fundamentals on boost (step up) convertersece.tamu.edu/~jose-silva-martinez/courses/ecen489/boost...

Jose Silva-Martinez 1Boost Converters

Fundamentals on Boost

(Step Up) Converters

Jose Silva-Martinez

Department of Electrical and Computer Engineering

Texas A&M University

May 28, 2017


Home Automation – Thermostat

5V3.3V/0.5A

AC24V

(un-regulated) Diode

Rectifier

Buck

Converter

BOOST

TPS61020

Control

System

including

Wifi

AA alkaline battery

Backup Power

Application of Texas Instruments TPS61020


Boost Converters: Applications

Fancy Electric Cars



Boost

Converter

12V or 24V

nominal VBATT

51V (125W)

1.25A

1.25A

Reservoir

cap

GND

Injector

coils

Full custom design resource: PMP8532

Boost

Converter


Output voltage is larger than Vin, but output current is smaller

Pin is approximately equal to Pout

If the MOSFET gate driver sticks in the “on” position, then there is

a short circuit to ground through the MOSFET

If the load is disconnected during operation when switch is in “off”

position, then L continues delivering current to the right and very

quickly charges C and may generate very large output voltage.

Boost Converters: PrinciplesVin

C

L

SN

High-Gain

Lowpass Filter

ZL

Vout

VREF

Ramp Generator

Pulse Width

Modulator

iDiL

iC

Vin

C

L

ZL

Vout

iD=0iL

iC

Diode is

reverse biased


Switch closed for DT seconds: Charging phase

Boost Converters: Principles

L

V

dt

di inL

If the switch stays closed for very long time, the current continues

increasing, only limited by the switch and/or battery!

Vin

C

L

ZL

Vout

iD=0iL

iC

Diode is

reverse biasediL(t1)

iL(t0)

IAVER

DTTime

sec/ AL

Vin


If the Switch remains opened for (1 − D)T

seconds

We assume continuous conduction; iL(t0) > 0.

In this graph, diode’s voltage drop is ignored; it can be easily

incorporated replacing Vout by Vout+VD when computing the

inductance’s current variation.

L

VV

dt

di outinL

Vin

C

L

ZL

Vout

iDiL

iC

Diode is

forward biased

iL(t1)

iL(t0)

IAVER

DT (1-D)TTime

sec/ AL

VV outin

sec/ AL

Vin


For f2 the switch is off, then


t0 t1 t2

iL(t1)

iL(t0)

IAVER

DT (1-D)T

Time

Vin=VB

C

L

SN

ZL

Vout

iDiL

iC



In the following derivation, it is assumed that Dvo is small

t0 t1 t2

iL(t1)

iL(t0)

IAVER

DT (1-D)T

Time


Vin

C

L

ZL

Vout

iDiL

iC

Diode is

forward biased

iout

sec/ AL

Vin

sec/ AL

VV outin

iL(t1)

iL(t0)

IAVER

DT (1-D)TTime

sec/ AL

VV outin

sec/ AL

Vin

Vin

C

L

ZL

Vout

iD=0iL

iC

Diode is

reverse biased

𝐼𝑅𝑀𝑆2 = 𝐼𝐴𝑉

2 +𝐼𝑝𝑘−𝑝𝑘

2

12

𝜆1 =2𝐼𝑝𝑘𝐷𝑇𝑐𝑘

; 𝜆2 = −2𝐼𝑝𝑘

1 − 𝐷 𝑇𝑐𝑘

𝝀𝟏 =𝒗𝑩𝑳; 𝝀𝟐 =

𝒗𝒊𝒏 − 𝒗𝐨𝐮𝐭𝑳



Vin

C

L

ZL

Vout

iDiL

iC

Diode is

forward biased

iout

iL_pk

0

IAVER_D

DT (1-D)T

IAVER_L

ID

𝑣out,𝐴𝑣𝑒𝑟𝑣𝑖𝑛

=1

1 − 𝐷 𝒗𝒊𝒏𝑰𝑳 = 𝒗𝐨𝐮𝐭_𝐀𝐯𝐞𝐫 ∙ 𝑰𝐨𝐮𝐭 𝒐𝒓𝒗𝒊𝒏

𝒗𝐨𝐮𝐭_𝐀𝐯𝐞𝐫=

𝑰𝐨𝐮𝐭

𝑰𝑳= 1-D

𝑰𝑨𝒗𝒆𝒓_𝑫 = 𝑰𝑨𝒗𝒆𝒓_𝑳 𝟏 − 𝑫

𝐼𝐴ver_𝐷𝐼𝐴𝑣𝑒𝑟_𝐿

= 1 − 𝐷

Average Current and Voltage


imax

imin

IAVER

DT (1-D)T Time

sec/ AL

VV outin

sec/ AL

Vin

DIL

iL_pk

0

IAVER_D

DT (1-D)T

IAVER_L

ID

𝑪𝒉𝒂𝒓𝒈𝒊𝒏𝒈 𝒑𝒉𝒂𝒔𝒆: 𝒊𝑪 = 𝒊𝑫 − 𝒊𝑹

𝐶∆𝑣0 = ∆𝑄 = 𝑰𝒍𝒐𝒂𝒅 𝑫 𝑇𝑐𝑘

𝑫𝒊𝒔𝒄𝒉𝒂𝒓𝒅𝒊𝒏𝒈 𝒑𝒉𝒂𝒔𝒆: 𝒊𝑪 = −𝒊𝒍𝒐𝒂𝒅

Therefore, the output ripple can be estimated as:

∆𝑣0 =𝑰𝒍𝒐𝒂𝒅_𝒂𝒗𝒆𝒓𝒂𝒈𝒆

𝑪𝑓𝑐𝑘𝑫

Vin

C

L

ZL

Vout

iD=0iL

iC

Diode is

reverse biased

Output Voltage Ripple


L

V

dt

diVv inL

inL ,

L

VV

dt

diVVv outinL

outinL

,

Closer look on inductor’s current

During charging phase

During the discharging phase

Current flowing through the inductor:

Worst case: DIL=2Iaverage

imax

imin

IAVER

DT (1-D)T Time

sec/ AL

VV outin

sec/ AL

Vin

DIL

22222

12

1

12

1LinppavgLrms IIIII D

2222

3

42

12

1inininLrms IIII D inLrms II

3

2

Vin=VB

C

L

SN

ZL

Vout

iDiL

iC


𝑣0𝑣𝑖𝑛

=1

1 − 𝐷

𝑣0𝜂𝑣𝐵

=1

1 − 𝐷𝜂 = 𝑝𝑜𝑤𝑒𝑟 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦

∆𝐼 = 2𝐼𝑝𝑘 =𝜂𝑣𝐵𝐿

𝐷𝑇𝑐𝑘

𝐼𝐿𝑚𝑎𝑥 = 𝐼𝐴𝑉 + 𝐼𝑝𝑘 = 𝐼𝐴𝑉 +𝜂𝑣𝐵2𝐿

𝐷𝑇𝑐𝑘

𝐼𝑆𝑊𝑚𝑎𝑥 = 𝐼𝐿𝑚𝑎𝑥 = 𝐼𝐷𝑚𝑎𝑥

𝑰𝑳𝒎𝒂𝒙 =𝑰𝟎

𝟏 − 𝑫+𝜼𝒗𝑩𝟐𝑳

𝑫𝑻𝒄𝒌 =𝒗𝟎

𝑹𝑳 𝟏 − 𝑫+𝜼𝒗𝑩𝟐𝑳

𝑫𝑻𝒄𝒌

Summary of results

Voltage Gain

Voltage gain considering power Efficiency

Inductance’s current ripple

Maximum current @ inductor

Switch current

Inductance’s current


The switch must be able to manage the maximum current

Design Considerations


Current Ripple:

LIN: Ratio of Inductance ripple to Average current

Load Capacitance

Computation of load capacitance

Design Considerations


Stability and “AC” Responsehttp://www.ti.com/lit/an/slva061/slva061.pdf

AC analysis is tricky!

System operates under large signal conditions and it is very

non-linear!

AC is for small signal sinewave waveforms! This system Is

not even close to that!


Stability and “AC” Response

http://www.ti.com/lit/an/slva061/slva061.pdf

Similarly to any AC analysis, we

have to assume small signal

variations, and define an

operating point!

Output of the PWM is duty cycle!

This chart includes resistive

losses, otherwise, D=0.7 should

result in V-D convertion ratio of

approximately 4.


Small signal model (TI)

20


22


23


Example: Maxim 668


Boost

Converter

+

Vin

-

Ro

+

Vout

-

iin

+

Vin

-

Req

iin

iout

out

outo

I

VR

o

out

out

out

out

in

ineq RD

I

VD

D

I

VD

I

VR

2211

1

1

D

VV in

out

1

inout IDI 1

Equivalent Impedance

Equivalent impedance

seen from source


Final Advice:

Check websites and look for tutorials of the following companies:

Texas Instruments

Maxim

Linear Technology

And many others

fundamentals on boost (step up) convertersece.tamu.edu/~jose-silva-martinez/courses/ecen489/boost...

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