EE155/255 Green Electronics

Quiz Review11/14/16

Prof. William DallyComputer Systems Laboratory

Stanford University

Course Logistics• Quiz is next Wednesday 11/16

– 7:00PM to 9:00PM– Room 200-203– Covers all material to date– One page of notes allowed

• Sign off last lab last week• Now in the project part of the course

– Checkpoint 1 today – send the course staff a progress report• What you have accomplished• Any problems you encountered and how you dealt with them• Any revisions to your original plan

– PCB review by checkpoint 2– Order parts and PCBs ASAP

YAH

Quiz Review

Periodic Steady State

Buck Converter

L

V2

+-

iLV1

+-

a

b

Dual-Source Buck

ab

c

Isolated Converters

Ideal Transformer

1:NIP

VP

IS

VS

+ +

62 Green Electronics

1:NIP

VP

IS

VS

+ +

Figure 10.1: An ideal transformer with turns ratio 1 : N maintains the relation-ships VS = NVP and IS = IP

N .

inductance in an ideal transformer. The transformer is a four-terminal device,the left two terminals are connected to the primary winding of the transformer.The right two terminals are connected to the secondary winding.

The dots in the symbol indicate the polarity of the windings. When a voltageVP is applied to the primary so that the terminal labeled with the dot is positive(as shown), a voltage VS will appear at the secondary with the dotted terminalpositive as well. When the primary is driven so that current IP flows into thedotted terminal, current IS flows out of the dotted terminal of the secondary.

A transformer is characterized by its turns ratio, NP : NS . Physically this isthe ratio of the number of turns in the primary and secondary windings. Elec-trically, this ratio determines the ratio of voltages and currents in the primaryand secondary. For an ideal transformer, the primary and secondary voltagesand currents are related by

VS =NS

NPVP (10.1)

IS =NP

NSVP (10.2)

An ideal transformer with a 1 : N turns ratio, as depicted in Figure 10.1 stepsup the voltage by N and steps down the current by N so that the power intothe primary is equal to the power out of the secondary

PS = VSIS =NS

NPVP

NP

NSIP = VP IP = PP (10.3)

A transformer does not have a preferred direction. Current, and power, canflow in either direction. If you reverse the primary and secondary of Figure 10.1you have a transformer with a N : 1 turns ratio.

Similarly the voltage of a transformer is not necessarily set by its primaryterminal, or even by a single set of terminals. A transformer always maintainsa voltage ratio between its terminals that is set by its turns ratio. However, in

Real Transformer

1:NiP

VP

iS

VS

+ +

LMiM

LL

VPR

+

ideal xfmr

iPR

Full-Bridge Converter

1:N

iP

VP

+

inverter

V1+

b b

aa

transformer

d d

c c

V2+L

iS

iL

rectifier output filter

C

VS

+

VR+

Full-Bridge Converter

UL UR

LL LR

EE 155/255 Lecture 8 - Isolated Converters

Periodic Steady State Analysis of Full Bridge

• Suppose V1 = 10V, transformer is 1:10, what duty factor gives V2=50V (ignoring leakage inductance)

• Now if primary referenced leakage inductance is 10µH iL = 10A, tcy = 10µs, what duty factor is needed?

Flyback Converter

1:N

VS

+

LM

a

+

C V2

b

V1

+

iM VP

+

LL iL

x

EA

iS

VPR

+

EE 155/255 Lecture 8 - Isolated Converters

Flyback ConverterWith LM and LL and Transformer Eliminated

LM

a

+

CNV2

V1

+

iM

LL iL

x

EA

iS

N

EE 155/255 Lecture 8 - Isolated Converters

Flyback Converter• Assume no leakage inductance• Magnitizing inductance is 50uH• Input voltage and output voltage both 100V• If 1A load current, 16us cycle time, and 1:1 transformer• How long an on period is required?

Motor Control

Motor Equations

V = |L|(v x B) V = KMw

F = i(L x B)t = KMi

t = IM(dw/dt)

Simultaneously both a motor and a generator

Back EMF

N

S

1

23

V(a) = -Kwsin(q-a)

Open-circuit voltage (back EMF) of each phase is a sinusoid that

Is proportional to w

Reaches peak value when rotor is 90 degrees from phase.

Torque

N

S

1

23

t = -(Ki)sin(q-ai)

Torque is proportional to current

Torque is maximum when rotor is 90 degrees behind phase

Torque from each phase is sinusoidal

Example• Suppose KM=1, RM=1, q=30deg, w=10rad/s, V=100V• What duty factor should be applied to each phase to give 1N-m torque?

Three-Phase Explained Graphically

0 50 100 150 200 250 300 350 400−1

−0.5

0

0.5

1

0 50 100 150 200 250 300 350 4000

0.2

0.4

0.6

0.8

1

0 50 100 150 200 250 300 350 400−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Sin(q-fi)

Sin2(q-fi)

SSin2(q-fi)

Current

Torque

Sum of Torque over phases

Control

Consider the Following Plants

1

s+ a

1

s2 + 2⇣!s+ !2

1

What type of controller do we need for each?

What should the parameters of our controller be?

Bode Plot for z=0.1 w=1000

−20

−10

0

10

20

30

40

Mag

nitu

de (d

B)

102 103 104−180

−135

−90

−45

0

Phas

e (d

eg)

Bode DiagramGm = Inf dB (at Inf rad/s) , Pm = 3.8 deg (at 3.31e+03 rad/s)

Frequency (rad/s)

Approach 1 – Integral Only – Low Gain

−140

−120

−100

−80

−60

−40

Mag

nitu

de (d

B)

102 103 104−270

−225

−180

−135

−90

Phas

e (d

eg)

Bode DiagramGm = 66 dB (at 1e+03 rad/s) , Pm = 90 deg (at 0.1 rad/s)

Frequency (rad/s)

Approach 2 – PDI – Large Gain Compensating Zero

−20

0

20

40

60

80

100

Mag

nitu

de (d

B)

10−1 100 101 102 103 104 105 106−135

−90

−45

0

45

Phas

e (d

eg)

Bode DiagramGm = Inf , Pm = 89.5 deg (at 1e+05 rad/s)

Frequency (rad/s)

Switches

DC I-V Characteristics of On Switches

V(d)0.0V 0.1V 0.2V 0.3V 0.4V 0.5V 0.6V 0.7V 0.8V 0.9V 1.0V 1.1V 1.2V 1.3V 1.4V 1.5V 1.6V 1.7V 1.8V 1.9V 2.0V

0A

5A

10A

15A

20A

25A

30A

35A

40A

45A

50AIx(m:1) Ix(h:2) Ix(i:C) I(Dd)

600V FETFCB36N60

600V IGBTFGH40N60

400V DiodeSTTH20R04

60V FETIRLB3036

FETs CharacterizedBy RON

IGBT Like a DiodeLittle Current Until ~0.7V

HV FET has high RONR ~ kV2

Diode and IGBTResistive at high Current

Transient Response of FET and IGBT

0ns 40ns 80ns 120ns 160ns 200ns 240ns 280ns 320ns0V

50V

100V

150V

200V

250V

300V

350V

400V

450V

500V

550V

0A

5A

10A

15A

20A

0V

50V

100V

150V

200V

250V

300V

350V

400V

450V

500V

550V

0A

5A

10A

15A

20A

0KW

1KW

2KW

3KW

4KW

5KW

6KW

7KW

8KW

9KW

10KW

V(di) Ix(i:C)

V(dm) Ix(h:2)

ix(i:c)*v(di) ix(h:2)*V(dm)

600V FETFCB36N60

InstantaneousPower

Turn On Turn Off

600V IGBTFGH40N60

36µJ 36µJ92µJ 428µJ

Turn-OnBuck w/ Diode

IP

ILQRR QD

ID

VDS

s

t1 t2 t3

t1 – ramp current to IL

t2 – diode reverse recovery

t3 – discharge drain capacitance

Current waveform in t2 and t3 may vary

Turn-OnBuck w/ Diode

t1 =ILs

E1 = 0.5VDDILt1 =0.5VDDIL

2

s

t2 =2QRR

sE2 =VDDt2 IL + 0.5t2s( )

t3 ≈2QD

IPE3 = 0.5VDDQD + 0.33VDDILt3

IP

ILQRR QD

ID

VDS

s

t1 t2 t3

Switches• Calculate the switching energy during turn-on for a MOSFET switching

10A through 100V in a buck configuration. Drain capacitance is 1nF, QRR is 25nC. Current ramp is 1A/ns.

PV• Suppose a single PV panel has

– VOC = 40V, ISC = 8A, and Pmax = 250W • You want to use this panel to provide a 5V supply with the maximum possible current(a) What converter topology should you use?(b) What variable should be regulated?(c) If losses are zero, what will be the maximum current out?(d) Will the panel be at its MPP when current reaches a maximum?(xc) What control law will provide maximum power beyond the point where 5V can be maintained?

(e) What would your answers be if the output were 100V?

IV Curve from SPICE Model

Typical String of 10 PV Panels

Magnetic Components

A Magnetic Circuit• Put N turns of wire on a magnetic core• Pass current i through resulting coil• Produces Magnetomotive Force

• F induces a flux f in the core

• Where R is the Reluctance of the core

• Change in f causes voltage

F = Ni

φ =FR

=NiR

V = N dφdt

Two Key Equations

Inductors• Suppose inductor for a converter with L=6.25uH and Imax=11A has• Ac = 1E-4m2

• Lc = 1E-2m• µ = 1000• f = 100kHz• Bmax = 0.1T

• What gap is needed? How many turns?

Transformers

V = Nd�

dt

V T = N�

V T = NBA

N =V T

BA

1

1:NiP

VP

iS

VS

+ +LM

iM

LL

VPR

+

ideal xfmr

iPR

Soft Switching

Soft Switching• Only switch a FET when:

– Zero voltage across it (ZVS)– Zero current through it (ZCS)– Both

• 4 Approaches– Phase-shifted full-bridge– Quasi-square wave– Quasi resonant– Active clamp

QSW Buck Converter

L

iLVS

+-

vC Load

+

-

C iLoadM2L

M1H

CX

D1

D2

Soft Switched Waveforms

Quasi-Resonant ZCS Buck

Lr

ir

VS

+- vC Load

+

-

C iLoad

M1

HCr D1

L

iL

D2

Initially M1 off D1 onM1 turns on ZCSir ramps linearly to iLD1 turns off ZCS & ZVSir starts charging Cr (tank circuit)Peak voltage at 2VSCr discharges through L

Pulse width fixed by iL and LrCrControl converter by varying frequency

0 2 4 6 8 100

50

100

150

200

v x (V)

0 2 4 6 8 1049.98

49.99

50

50.01

50.02

v c (V)

0 2 4 6 8 100

10

20

i r (A)

0 2 4 6 8 106

8

10

12

14

i L (A)

Inverters and Power Factor

Definition

�I = (Vin �DhVd)tcyL

i(s) = �dh(s)hVdisL

� vd(s)hDhisL

�Vd =

✓DhI �

Vd

R

◆tcyC

vd(s) =hDhii(s)

sC� vd(s)

sRC

vd(s)

✓1 +

1

sRC

◆=

hDhii(s)sC

vd(s) (sRC + 1) = hDhiRi(s)

vd(s) =hDhiRi(s)

sRC + 1=

hDhii(s)sC + 1

R

vd(s) = �hDhihVdidh(s) + hDhi2vd(s)s2LC + sL

R

vd(s)

✓s2LC + s

L

R+ hDhi2

◆= �hDhihVdidh(s)

vd(s)

dh(s)= � hDhihVdi

s2LC + sLR + hDhi2

vd(s)

dh(s)= �

hDhihVdiLC

s2 + s 1RC + hDhi2

LC

PF =Real Power

Apparent Power

PF =Pp

P 2 +Q2

PF =1p

1 + THD2=

I1,rms

Irms

1

Apparent Power = Vrms x Irms

Make a PWM Sine Wave and Filter

PWM Synthesis

0 2 4 6 8 10 12 14 16−1

−0.5

0

0.5

1

1.5

V AC, V

Saw (V

)

0 2 4 6 8 10 12 14 16−1

−0.5

0

0.5

1

V PWM

(V)

0 2 4 6 8 10 12 14 16−1.5

−1

−0.5

0

0.5

1

1.5

V Out

(V)

0 2 4 6 8 10 12 14 16−1

−0.5

0

0.5

1

I L (A)

t (ms)

x = sine > sawy = -sine > saw

Digitally generate sine with quarter-wave table

Spectrum of 100kHz PWM Signal

101 102 103 104 105 1060

1

2

3

4

5

6

7

8

9x 105

f (Hz)

Mag

Batteries

Batteries

LB RB

VBS VB

+

-

Voltage Surface V(x,T,I)

/LWKLXP�LRQ5HFKDUJHDEOH��EDWWHU\

3.5

4.0

4.5

/ V

Discharge Temperature Characteristics for NCR18650B1Scell-1 Charge:CC-CV:1.625A-4.20V(65.0mA cut)

Discharge:CC:3.25A(E.V.:2.50V)1.-20Υ 2.-10Υ 3.0Υ 4.25Υ 5.40Υ 6.45Υ 7.60Υ

Discharge Rate Characteristics for NCR18650BCharge:CC-CV:1.625A-4.2V (65.0mA cut)Discharge:CC:Variable Current (E.V.:2.50V)Temp:25Υ

2G23X0KYKU

0 500 1000 1500 2000 2500 3000 3500 4000Discharge Capacity / mAh

2.0

2.5

3.0

3.5

Cel

l Vo

ltage

2.0CA 1.0CA 0.5CA 0.2CA

/LWKLXP�LRQ5HFKDUJHDEOH��EDWWHU\

3.5

4.0

4.5e

/ V

Discharge Temperature Characteristics for NCR18650B1Scell-1 Charge:CC-CV:1.625A-4.20V(65.0mA cut)

Discharge:CC:3.25A(E.V.:2.50V)1.-20Υ 2.-10Υ 3.0Υ 4.25Υ 5.40Υ 6.45Υ 7.60Υ

Discharge Temperature Characteristics for NCR18650B

2G23X0KYKU

0 500 1000 1500 2000 2500 3000 3500 4000Discharge Capacity / mAh

2.0

2.5

3.0

3.5

Cel

l V

olta

ge

40Υ 25Υ 0Υ -10Υ -20Υ

Charge Cycle /LWKLXP�LRQ5HFKDUJHDEOH��EDWWHU\

3.5

4.0

4.5

e /

V

3000

4000

5000

mA

3000

4000

mA

h

3.5

4.0

4.5

e /

V

3000

4000

5000

mA

Charge Characteristics for NCR18650B1S

No.1 Charge:CC-CV:1.625A-4.20V(65.0mA cut)

Cell Voltage

Charge Characteristics for NCR18650B

2G23X0KYKU

0 60 120 180 240Charge Time / min

2.0

2.5

3.0

3.5

Cel

l Vo

ltage

0

1000

2000

3000

Cur

rent

/ m

1000

2000

3000

Cap

acity

/ m

0 60 120 180 240Charge Time / min

2.0

2.5

3.0

3.5

Cel

l Vo

ltage

0

1000

2000

3000

Cur

rent

/ m

Capacity

45Υ 25Υ 0Υ

Current

YAH