4

 2006 Semtech Corp.

www.semtech.com

SC1480

POWER MANAGEMENT

Notes:

(1) When the inductor is in continuous conduction mode, the output voltage will have a DC regulation level higher

than the error-comparator threshold by 50% of the ripple voltage.

(2) Using a current sense resistor, this measurement relates to PGND minus the voltage of the source on the

low-side MOSFET.

(3) This device is ESD sensitive. Use of ESD handling precautions is required.

(4) Measured in accordance with JESD51-1, JESD51-2 and JESD51-7.

Electrical Characteristics Cont.

r

e

t

e

m

a

r

a

Ps

n

o

i

t

i

d

n

o

CC

°

5

2C

°

5

2

1

o

t

C

°

0

4

-s

t

i

n

U

n

i

Mp

y

Tx

a

Mn

i

Mx

a

M

y

a

l

e

D

tl

u

a

F

D

O

O

G

PD

O

O

G

P

e

d

i

s

t

u

o

d

e

c

r

o

f

B

F

.

w

o

d

n

i

w

0

.

2s

µ

d

l

o

h

s

e

r

h

T

e

g

a

tl

o

V

r

e

d

n

U

A

C

C

Vv

m

0

0

1

s

i

s

e

r

e

t

s

y

H

e

g

d

E

g

n

i

s

i

R0

.

47

.

33

.

4V

d

l

o

h

s

e

r

h

T

e

g

a

tl

o

V

r

e

d

n

U

P

D

D

V

3

.

30

.

35

7

.

3V

s

i

s

e

r

e

t

s

y

H

e

g

a

tl

o

V

r

e

d

n

U

P

D

D

V

0

5

2V

m

t

u

o

k

c

o

L

e

r

u

t

a

r

e

p

m

e

T

r

e

v

Os

i

s

e

r

e

t

s

y

H

C

°

0

15

6

1C

t

r

a

t

S

t

f

o

S

e

m

i

T

p

m

a

R

t

r

a

t

S

-

tf

o

S.t

i

m

il

t

n

e

r

r

u

c

ll

u

f

o

t

h

g

i

h

N

I

F

E

R6

.

1s

m

e

m

i

T

k

n

a

l

B

e

g

a

tl

o

V

-

r

e

d

n

Un

O

-

n

r

u

T

S

P

M

S2

s

m

s

r

e

v

i

r

D

e

t

a

G

e

m

i

T

d

a

e

Dg

n

i

s

ir

L

D

r

o

H

D0

3s

n

e

c

n

a

t

s

i

s

e

R

n

w

o

D

-l

l

u

P

L

Dw

o

l

L

D8

.

06

.

1

Ω

e

c

n

a

t

s

i

s

e

R

p

U

-l

l

u

P

L

Dh

g

i

h

L

D2

4

Ω

e

c

n

a

t

s

i

s

e

R

n

w

o

D

-l

l

u

P

H

DV

5

=

X

L

-

T

S

B

,

w

o

l

H

D2

4

Ω

e

c

n

a

t

s

i

s

e

R

p

U

-l

l

u

P

H

DV

5

=

X

L

-

T

S

B

,

h

g

i

h

H

D2

4

Ω

t

n

e

r

r

u

C

k

n

i

S

L

DV

5

.

2

=

D

N

G

P

-

L

D2

A

t

n

e

r

r

u

C

e

c

r

u

o

S

L

DV

5

.

2

=

L

D

-

P

D

D

V1

A

t

n

e

r

r

u

C

k

n

i

S

H

DV

5

=

X

L

-

T

S

B

,

V

5

.

2

=

X

L

-

H

D1

A

t

n

e

r

r

u

C

e

c

r

u

o

S

H

D,

V

5

.

2

=

H

D

-

T

S

B

V

5

=

X

L

-

T

S

B

1A

Test Conditions: V