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CS51031YD8

Fast PFET Buck Controller Does Not Require Compensation

Cherry semiconductor 

Applications Information: continued

where VSAT = Rds(on) ´ IOUT max. and Rds(on)is the value at

TJ 100ûC.

If VF = 0.60V and VSAT = 0.60V then the above equation

becomes:

DMAX =

5.6

9

= 0.62

DMIN =

5.6 = 0.40

13.8

2) Switching frequency and on and off time calculations

Given that fSW = 200kHz and DMAX = 0.80

T = 1 = 5µs

fSW

TON(max) = T ´ DMAX = 5µs ´ 0.62 @ 3µs

TON(min) = T ´ DMIN = 5µs ´ 0.40 = 2µs

TOFF(max) = TON(min) = 5µs - 2µs = 3µs

3) Oscillator Capacitor Selection

The switching frequency is set by COSC, whose value is

given by:

95 ´ 10-6

COSC in pF =

( ( ) Fsw

1+

3

Fsw

´ 10

6

-

30 ´ 10 3

Fsw

2

)

5) Output capacitor

The output capacitor and the inductor form a low pass fil-

ter. The output capacitor should have a low ESL and ESR.

Low impedance aluminum electrolytic, tantalum or organ-

ic semiconductor capacitors are a good choice for an out-

put capacitor. Low impedance aluminum are less expen-

sive. Solid tantalum chip capacitors are available from a

number of suppliers and are the best choice for surface

mount applications.

The output capacitor limits the output ripple voltage. The

CS51031 needs a maximum of 20mV of output ripple for

the feedback comparator to change state. If we assume that

all the inductor ripple current flows through the output

capacitor and that it is an ideal capacitor (i.e. zero ESR), the

minimum capacitance needed to limit the output ripple to

50mV peak to peak is given by:

C=

ÆI

=

0.6A

= 7.5µF

8 ´ fSW ´ ÆV 8 ´ (200 ´ 103Hz) ´ (50 ´ 10-3V)

The minimum ESR needed to limit the output voltage rip-

ple to 50mV peak to peak is:

ESR = ÆV = 50 ´ 10-3 = 83m½

ÆI

0.6A

The output capacitor should be chosen so that its ESR is

less than 83m½.

During the minimum off time, the ripple current is 0.4A

and the output voltage ripple will be:

ÆV=ESR ´ ÆI = 83m½ ´ 0.4 = 33mV.

4) Inductor selection

The inductor value is chosen for continuous mode opera-

tion down to 0.3Amps.

The ripple current ÆI = 2 ´ IOUTmin = 2 ´ 0.3A = 0.6A.

Lmin =

(VOUT + VD) ´ TOFF(max)

ÆI

=

5.6V ´ 3µs

0.6A

=28µH

This is the minimum value of inductor to keep the ripple

current to <0.6A during normal operation.

A smaller inductor will result in larger ripple current.

Ripple current at a minimum off time is

ÆI = (VOUT + VF) ´ TOFF(min) = 5.6V ´ 2µs =0.4A

LMIN

28µH

The core must not saturate with the maximum expected

current, here given by:

IMAX = IOUT + ÆI/2 = 3A+0.4A/2 = 3.2A

6) VFB divider

( ) ( ) VOUT = 1.25V

R1 + R2

R2

= 1.25V

R1

R2

+1

The input bias current to the comparator is 4µA. The resis-

tor divider current should be considerably higher than this

to ensure that there is sufficient bias current. If we choose

the divider current to be at least 250 times the bias current

this permits a divider current of 1mA and simplifies the

calculations.

5V = R1 + R2 = 5k½

1mA

Let R2 = 1K

Rearranging the divider equation gives:

( ) ( ) R1 = R2 VOUT - 1 = 1k½ 5V - 1 = 3k½

1.25

1.25

6

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