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EV0034

2A, 1.3MHz Step-Up Converter

Monolithic Power Systems 

Monolithic Power Systems

at the switching frequency, and so the output

ripple is calculated as:

VRIPPLE

≈

(1 −

VIN

VOUT

) × ILOAD

C2 × fSW

+ ILOAD × RESR × VOUT

VIN

Where RESR is the equivalent series resistance

of the output capacitors.

Choose an output capacitor to satisfy the

output ripple and load transient requirements

of the design. A 4.7µF-22µF ceramic capacitor

is suitable for most applications.

Selecting the Inductor

The inductor is required to force the higher

output voltage while being driven by the input

voltage. A larger value inductor results in less

ripple current that results in lower peak

inductor current, reducing stress on the

internal n-channel.switch. However, the larger

value inductor has a larger physical size,

higher series resistance, and/or lower

saturation current.

A 4.7µH inductor is recommended for most

applications. However, a more exact

inductance value can be calculated. A good

rule of thumb is to allow the peak-to-peak

ripple current to be approximately 30-50% of

the maximum input current. Make sure that the

peak inductor current is below 75% of the

current limit at the operating duty cycle to

prevent loss of regulation due to the current

limit. Also make sure that the inductor does not

saturate under the worst-case load transient

and startup conditions. Calculate the required

inductance value by the equation:

L = VIN × (VOUT - VIN )

VOUT × fSW × ∆I

IIN(MAX)

=

VOUT

×ILOAD(MAX)

VIN × η

∆I = (30% − ) 50% IIN(MAX)

Where ILOAD(MAX) is the maximum load current, ∆I

is the peak-to-peak inductor ripple current, and η

is efficiency.

MP1527

2A, 1.3MHz

Step-Up Converter

Selecting the Diode

The output rectifier diode supplies current to the

inductor when the internal MOSFET is off. To

reduce losses due to diode forward voltage and

recovery time, use a Schottky diode with the

MP1527. The diode should be rated for a

reverse voltage equal to or greater than the

output voltage used. The average current

rating must be greater than the maximum load

current expected, and the peak current rating

must be greater than the peak inductor current.

Compensation

The output of the transconductance error

amplifier (COMP) is used to compensate the

regulation control system. The system uses

two poles and one zero to stabilize the control

loop. The poles are fP1 set by the output

capacitor and load resistance and fP2 set by

the compensation capacitor C3. The zero fZ1

is set by the compensation capacitor C3 and

the compensation resistor R3. These are

determined by the equations:

fP1 = 1 / (π*C2*RLOAD)

fP2 = GEA / (2π*AVEA*C3)

fZ1 = 1 / (2π*C3*R3)

Where RLOAD is the load resistance, GEA is the

error amplifier transconductance, and AVEA is

the error amplifier voltage gain.

The DC loop gain is:

AVDC = AVEA*GCS*(VIN / VOUT)*RLOAD*(VFB / VOUT)

or

AVDC = AVEA*GCS*VIN*VFB*RLOAD /(VOUT)2

Where GCS is the current sense gain, VIN is the

input voltage, VFB is the feedback regulation

threshold, and VOUT is the regulated output

voltage.

MP1527 Rev 1.8_8/31/05

Monolithic Power Systems, Inc.

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