation of ac flux density and core loss Fitting an equation to the plot at right P fe = K fe (ΔB) β A c l m = slope K fe = constant that depends on f s A c l m = core volume

22 Calculation of ac flux density and core loss Fitting an equation to the plot at right P fe = K fe (ΔB) β A c l m = slope K fe = constant that depends on f s A c l m = core volume At 60 C: = 2.6 K fe = 16 (50 khz), 40 (100 khz) with P fe in watts, A c l m in cm 3, B in Tesla From previous slide: B = DT s 2n 1 A c ECEN More turns less B less core loss

23 Copper loss Power loss in resistance of wire Must allocate the core window area between the various windings Winding 1 allocation α 1 W A Winding 2 allocation α 2 W A etc. { { 0<α j <1 α 1 α 2 α k =1 Total window area W A Optimum choice: α m = n mi m Σ n j I j n =1 (leads to minimum total copper loss) The resulting total copper loss is: P cu = ρ(mlt)n2 1I tot with I W A K tot = u 2 k Σ j =1 n j n 1 I j Choose wire gauges: A w1 α 1K u W A n 1 A w2 α 2K u W A n 2 ECEN More turns more resistance more copper loss

24 Total power loss P tot = P cu P fe There is a value of B (or n 1 ) that minimizes the total power loss P tot = P fe P cu P fe = K fe (ΔB) β A c l m P cu = ρ(mlt)n2 2 1I tot W A K u B = DT s 2n 1 A c Power loss Copper loss P cu P tot Optimum ΔB Core loss P fe ΔB Prelab assignment for next week: use a spreadsheet or other computer tool to compute P tot vs. n 1, and find the optimum n 1. Then design your flyback transformer. ECEN

25 Effect of transformer leakage inductance L M v T i g L l v l ir on Transformer model i Q 1 Voltage spike caused by leakage inductance v l = L l di l dt ECEN :n v T { DT s D 1 C R v/n v t Leakage inductance L l is caused by imperfect coupling of primary and secondary windings Leakage inductance is effectively in series with transistor Q 1 When MOSFET switches off, it interrupts the current in L l L l induces a voltage spike across Q 1 If the peak magnitude of the voltage spike exceeds the voltage rating of the MOSFET, then the MOSFET will fail.

26 Protection of Q1 using a voltage-clamp snubber i g R s Snubber { v s Usually, C s is large C s Q 1 Flyback transformer 1:n v T D 1 C R v Snubber provides a place for current in leakage inductance to flow after Q 1 has turned off Peak transistor voltage is clamped to v s v s > V/n Energy stored in leakage inductance (plus more) is transferred to capacitor C s, then dissipated in R s Decreasing R s decreases the peak transistor voltage but increases the snubber power loss See supplementary flyback notes for an example of estimating C s and R s ECEN

Lecture 4 ECEN 4517/5517

Lecture 4 ECEN 4517/5517 Experiment 3 weeks 2 and 3: interleaved flyback and feedback loop Battery 12 VDC HVDC: 120-200 VDC DC-DC converter Isolated flyback DC-AC inverter H-bridge v ac AC load 120 Vrms