Tài liệu Circuits & Electronics P24 pptx

Power Conversion Circuits PCCPower efficiency of converter important, so use lots of devices: MOSFET switches, clock circuits, inductors, capacitors, op amps, diodes PCC 110V 60Hz + – 5V

6.002 Fall 2000 Lecture 24 1

6.002 CIRCUITS AND

ELECTRONICS

Power Conversion Circuits

and Diodes

Power Conversion Circuits (PCC)

Power efficiency of converter important,

so use lots of devices:

MOSFET switches, clock circuits,

inductors, capacitors, op amps, diodes

PCC

110V

60Hz

+ – 5V DC

solar cells,

DC-to-DC UP converter

R

6.002 Fall 2000 Lecture 24 3

First, let’s look at the diode

Can use this exponential model with

analysis methods learned earlier

„ analytical „ graphical „ incremental

(Our fake expodweeb was modeled after this device!)

D

v

D

i

D

v

D

i

S

I

D

v

+ –

D

i

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛

−

= I e 1

D

V v

S D

A 10

IS = −12

V 025

0

VT =

q

T

k

VT = Boltzmann’s constanttemperature in Kelvins

charge of an electron

Another analysis method:

piecewise–linear analysis

P–L diode models:

D

v

D

i

0

Ideal diode model

Æ i D = 0

“open”

or off

v D < 0

Æ v D = 0

“short”

or on

i D ≥ 0

6.002 Fall 2000 Lecture 24 5

D

v

D

i

V 6 0

0

vD =

0

iD =

“Practical” diode model

ideal with offset

V 6 0

+ –

Another analysis method:

piecewise–linear analysis

Open segment

Short segment

Another analysis method:

piecewise–linear analysis

„ Replace nonlinear characteristic with

linear segments.

„ Perform linear analysis within each

segment.

Piecewise–linear analysis method

6.002 Fall 2000 Lecture 24 7

(We will build up towards an AC-to-DC converter)

R vO

+

–

+ –

I

v

V 6

0

+ –

Example

Consider

v I is a sine wave

vO =

0

i D =

“Open segment”:

R

+ – vI

+

–

+ –

V 6

0 6

.

0

v I <

R vO

+

–

+ –

I

v

(v 0 6)/ R

i D = I −

6 0 v

vO = I −

“Short segment”:

R

+ –

+

–

+ –

V 6 0

6

.

0

v I ≥ vI

Example

V 6

0

circuit

6.002 Fall 2000 Lecture 24 9

Example

t

6

.

0

I

v

O

v

Now consider — a half-wave rectifier

I

O

v

+ –

+ –

V 6

0

+–

C

6.002 Fall 2000 Lecture 24 11

A half-wave rectifier

t

diode on diode off

C

current pulses charging capacitor

MIT’s supply shows

“snipping” at the peaks

(because current drawn

at the peaks)

I

v

O

v

Demo

DC-to-DC UP Converter

The circuit has 3 states:

I S is on, diode is off

i increases linearly

II S turns off, diode turns on

C charges up, v O increases

III S is off, diode turns off

C holds v (discharges into load)

t

S

v

S

closed S

open

T

p

T

O

v

+

–

+ –

DCI

V

C

S

i

switch

S

Do not use resistive elements!

6.002 Fall 2000 Lecture 24 13

More detailed analysis

I Assume i(0) = 0, vO(0) > 0

S on at t = 0, diode off

+ –

I

i

L

O

v

t

i

L

T

V

T

i ( ) = I

T

dt

di L

VI =

i is a ramp

L

VI

= slope

2

) T (

Li 2

1 : T t

at stored energy

Δ

L

T

V

E I

2

2

2

= Δ

II S turns off at t = T

diode turns on (ignore diode voltage drop)

+ –

I

L

O

v

S

i

Diode turns off at T′ when i tries to go negative.

t

i

T 0

L

T

VI

LC

O

1

=

ω

T ′ TP

State III starts here

6.002 Fall 2000 Lecture 24 15

II S turns off at t = T, diode turns on

Diode turns off at T′ when I tries to go negative.

LC

O

1

=

ω

ignore

diode

drop vO(T )

O

v

T

0

Capacitor voltage

P

T

O

v

Δ

t

i

T 0

L

T

VI

T ′ TP

LC

O

1

=

ω

III.

Let’s look at the voltage profile

II S turns off at t = T, diode turns on

LC

O

1

=

ω

ignore

diode

drop vO(T )

O

v

T

0

Capacitor voltage

P

T

O

v

Δ

t

i

T 0

L

T

VI

T ′ TP

LC

O

1

=

ω

III.

Let’s look at the voltage profile

6.002 Fall 2000 Lecture 24 17

III S is off, diode turns off

C holds v O after T′

i is zero

+ –

I

+

–

Eg, no load

O

v

0

Capacitor voltage

P

T

III S is off, diode turns off

C holds v O after T′

i is zero

until S turns ON at T P, and cycle repeats

I II III I II III …

Thus, v O increases each cycle, if there is no load

O

v

)

(n

vO

+ –

I

+

–

Eg, no load

6.002 Fall 2000 Lecture 24 19

What is vO after n cycles Æ vO(n) ?

Use energy argument … (KVL tedious!)

Each cycle deposits ∆E in capacitor.

2

) T t

( i

L 2

1

E = = Δ

2 I

L

T

V L 2

1

⎟

⎠

⎞

⎜

⎝

⎛

=

L

T

V 2

1 E

2

2 I

= Δ

After n cycles, energy on capacitor

L 2

T

nV E

n

2

2 I

= Δ

This energy must equal 2

O( n )

Cv 2 1

or

LC

T

nV )

n ( v

2

2 I

LC

1

O =

ω

n T

V )

n (

so,

L 2

T

nV )

n (

Cv 2

1 2 I 2 2

How to maintain vO at a given value?

recall

L

T

V

E I

2

2

2

= Δ

Another example of negative feedback:

↓

↑

−

T v

v

T v

vO ref

then if

then if

O

v

+

–

+

–

I

control

change T

T

p

T

pwm

+ vref

compare –

O

v