AIC1639

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

SPEC NO: DS-1639-00 05/26/00

ANALOG INTEGRATIONS CORPORATION

www.analog.com.tw

4F, 9 Industry E. 9th Road, Science-based Industrial Park , Hsinchu, Taiwan, R.O.C.

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

9

t

Continuous Conduction Mode

At the boundry between continuous and discontinu-

ous mode, output current (IOB) is determined by

( )

x

1

Ton

L

Vin

2

1

Vd

Vout

Vin

−













+

where Vd is the diode drop,

x = (Ron+Rs)Ton/L.

Ron= Switch turn on resistance, Rs= Inductor DC re-

sistance

Ton = Switch ON time

In the discontinuous mode, the switching frequency

(fsw) is

fsw =

)

x

(1

Ton

Vin

Vin)(Iout)

Vd

2(L)(Vout

2

2

+

×

−

+

In the continuous mode, the switching frequency is

fsw=

(

)

)]

Vsw

Vd

Vout

Vsw

Vin

(

2

x

[1

Vsw)

Vd

(Vout

Vin

Vd

Vout

Ton

1

−

+

−

+

−

+

−

+













−

+

−

+

≅

Vsw

Vd

Vout

Vin

Vd

Vout

Ton

1

where Vsw = switch drop and proportion to output cur-

rent.

INDUCTOR SELECTION

To operate as an efficient energy transfer element, the

inductor must fulfill three requirement. First, the induc-

tance must be low enough for the inductor to store

adquate energy under the worst case condition of

minimum input voltage and switch ON time. Second,

the inductance must also be high enough so maxi-

mum current rating of AIC1639 and inductor are not

exceed at the other worst case cindition 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 unfo-

trunately this is inversely related to physical size.

Minimum and Maximum input voltage, output voltage

and output current must be established before and in-

cuctor can be selected.

In discontinuous mode operation, at the end of the

switch ON time, peak current and energy in the induc-

tor build according to













+

−

−













+

=

Ton)

L

Rs

Ron

exp(

1

Rs

Ron

Vin

( )











 −













≅

2

x

1

Ton

L

Vin

Ton

L

Vin

≅

(simple lossless equation),

where x=(Ron+Rs)Ton/L

2

Ipk

L

2

1 ×

Power required from the inductor per cycle must be

equal or greater than

)

fsw

1

(

Vin)(Iout)

Vd

(Vout

/f

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

Ton

2L

Vsw

Vin

Iout

2

x

Vsw

Vin

Vsw

Vd

Vout