Forward and Flyback (Converters with isolation)

(1)

Forward and Flyback

(Converters with isolation

)

4.1 Transfer of DC current via transformer

4.2 Forward

4.2.1 Voltage transfer function

4.2.2 Magnetization inductance problem

4.2.3 Transformer reset

4.2.4 Reset of forward

4.3 Coupled inductors

4.4 Flyback

4.4.1 Voltage transfer function

4.4.2 Flyback with multiple outputs

4.4.3 Characteristics of Flyback

Prof. S. Ben-Yaakov , DC-DC Converters [4- 2]

Is transfer of DC current possible?

I

_o

t

s

₁

t

L in

R

n

V

⋅

s

₂

I

_o

V

_in

1:n

R

_L

V

_o

S

₁

S

2

(2)

Replacing S

₂

by a Diode

I

_o

V

_in

D

1:n

R

_L

V

_o

S

₁

D

₁

D

₂

L

I

_L

I

_C

I

_R

C

R

1:n

V

_in

S

V

_o

T

Forward :

z

T is a transformer

z

Output section: Buck

z

Buck derived

(3)

t

S

ON

t

V

_X

t

V

₂

V

in

⋅

n

1 D 2

V

−

T

_s

ON

D

₁

D

₂

L

I

_L

I

_C

I

_R

C

R

1:n

V

_in

S

V

_o

V

_X

V

₂

At steady state, over one switching cycle:

;

L

V

=

0

on in o

_nD

V

0 S

S

₊

+

₋

=

⇒

=

;

t

)

V

nV

(

S

₊

≈

_in

−

_o

⋅

_on

;

t

)

V

(

S

₋

≈

−

_o

⋅

_off

Voltage transfer function

Magnetization Inductance Problem

V

_Lm

t

?

V

_in

S

V

_in

1:n

L

_m

ideal

(4)

Transformer Reset

V

_in

t

_off

t

_on

V

_in

t

_on

)

V

(

_reset

−

_in

V

_L

V

_L

off

in

reset

on

in

D

(

V

)

D

V

=

−

S

V

_in

L

_m

V

_reset

Applying a Reset Winding

on

D

n

in

V

n

reset

V

off

D

1

3 ≥

Reset Requirements

V

_reset

S

V

_in

n

₁

n

₂

n

₃

L

_m1

V

_in

I

_L

V

_reset

ON

OFF

(5)

Applying same source

on

D

n

in

V

off

D

n

in

V

1

3 >

D

_off

on

D

n

<

1

3 V

_reset

S

V

_in

n

₁

n

₂

n

₃

Assignment

Given: 0.1<D

_on

<0.7

What will be the voltage stress on S?

V

_reset

S

V

_in

n

₁

n

₂

(6)

Reset of Forward

Calculation of n

₃

z

Calculation can be done by looking

at any of the windings n

₁

, n

₂

, n

₃

D

₁

D

₂

L

I

_L

I

_C

I

_R

C

R

D

₃

n

1

n

₂

n

₃

V

_in

S

Fundamental requirement:

n

₁

L

_m

n

2

n

₃

o

v

_winding

=

z

Example: Looking at n

₁

S

₁

S

₂

t

1 n

V

t

_on

t

_off

S

₁

S

₂

t

_on

t

_off

t

1 n

V

in

V

in

V

3 1 in

_n

n

V

S

₁

=S

₂ Lm

I

1

2

Lm

I

(7)

on in 3 1 off in

V

D

n

D

V

≥

on off 1 3

_D

D

n

≤

off off 1 3

D

1 D

n

−

≤

S

₁

S

₂

t

1 n

V

t

_on

t

_off

S

₁

S

₂

t

_on

t

_off

t

1 n

V

in

V

in

V

3 1 in

_n

n

V

S

₁

=S

₂ Lm

I

1

2

Lm

I

VT

Voltage on switch when “off”

in 3 1 in s

V

n

V

=

+

]

D

1 D

1 [

V

max on max on in s

=

+

₋

t

1 n

V

D

_on

D

_off in

V

3 1 in

_n

n

V

VT

(8)

2 1

n

=

z

primary current is a reflection

of I

₂

plus I

_m

t

_on

_t

off

T

_S

I

_pk

I

₂

t

_off

I

₁

n

I

pk pk 2

I

n

Switch Current

D

₁

D

₂

L

I

_L

I

_C

I

_R

C

R

D

₃

n

1

n

₂

n

₃

V

_in

I

₁

I

₂

S

Reset winding

Taking into account Lm

D

₁

D

₂

L

C

_R

D

₃

n

₁

n

₂

n

₃

V

_in

I

_Lm

I

₂

S

L

_m

t

I

_Lm

I

₂

n

I

_S

_I

₂

_n+I

_Lm

(9)

Question 1: How does I

_n3

look ?

t

T

_S 3 n

I

_I

pk

?

t

_off

Question 2: If D

_max

=0.3

2.1 Calculate n

₁

/n

₃

2.2 Find maximum voltage on switch

on on in 3 1 in reset on on 3 1

D

1 D

V

n

V

D

1 D

n

−

=

−

=

Reset Current

Coupled inductor

L

₁

L

₂

L

₁

L

₂

n

₁

n

₂ 2 2 1 2 1

n

L













=

(10)

S

₁

S

₂

t

S

₁

S

₂

V

₁

I

1

I

2

V

₂

1:1

Current in a Winding CAN be Interrupted !

S

₁

S

₂

I

₁

I

₂ 1 pk

I

L

V

₂

L

V

₁ 2 pk

I

t

_S

t

At transition

2 1 2 1 pk 2 pk 1 pk pk

I

N

I

N

I

=

S

₁

S

₂

V

₁

I

1

I

2

V

₂

1:1

(not in wires)

Energy stored in core

(11)

Problem: Leakage inductance

(will be discussed later)

S

₁

S

₂

I

₁

I

₂ 1 pk

I

L

V

1 2 pk

I

t

_S

t

_S

t

L

n

V

22

n

I

n

I

1 I

2 1 2 1

pk

=

⋅

=

⋅

S

₁

S

₂

V

₁

I

1

I

2

V

₂

1:n

L

Coupled windings

1

10 n

V

=

⋅

2

20 _n

1 V

V

=

(negative voltage)

S

₁

V

₁

1:n

S

₁

V

₁

V

₂

1:n

V

₂₀

S

₂

V

₁

_V

V

₂ 10

1:n

(12)

Buck Boost

Polarity Reversal

0 D

V

D

V

_in

_on

+

_o

_off

=

off

on

in

o

D

V

−

=

S

V

_in

D

L

C

R

V

_o

Flyback

Buck Boost

Isolated Back-Boost

Flyback

S

V

_in

D

L

C

R

V

_o

S

V

_in

D

C

R

V

_o

(13)

Flyback converter

t

_on

t

_off

D

₁

C

R

1:n

V

_in

S

V

_o

C

R

1:n

V

_in

S

V

_o

nV

_in

C

R

1:n

V

_in

S

V

_o

V

_o

n

V

_o

n

V

_in

I

Voltage transfer function

The average voltage method

Voltage across primary

V

₁

t

_off

t

-V

_in

t

_on

T

_s

n

V

_o

off

o

on

in

t

n

V

t

V

⋅

=

⋅

off

o

on

in

D

V

D

V

⋅

=

⋅

o

on

D

n

V

₌

D

₁

C

R

1:n

V

_in

S

V

_o

V

₁

(14)

Voltage transfer function

The

∆

I method

D

₁

C

R

V

_in

S

V

_o

I

₁

I

₂

L

₁

n

₁

L

₂

n

₂

t

_on

T

_s

t

I

₁ 1

I

∆

1

I

∆

2 1 2

_I

n

;

t

L

V

n

t

L

V

n

I

n

I

n

_off 2 o 2 on 1 in 1 2 2 1 1

∆

=

∆

=

off on off on in o 2 1 1 2

D

t

V

L

n

=

off on in o 2 2 1 1 2

D

V

n

=













off on 1 2 in o

D

n

V

=

Flyback with multiple outputs

V

_in

S

D

₁

V

o1

D

₂

V

o2

C

₁

R

₁

R

₂

C

₂

n

₁

n

₂

(15)

t

_on

t

_off 2 1 2 o o 1

V

n

V

n

=

V

_in

_V

o1

V

_o2

C

₁

R

₁

R

₂

C

₂

n

₁

n

₂

V

_in

_V

o1

V

_o2

C

₁

R

₁

R

₂

C

₂

n

₁

n

₂

Flyback

Characteristics of Flyback

z

Isolation

z

Step-up or Step-down

z

Discontinuous current at input and

output (Buck-Boost Derived)

(16)

Exercise

V2 TD = 0 TF = 0.01u PW = 10u PER = 20u V1 = 0 TR = 0.01u V2 = 15 Dbreak D1 out_gnd 0 _gate L2 {L1*(n*n)} 0 RL {Load} drain V1 {Vin} R4 10meg out 0 C1 220u IC = 6 KK1 COUPLING = 1 K_Linear L1 = l1 L2 = l2 PARAM ET ERS: n = 0.5 Vin = 12 Load = 10 L1 = 300u + -+ -Sbreak S1 L1 {L1}