LTC2927

12

2927fb

Figure 15. Supply Sequencing (from Figure 16)

Figure 16. Supply Sequencing Example

1V/DIV

10ms/DIV

MASTER

SLAVE1

SLAVE2

2927 F15

10ms/DIV

1V/DIV

MASTER

3.3V

138k

100k

ON

RAMP

SLAVE1

1.8V

3.3V

3.3V

LTC2927

GND

SDO

RUN/SS

FB

IN

DC/DC

FB = 1.235V

OUT

RAMPBUF

TRACK

0.1μF

1.65k

16.5k

35.7k

3.48k

ON

RAMP

SLAVE2

2.5V

2927 F16

3.3V

EARLY

3.3V

LTC2927

GND

SDO

RUN/SS

FB

IN

DC/DC

FB = 0.8V

OUT

RAMPBUF

TRACK

0.1μF

0.1μF

88.7k

887k

412k

36.5k

In Figure 15, the slave 1 supply and the slave 2 supply are

sequenced instead of tracking. The 3.3V master ramps up

at 100V/s. The 1.8V slave 1 supply ramps up at 1000V/s

beginning 10ms after the master signal starts to ramp up.

The 2.5V slave 2 supply ramps up at 1000V/s beginning

25ms after the master signal begins to ramp up. Note

that not every combination of ramp rates and delays is

possible. Small delays and large ratios of slave ramp rate

to master ramp rate may result in solutions that require

negative resistors. In such cases, either the delay must

be increased or the ratio of slave ramp rate to master

ramp rate must be reduced. In this example, solving for

the slave supply yields:

1. Set the ramp rate of the master signal.

From Equation 1:

C

A

Vs

F

RAMP

==

μ

10

100

01

μ

/

.

2. Solve for the pair of resistors that provide the desired

slave supply behavior, assuming no delay.

Rk

Vs

Vs

k

TB

=•

=

16 5

100

1000

165

.

/

/

.

ΩΩ

From Equation 3:

R

V

V

k

V

k

V

k

k

TA

′ =

+−

=−

08

1 235

16 5

1 235

35 7

08

165

213

.

.

.

.

.

.

.

.

ΩΩ

Ω

Ω

From Equation 4:

R

Vk

ms

V s

k

TA

″ =

•

•

=

08

165

10

100

132

..

/

.

Ω

Ω

From Equation 5:

Rk

k

k

TA

=−

=

213

1 32

348

..

.

ΩΩ

Ω

Supply Sequencing Example

APPLICATIO S I FOR ATIO