ADP2114 Analog Devices, ADP2114 Datasheet - Page 30

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ADP2114

Manufacturer Part Number
ADP2114
Description
Configurable, Dual 2 A/Single 4 A, Synchronous Step-Down DC-to-DC Regulator
Manufacturer
Analog Devices
Datasheet

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ADP2114
DESIGN EXAMPLE
The external component selection procedure from the Control
Loop Compensation section is used for this design example.
Table 9. 2-Channel Step-Down DC-to-DC Converter
Requirements
Parameter
Input Voltage, V
Channel 1, V
Channel 2, V
Pulse-Skip Feature
CHANNEL 1 CONFIGURATION AND COMPONENTS
SELECTION
Complete the following steps to configure Channel 1:
1.
2.
For the target output voltage, V
V1SET pin through a 47 kΩ resistor to GND (see Table 4).
Because one of the fixed output voltage options is chosen,
the feedback pin (FB1) must be directly connected to the
output of Channel 1, V
Estimate the duty-cycle, D, range. Ideally,
That gives the duty cycle for the 3.3 V output voltage and
the nominal input voltage of D
The minimum duty cycle, D
voltage (10% above the nominal) is D
maximum = 5.5 V
The maximum duty cycle, D
voltage (10% less than nominal) is D
minimum = 4.5 V.
However, the actual duty cycle is larger than the calculated
values to compensate for the power losses in the converter.
Therefore, add 5% to 7% at the maximum load.
Based on the estimated duty-cycle range, choose the
switching frequency according to the minimum and
maximum duty-cycle limitations, as shown in Figure 72.
For the Channel 1 V
choose f
This frequency option provides the smallest sized solution.
If a higher efficiency is required, choose the 300 kHz option.
However, the PCB footprint area of the converter will be
larger because of the bigger inductor and output capacitors.
D =
V
V
OUT
IN
OUT1
OUT2
SW
IN
= 600 kHz with a maximum duty cycle of 0.8.
Specification
5.0 V ±10%
3.3 V, 2 A, 1% V
ripple (p-p)
1.8 V, 2 A, 1% V
ripple (p-p)
Enabled
IN
= 5 V and V
OUT1
.
MIN
MAX
NOM
, for the maximum input
OUT
OUT
OUT
, for the minimum input
OUT
= 0.66 at V
= 3.3 V, connect the
= 3.3 V combination,
MAX
MIN
Additional
Requirements
None
Maximum load step:
1 A to 2 A, 5% droop
maximum
Maximum load step:
1 A to 2 A, 5% droop
maximum
None
= 0.73 at V
= 0.60 at V
IN
= 5.0 V.
IN
IN
(20)
Rev. 0 | Page 30 of 40
3.
4.
5.
Select the inductor by using Equation 5.
In Equation 5, V
and f
Therefore, when L = 3.3 μH (the closest standard value) in
Equation 3, ΔI
Although the maximum output current required is 2 A, the
maximum peak current is 3.3 A under the current limit
condition (see Table 7). Therefore, the inductor should be
rated for 3.3 A of peak current and 3 A of average current
for reliable circuit operation.
Select the output capacitor by using Equation 8 and
Equation 9.
Equation 8 is based on the output ripple (ΔV
Equation 9 is for capacitor selection based on the transient
load performance requirements that allow, in this case, 5%
maximum deviation. As previously mentioned, perform
these calculations and choose whatever equation yields the
larger capacitor size.
In this case, the following values are substituted for the
variables in Equation 8 and Equation 9:
ΔI
f
ΔV
ESR = 3 mΩ (typical for ceramic capacitors)
ΔI
ΔV
The output ripple based calculation (see Equation 8) dictates
that C
calculation (see Equation 9) dictates that C
meet both requirements, choose the latter. As previously
mentioned in the Control Loop Compensation section, the
capacitor value reduces with applied dc bias; therefore, select a
higher value. In this case, the next higher value is 47 μF
with a minimum voltage rating of 6.3 V.
Calculate the feedback loop, compensation component
values by using Equation 15.
H(s) = g
L
C
C
SW
OUT_MIN
OUT_MIN
=
L
OUT_STEP
= 600 kHz
RIPPLE
DROOP
= 0.566 A
(
SW
V
Δ
OUT
IN
= 600 kHz, which results in L = 3.11 μH.
I
M
L
= 33 mV (1% of 3.3 V)
= 0.165 V (5% of 3.3 V)
= 4.0 μF, whereas the transient load based
× G
×
= 1 A
V
8
ΔI
f
OUT
×
SW
CS
OUT_STEP
L
f
SW
IN
×
= 0.566 A.
)
×
= 5 V, V
×
V
V
V
(
V
OUT
REF
ΔV
OUT
IN
×
RIPPLE
ΔI
× Z
OUT
f
L
SW
COMP
-
= 3.3 V, ΔI
ΔI
×
ΔV
L
3
(s) × Z
×
DROOP
ESR
L
)
FILT
= 0.3 × I
OUT
(s)
RIPPLE
= 30 μF. To
L
= 0.6 A,
), and

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