6

Applications Information: continued

D =

2) Switching Frequency and on and off time calculations.

95

T =

= 5µs

3) Inductor selection

Pick the inductor value to maintain continuous mode opera-

tion down to 0.3 Amps.

The CS51033 will operate with almost any value of inductor.

With larger inductors the ripple current is reduced and the

regulator will remain in a continuous conduction mode for

lower values of load current. A smaller inductor will result

in larger ripple current. The core must not saturate with the

maximum expected current, here given by:

= 3A + 0.6A/2 = 3.3A

4) Output Capacitor

The output capacitor limits the output ripple voltage. The

CS51033 needs a maximum of 15mV of output ripple for the

feedback comparator to change state. If we assume that all

the inductor ripple current flows through the output capaci-

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

mum capacitance needed to limit the output ripple to 50mV

peak to peak is given by:

=

≅ 11.4µF

The minimum ESR needed to limit the output voltage ripple

to 50mV peak to peak is:

ESR =

=

= 55mΩ

The output capacitor should be chosen so that its ESR is at

least half of the calculated value and the capacitance is at

least ten times the calculated value. It is often advisable to

use several capacitors in parallel to reduce the ESR.

Low impedance aluminum electrolytic, tantalum or organic

semiconductor capacitors are a good choice for an output

capacitor. Low impedance aluminum are the cheapest but

are not available in surface mount at present. Solid tantalum

chip capacitors are available from a number of suppliers

and offer the best choice for surface mount applications. The

capacitor working voltage should be greater than the output

voltage in all cases.

The input bias current to the comparator is 4µA. The resistor

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

gives a divider current of 1mA and simplifies the calcula-

tions.

= R1+R2 = 1.5KΩ

Let R2 = 1K

Rearranging the divider equation gives:

R1 = R2

(

-1

6) Divider bypass capacitor Crr

Since the feedback resistors divide the output voltage by a

factor of 4, i.e. 5V/1.25V= 4 it follows that the output ripple

is also divided by four. This would require that the output

ripple be at least 60mV (4 × 15mV) to trip the feedback com-

pactor. We use a capacitor Crr to act as an ac short so that

the output ripple is not attenuated by the divider network.

The ripple voltage frequency is equal to the switching fre-

quency so we choose Crr so that:

is negligible at the switching frequency.

C =

≅ 0.265µF

dead time for load transients so that the IC does not enter a

fault mode every time the load changes abruptly. Secondly

it disables the fault circuitry during startup, it also provides

soft start by clamping the reference voltage during startup

1

2πf3

1

2πfC

1.5V

1.25

1.25

1.5V

1mA

R1

R2

R1 + R2

R2

0.6A

∆V

∆I

0.6A

∆I

2

2.1V × 4.3µs

0.6A

∆I

1