ADP2116-EVALZ AD [Analog Devices], ADP2116-EVALZ Datasheet - Page 30

• Manufacturer Part Number: ADP2116-EVALZ
• Description: Configurable, Dual 3 A/Single 6 A, Synchronous, Step-Down DC-to-DC Regulator
• Manufacturer: AD [Analog Devices]

ADP2116
5.
Table 10. Channel 1 Circuit Settings
Circuit Parameter
Output Voltage, V
Reference Voltage, V
Error Amplifier Transconductance, g
Current-Sense Gain, G
Switching Frequency, f
Crossover Frequency, f
Zero Frequency, f
Output Inductor, L
Output Capacitor, C
Compensation Resistor, R
Compensation Capacitor, C
CHANNEL 2 CONFIGURATION AND COMPONENTS
SELECTION
Complete the following steps to configure Channel 2:
1.
2.
Calculate the compensation component values of the
feedback loop by using the following equation:
where:
g
G
V
V
C
for dc bias).
Therefore, from Equation 18,
Substituting R
For a target output voltage (V
V2SET pin through a 4.7 kΩ resistor to GND (see Table 4).
Because one of the fixed output voltage options is chosen,
the feedback pin (FB2) must be directly connected to the
output of Channel 2, V
Estimate the duty cycle (D) range. Ideally,
Therefore, for an output voltage of 1.2 V and a nominal
input voltage (V
is 0.24. Using the maximum input voltage (10% greater than
the nominal, or 5.5 V) results in the minimum duty cycle
(D
(10% less than the nominal, or 4.5 V) results in the maximum
duty cycle (D
m
REF
OUT
OUT
CS
= 550 μS.
MIN
= 4 A/V.
= 0.6 V.
R
= 0.8 × 69 μF (capacitance derated by 20% to account
R
= 2.5 V.
D =
) of 0.22, whereas using the minimum input voltage
COMP
COMP
V
V
= 30 kΩ.
OUT
ZERO
IN
OUT
=
OUT
MAX
OUT
0
COMP
REF
9 .
CS
IN
SW
CROSS
) of 0.27.
×
) of 5.0 V, the nominal duty cycle (D
⎛
⎜
⎜
⎝
in Equation 19 yields C
COMP
2 (
COMP
) π
g
OUT2
m
f
G
CROSS
.
m
CS
OUT
⎞
× ⎟ ⎟
⎠
Setting
See Step 1
Fixed, typical
Fixed, typical
Fixed, typical
See Step 2
1/12 f
1/8 f
See Step 3
See Step 4
See Equation 18
See Equation 19
) of 1.2 V, connect the
⎛
⎜
⎜
⎝
C
CROSS
SW
OUT
V
REF
V
COMP
OUT
⎞
⎟
⎟
⎠
= 820 pF.
3.3 μH
Value
2.5 V
0.6 V
550 μS
4 A/V
600 kHz
50 kHz
6.25 kHz
(47 + 22) μF
30 kΩ
820 pF
NOM
Rev. 0 | Page 30 of 36
)
3.
4.
However, the actual duty cycle will be larger than the
calculated values to compensate for the power losses in the
converter. Therefore, add 5% to 7% to the value calculated
for the maximum load.
The switching frequency (f
based on the Channel 1 requirements, meets the duty cycle
ranges that were previously calculated. Therefore, this
switching frequency is acceptable.
Select the inductor by using the following equation:
In this equation, V
and f
Therefore, when L = 2.2 μH (the closest standard value) in
Equation 5, ΔI
Although the maximum output current required is 3 A,
the maximum peak current is 4.5 A for the current-limit
condition (see Table 7). Therefore, the inductor should be
rated for a peak current of 4.5 A and an average current of
3 A for reliable circuit operation in all conditions.
Select the output capacitor by using the following equations:
The first equation is based on the output ripple (ΔV
whereas the second equation is based on the transient load
performance requirements that allow, in this case, 5% maxi-
mum deviation. As previously mentioned, perform these
calculations and then choose a capacitor based on the larger
calculated capacitor size.
In this case, the following values are used:
ΔI
f
ΔV
ESR = 3 mΩ (typical for ceramic capacitors)
ΔI
ΔV
The output ripple based calculation dictates that C
whereas the transient load based calculation dictates that
C
choose a capacitor. As previously mentioned in the Output
Capacitor Selection section, the capacitance value decreases
when dc bias is applied; therefore, select a higher value. In
this case, choose a 47 μF, 6.3 V capacitor and a 100 μF,
6.3 V capacitor in parallel to meet the requirements.
SW
OUT
L
OUT_STEP
= 600 kHz
RIPPLE
DROOP
= 0.69 A
SW
C
C
L
= 125 μF. To meet both requirements, use the latter to
OUT_MIN
OUT_MIN
=
= 600 kHz, which results in L = 1.67 μH.
= 12 mV (1% of 1.2 V)
= 0.06 V (5% of 1.2 V)
(
V
= 1.5 A
Δ
IN
I
L
≅
−
L
×
≅
V
= 0.69 A.
8
ΔI
f
OUT
SW
×
IN
OUT_STEP
= 5 V, V
f
SW
)
×
×
V
V
(
OUT
ΔV
IN
SW
OUT
×
) of 600 kHz, which is chosen
RIPPLE
⎛
⎜
⎜
⎝
ΔI
= 1.2 V, ΔI
f
L
SW
-
×
ΔI
ΔV
3
L
×
DROOP
L
ESR
= 0.3 × I
)
⎞
⎟
⎟
⎠
OUT
L
= 20 μF,
= 0.9 A,
RIPPLE
),