A8650 datasheet(19/31 Pages) ALLEGRO

19 / 31 page

Low Input Voltage, Adjustable Frequency

2 A Synchronous Buck Regulator with EN/SYNC and Power OK

A8650

Allegro MicroSystems, LLC

115 Northeast Cutoff

Worcester, Massachusetts 01615-0036 U.S.A.

1.508.853.5000; www.allegromicro.com

frequency (fOSC), the FSET resistor can be calculated as follows:

– 1.7

fOSC

24900

RFSET

(2)

where fOSC is in kHz and RFSET is in kΩ.

The oscillator frequency will be the final output switching fre-

quency (unless the synchronization function is applied), so when

the oscillator frequency is being chosen, the designer should be

aware of the minimum controllable on-time, tON(MIN), of the

A8650. If the system required on-time is less than the A8650

minimum controllable on-time, then switch node jitter will occur

and the output voltage will have increased ripple or oscillations.

The minimum oscillator frequency required should be calculated

as follows:

= t

ON(MIN) ×VIN(MAX)

VOUT

fSWMAX

(3)

where

VOUT is the output voltage,

tON(MIN) is the minimum controllable on-time of the A8650

(worst case 105 ns), and

VIN(MAX) is the maximum required operational input voltage

(not the peak surge voltage).

If the A8650 PWM synchronization function is employed, then

the oscillator frequency should be chosen such that jitter will not

result at the maximum synchronized switching frequency, deter-

mined from equation 3:

1.5 × fSW < fSWMAX

(4)

Output Inductor (LO)

When considering a peak current mode regulator, it is common

knowledge that, without adequate slope compensation, the system

will become unstable when the duty cycle is near or above 50%.

However, the slope compensation in the A8650 is a fixed value

(SE). Therefore, it is important to calculate an inductor value so

the falling slope of the inductor current (SF) will work well with

the A8650 slope compensation.

Equations 5 and 6 can be used to calculate a range of values for

the output inductor based on the well known approach of provid-

ing slope compensation that matches 50% to 100% of the falling

slope of the inductor current:

≤

≤ LO

2 × SE

VOUT

(5)

where LO is in µH.

In equation 5, the slope compensation (SE ) is a function of

switching frequency, as follows:

SE = 1.175 × fOSC

(6)

where SE is in A/µs and fSW is in MHz.

More recently, Dr. Raymond Ridley presented a formula to calcu-

late the amount of slope compensation required to critically damp

the double poles at half the PWM switching frequency:

≥

IN(min)

OUT

1– 0.18 ×

(7)

This formula allows the inclusion of the duty cycle (D), which

should be calculated at the minimum input voltage to ensure

optimal stability. Also, to avoid dropout (that is, saturation of the

buck regulator), VIN(min) must be approximately 0.75 to 1.0 V

above VOUT when calculating the inductor value with equation 7.

If equations 6 or 7 yield an inductor value that is not a standard

value, then the next closest available value should be used. The

final inductor value should allow for 10% to 20% of initial toler-

ance and 10% to 20% of inductor saturation.

The saturation current of the inductor should be higher than the

peak current capability of the A8650. Ideally, for output short

circuit conditions, the inductor should not saturate even at the

highest cycle-by-cycle current limit at minimum duty cycle

(ILIM(5%); 4.7 A(max)). This may be too costly. At the very least,

the inductor should not saturate given the peak operating current

according to the following equation:

IPEAK 4.1 –

SE × (VOUT )

1.15 × fSW × VIN(max)

(8)

where VIN(max) is the maximum continuous input voltage, such

as 5.5 V.

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