MCP1612T-ADJI/MF(2005) 查看數據表（PDF） - Microchip Technology

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MCP1612T-ADJI/MF
(Rev.:2005)

Single 1A, 1.4 MHz Synchronous Buck Regulator

Microchip Technology 

MCP1612

5.2.2 BUCK INDUCTOR

There are many requirements that need to be satisfied

when selecting the buck inductor. The application,

physical size, current rating, resistance, mounting

method, supplier, temperature range, minimum

inductance and cost all need to be considered.

Many suppliers specify the maximum peak current that

an inductor can handle before magnetic saturation

occurs. The peak current is equal to the maximum DC

output current, plus one-half the peak-to-peak AC

ripple current.

When the P-channel MOSFET is on, the current in the

buck inductor is ramped up. The voltage across the

inductor, the inductance and the MOSFET on-time are

required to determine the peak-to-peak ripple current.

When operating in Continuous Current mode, the on-

time of the P-channel MOSFET is determined by

multiplying the duty cycle by the switching period. The

following equation can be used to determine the duty

cycle.

EQUATION 5-2:

DutyCycle = -V---O----U----T-

VIN

The on-time is then defined as follows.

EQUATION 5-3:

TON

=

DutyCycle × ----1-----

FSW

Where:

FSW = switching frequency

The AC ripple current in the inductor can be calculated

by the following relationship.

EQUATION 5-4:

VL

=

L

×

Δ-----I--L-

Δt

Solving for ΔIL yields:

EQUATION 5-5:

ΔIL

=

-V---L- × Δt

L

Where:

VL = voltage across the inductor

(VIN – VOUT)

Δt = on-time of the P-channel MOSFET

The value of the buck inductor is chosen to be 3.3 µH.

The AC ripple current is controlled by the size of the

buck inductor. The value of the inductor will therefore

need to be raised so that the converter operates in

Continuous Conduction mode. Calculation of the buck

inductor current rating follows.

VIN = 3.3V

VOUT = 1.2V

FSW = 1.4 MHz

IOUT(MAX) = 1A

TON = (1.2V/3.3V) x (1/1.4 MHz)

TON = 260 ns

VL = (3.3V – 1.2V) = 2.1V

ΔIL = (2.1V/3.3 µH) x 260 ns

ΔIL = 165 mA

IL(PEAK) = IOUT(MAX) + 1/2 ΔIL

IL(PEAK) = 1A + (165 mA)/2

IL(PEAK) = 1.08A

The inductor selected must have an inductance of

3.3 µH at a peak current rating of 1.08A. The DC resis-

tance of the inductor should be as low as is feasibly

possible. Extremely low DC resistance inductors are

available, though a trade-off between size and cost

should be considered.

5.2.3 OUTPUT CAPACITOR

The output capacitor is used to filter the inductor AC

ripple current and provide storage for load transients.

The size and Equivalent Series Resistance (ESR) of

the output capacitor determines the amount of ripple

voltage present at the output of the converter. When

selecting the output capacitor, a design trade-off has to

be made between the acceptable ripple voltage and the

size/cost of the output capacitor. Ceramic capacitors

have very low ESR, but increase in cost with higher

values. Tantalum and electrolytic capacitors are

relatively inexpensive in higher values, but they also

have a much higher ESR.

The amount of capacitance needed to obtain the

desired ripple voltage is calculated by using the

following relationship.

EQUATION 5-6:

IC = C × Δ---Δ--V--t-C--

DS21921B-page 12

© 2005 Microchip Technology Inc.

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