AIC1638

3-PIN ONE-CELL STEP-UP DC/DC CONVERTER

SPEC NO: DS-1638-00 03/01/00

ANALOG INTEGRATIONS CORPORATION

www.analog.com.tw

4F,9IndustryE.9th Road,Science-basedIndustrial Park , Hsinchu,Taiwan, R.O.C.

TEL: (8863)577-2500 FAX:(8863)577-2510

13

At

the

boundary

between

continuous

and

discontinuous mode, output current (IOB) is

determined by

()

x

1

T

L

V

2

1

V

V

V

I

ON

IN

D

OUT

IN

OB

−













+

=

where V

x = (R

R

resistance

T

In the discontinuous mode, the switching frequency

(fsw) is

x

(1

2

T

2

V

)

)(I

V

V

2(L)(V

fsw

ON

IN

OUT

IN

D

OUT

+

×

−

+

=

In the continuous mode, the switching frequency is

()

)]

V

V

V

V

V

(

2

x

[1

)

V

V

(V

V

V

V

T

1

fsw

SW

D

OUT

SW

IN

SW

D

OUT

IN

D

OUT

ON

−

+

−

+

−

+

−

+

=













−

+

−

+

≅

SW

D

OUT

IN

D

OUT

ON

V

V

V

V

V

V

T

1

where Vsw = switch drop and proportion to output

current.

INDUCTOR SELECTION

To operate as an efficient energy transfer element,

the inductor must fulfill three requirements. First, the

inductance must be low enough for the inductor to

store adequate energy under the worst case

condition of minimum input voltage and switch ON

time. Second, the inductance must also be high

enough so maximum current rating of AIC1638 and

inductor are not exceed at the other worst case

condition of maximum input voltage and ON time.

Lastly, the inductor must have sufficiently low DC

resistance so excessive power is not lost as heat in

the windings. But unfortunately this is inversely

related to physical size.

Minimum and Maximum input voltage, output voltage

and output current must be established before and

inductor can be selected.

In discontinuous mode operation, at the end of the

switch ON time, peak current and energy in the

inductor build according to













+

−

−













+

=

Ton)

L

Rs

Ron

exp(

1

Rs

Ron

Vin

()











 −













≅

2

x

1

T

L

V

ON

IN

ON

IN

T

L

V

≅

(Simple losses equation),

2

Ipk

L

2

1

EL

×

=

Power required from the inductor per cycle must be

equal or greater than

)

f

1

)(

)(I

VI

V

(V

/f

P

SW

OUT

N

D

OUT

SW

L

−

+

=

in order for the converter to regulate the output.

When loading is over IOB, PFM controller operates

in continuous mode. Inductor peak current can be

derived from











 −













−

+













−

−

−

+

=

2

x

1

T

2L

V

V

I

2

x

V

V

V

V

V

I

ON

SW

IN

OUT

SW

IN

SW

D

OUT

PK

Valley current (Iv) is











 −

×













−

−













−

−

−

+

=

2

x

1

T

2L

V

V

I

2

x

V

V

V

V

V

Iv

ON

SW

IN

OUT

SW

IN

SW

D

OUT