ECEN 4517 1

Lecture 6ECEN 4517/5517

Step-up dc-dc converterwith isolation (flyback)

Feedback controller toregulate HVDC

Experiment 4: inverter system

DC-AC inverter (H-bridge)

12 VDC HVDC: 120 - 200 VDC

AC load120 Vrms60 Hz

Battery

DC-ACinverter

H-bridge

DC-DCconverter

Isolatedflyback

+

d(t)

Feedbackcontroller

Vref Digitalcontroller

d(t)

+

vac(t)

ECEN 4517 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. 19-21):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. 26-28):Definitely finish Exp. 3, and begin Exp. 4

The following week in lab (Mar. 4-6):Exp. 3 final report due

ECEN 4517 3

Goals in upcoming weeksExp. 4: A three-week experiment

Exp. 4 Part 1:Design and fabrication offlyback transformer

Snubber circuitDemonstrate flyback

converter power stageoperating open loop

Exp. 4 Part 2:Design feedback loopMeasure loop gain, compare with simulation and theoryDemonstrate closed-loop control of converter output voltage

snubber

PWM Compensator +

Vref

Vbatt

vHVDC

ECEN 4517 4

Exp. 4, Part 3H-bridge inverter, off grid

IR3101 IR3101

vHVDC

AC load120 Vrms

60 Hz

+ vac(t)

Digitalcontroller

iac(t)

Filtering of ac output not explicitly shown

IR 3101 half-bridge modules withintegrated drivers

Grid-tied: control iac(t)

Off-grid: control vac(t)

Exp. 4 Part 3: off-grid inverter

Demonstrate modified sine-wave inverter(required)

Demonstrate PWM inverter (extra credit)

ECEN 4517 5

Modified Sine-Wave Inverter

vac(t) has arectangularwaveform

Inverter transistorsswitch at 60 Hz,T = 8.33 msec

T/2

DT/2+ VHVDC

VHVDC

vac(t)

RMS value of vac(t) is:

Vac,RMS =1T vac

2 t dt0

T

= D VHVDC

Choose VHVDC larger thandesired Vac,RMS

Can regulate value ofVac,RMS by variation of D

Waveform is highlynonsinusoidal, withsignificant harmonics

ECEN 4517 6

PWM Inverter

Average vac(t) has asinusoidal waveform

Inverter transistorsswitch at frequencysubstantially higherthan 60 Hz

Choose VHVDC larger than desired Vac,peak

Can regulate waveshape and value of Vac,RMS by variation of d(t)

Can achieve sinusoidal waveform, with negligible harmonics

Higher switching frequency leads to more switching loss andneed to filter high-frequency switching harmonics and common-mode currents

t

vac(t)

ECEN 4517 7

The buck-boost converter

Subinterval 1 Subinterval 2

+

+

V

Vg

iL

+

+

V

Vg

iL

+

+

V

1 2

VgiL

VVg

= D1 D

Switch in position 1: Vg chargesinductor

Switch in position 2: energy stored ininductor is transferred to output

Conversion ratio:

ECEN 4517 8

The flyback converter:A transformer-isolated buck-boost converter

buck-boost converter:

construct inductor

winding using two

parallel wires:

+ L

V

+

Vg

Q1 D1

+ L

V

+

Vg

Q1 D1

1:1

See also:supplementarynotes on Flybackconverter, Exp. 4web page

ECEN 4517 9

Derivation of flyback converter, cont.

Isolate inductor

windings: the flyback

converter

Flyback converter

having a 1:n turns

ratio and positive

output:

+ LM

V

+

Vg

Q1 D1

1:1

+

LM

+

V

Vg

Q1

D11:n

C

ECEN 4517 10

A simple transformer model

Multiple winding transformer Equivalent circuit model

n1 : n2

: n3

+

v1(t)

+

v2(t)

+

v3(t)

i1(t) i2(t)

i3(t)

n1 : n2

: n3

+

v1(t)

+

v2(t)

+

v3(t)

i1(t) i2(t)

i3(t)

Idealtransformer

i1'(t)

LM

iM(t)

v1(t)n1

=v2(t)n2

=v3(t)n3

= ...

0 = n1i1' (t) + n2i2(t) + n3i3(t) + ...

ECEN 4517 11

The magnetizing inductance LM

Transformer core B-H characteristic 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 iM

is too large

B(t) v1(t) dt

H(t) iM(t)

slope LM

saturation

ECEN 4517 12

Volt-second balance in LM

The magnetizing inductance is a real inductor,

obeying

integrate:

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, iM(T

s) = i

M(0), and hence

n1 : n2

: n3

+

v1(t)

+

v2(t)

+

v3(t)

i1(t) i2(t)

i3(t)

Idealtransformer

i1'(t)

LM

iM(t)v1(t) = L MdiM(t)

dt

iM(t) iM(0) =1

L Mv1()d

0

t

0 = 1Tsv1(t)dt

0

Ts

ECEN 4517 13

The flyback transformer

A two-winding inductor Symbol is same as

transformer, but functiondiffers significantly fromideal transformer

Energy is stored inmagnetizing inductance

Magnetizing inductance isrelatively 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

+

LM

+

v

Vg

Q1

D11:n

C

Transformer model

iig

R

iC+

vL

ECEN 4517 14

Subinterval 1

CCM: small ripple

approximation leads to

+

LM

+

v

Vg

1:n

C

Transformer model

iig

R

iC+

vL

vL = VgiC =

vR

ig = i

vL = Vg

iC = VR

ig = I

Q1 on, D1 off

ECEN 4517 15

Subinterval 2

CCM: small ripple

approximation leads to

vL = vn

iC =in

vR

ig = 0

vL = Vn

iC =In

VR

ig = 0

+

+

v

Vg

1:n

C

Transformer model

i

R

iC

i/n

v/n

+

+

vL

ig= 0

Q1 off, D1 on

ECEN 4517 16

CCM Flyback waveforms and solution

Volt-second balance:

Conversion ratio is

Charge balance:

Dc component of magnetizing

current is

Dc component of source current is

vL

iC

ig

t

Vg

0

DTs D'TsTs

Q1 D1Conducting

devices:

V/n

V/R

I/n V/R

I

vL = D Vg + D' Vn = 0

M(D) = VVg= n DD'

iC = D VR + D'

In

VR = 0

I = nVD'R

Ig = ig = D I + D' 0

ECEN 4517 17

Equivalent circuit model: CCM Flyback

+

+ R

+

V

VgD'In

D'Vn

+

DVgDI

IIg

+ R

+

V

Vg

IIg

1 : D D' : n

vL = D Vg + D' Vn = 0

iC = D VR + D'

In

VR = 0

Ig = ig = D I + D' 0

ECEN 4517 18

Step-up DC-DC flyback converter

Need to step up the 12 V battery voltage to HVDC (120-200 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 F: 2.88 A

35 V 2200 F: 3.45 A

Snubber loss

Need to choose turns ratio, as well as D, fs, to minimize peak currents

Possible project for expo: build a better (and more complex) step-up dc-dcconverter

ECEN 4517 19

Design of CCM flyback transformer

+

LM

+

V

Vg

Q1

D1

n1 : n2

C

Transformer model

iMi1

R

+

vM

i2

vM(t)

0

Vg

DTs

iM(t)

IM

0

iM

i1(t)

IM

0i2(t)

IM

0

n1n2

ECEN 4517 20

Approach

Use your PQ 32/20 core

Choose turns ratio n2/n1, LM, D, and fs (choose your ownvalues, dont use values in supplementary notes)

Select primary turns n1 so that total loss Ptot in flybacktransformer is minimized:Ptot = Pfe + Pcu = core loss plus copper loss

Determine air gap length

Determine primary and secondary wire gauges

Make sure that core does not saturate

ECEN 4517 21

Core lossCCM flyback example

dB(t)dt

=vM (t)n1Ac

dB(t)dt

=Vg

n1Ac

B(t)

Hc(t)

Minor BH loop,CCM flybackexample

BH loop,large excitation

Bsat

BBmax

vM(t)

0

Vg

DTs

B(t)

Bmax

0

B

Vgn1Ac

B-H loop for this application: The relevant waveforms:

B(t) vs. applied voltage,from Faradays law:

For the firstsubinterval:

Solve for B: B =VgDTs2n1Ac

ECEN 4517 22

Calculation of ac flux densityand core loss

B =VgDTs2n1Ac

Pfe = K fe(B)Ac lm = slope

Kfe = constant that depends on fsAclm = core volume

Fitting an equation to the plot at right

At 60C:

= 2.6

Kfe = 16 (50 kHz), 40 (100 kHz)

with Pfe in watts, Aclm in cm3, B in Tesla

More turns less B less core loss

From previous slide:

ECEN 4517 23

Copper lossPower loss in resistance of wire

Total windowarea WA

Winding 1 allocation1WA

Winding 2 allocation2WA

etc.

{{

0 < j < 11 + 2 + + k = 1

Must allocate thecore window areabetween the variouswindings

m =nmIm

n jI jn = 1

Optimumchoice: (leads to minimum total copper loss)

The resulting total copper loss is: Pcu =(MLT )n12I tot

2

WAK uI tot =

n jn1

I jj = 1

k

with

Aw1 1KuWA

n1

Aw2 2K uWA

n2

Choose wire gauges: More turns more resistance more copper loss

ECEN 4517 24

Total power lossPtot = Pcu + Pfe

Ptot = Pfe + Pcu

There is a value of B(or n1) that minimizesthe total power loss

Pfe = K fe(B)Ac lm

B

Powerloss

Ptot

Copper loss P

cu

Cor

e lo

ss P

fe

Optimum B

Pcu =(MLT )n12I tot

2

WAK u

B =VgDTs2n1Ac

Prelab assignment for next week: use aspreadsheet or other computer tool to computePtot vs. n1, and find the optimum n1.

Then design your flyback transformer.

ECEN 4517 25

Effect of transformer leakage inductance

+

LM

+

v

Vg

Q1

D11:n

C

Transformer model

iig

R

Ll

+ vl

+

vT(t)

Leakage inductance Ll is caused by

imperfect coupling of primary andsecondary windings

Leakage inductance is effectively inseries with transistor Q1

When MOSFET switches off, itinterrupts the current in L

l

Ll induces a voltage spike across Q1

t

Vg + v/n

vT(t)

iRon

DTs

{Voltage spikecaused byleakageinductancevl = L l

dildt

If the peak magnitude of thevoltage spike exceeds thevoltage rating of the MOSFET,then the MOSFET will fail.

ECEN 4517 26

Protection of Q1using a voltage-clamp snubber

+

+

v

Vg

Q1

D11:n

C

Flyback transformer

ig

R

+

vT(t)

CsRs

vs

+

Snubber{ Snubber provides a placefor current in leakageinductance to flow afterQ1 has turned off

Peak transistor voltage isclamped to Vg + vs

vs > V/n

Energy stored in leakageinductance (plus more) istransferred to capacitorCs, then dissipated in RsUsually, Cs is large

Decreasing Rs decreases the peak transistor voltage but increases thesnubber power loss

See supplementary flyback notes for an example of estimating Cs and Rs