MP3410 — 1.3A, 550kHz SYNCHRONOUS RECTIFIED STEP-UP CONVERTER
MP3410 Rev. 1.12 www.MonolithicPower.com 7
6/30/2011 MPS Proprietary Information. Patent Protected. Unauthorized Photocopy and Duplication Prohibited.
© 2011 MPS. All Rights Reserved.
APPLICATION INFORMATION
COMPONENT SELECTION
Setting the Output Voltage
Set the output voltage by selecting the resistive
voltage divider ratio. The voltage divider drops
the output voltage to the 1.19V feedback
voltage. Use a 100k resistor for R2 of the
voltage divider. Determine the high-side resistor
R1 by the equation:
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
=
2R
V
VV
1R
FB
FBOUT
Where V
OUT
is the output voltage, V
FB
is the
1.19V feedback voltage and R2=100k.
Selecting the Input Capacitor
An input capacitor is required to supply the AC
ripple current to the inductor, while limiting noise
at the input source. Multi-layer ceramic
capacitors are the best choice as they have
extremely low ESR and are available in small
footprints. Use an input capacitor value of 4.7F
or greater. This capacitor must be placed
physically close to the device.
Selecting the Output Capacitor
A single 4.7µF to 10µF ceramic capacitor
usually provides sufficient output capacitance
for most applications. Larger values up to 22µF
may be used to obtain extremely low output
voltage ripple and improve transient response.
The impedance of the ceramic capacitor at the
switching frequency is dominated by the
capacitance, and so the output voltage ripple is
mostly independent of the ESR. The output
voltage ripple V
RIPPLE
is calculated as:
SW
UT
O
IN
UT
OLOAD
RIPPLE
f2CV
VVI
V
××
−
=
Where V
IN
is the input voltage, I
LOAD
is the load
current, C2 is the capacitance of the output
capacitor and f
SW
is the 550kHz switching
frequency.
Selecting the Inductor
The inductor is required to force the output
voltage higher while being driven by the lower
input voltage. A good rule for determining the
inductance 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 the minimum
current limit at the duty cycle used (to prevent
loss of regulation due to the current limit
variations).
Calculate the required inductance value L using
the equations:
IfV
) V- (VV
L
SWOUT
INOUTIN
Δ××
=
η×
×
=
IN
)MAX(
LOADOUT
)MAX(IN
V
IV
I
)
)MAX(IN
I%50%30I −
Where I
LOAD(MAX)
is the maximum load current, I
is the peak-to-peak inductor ripple current and
is efficiency. For the MP3410, typically, 4.7µH is
recommended for most applications. Choose an
inductor that does not saturate at the peak
switch current as calculated above with
additional margin to cover heavy load transients
and extreme startup conditions.
