Datasheet

ADP2116 Data Sheet
Rev. A | Page 30 of 36
5. Calculate the compensation component values of the
feedback loop by using the following equation:








×








×=
REF
OUTOUT
CS
m
CROSS
COMP
V
VC
Gg
f
R
)π2(
9.0
where:
g
m
= 550 µS.
G
CS
= 4 A/V.
V
REF
= 0.6 V.
V
OUT
= 2.5 V.
C
OUT
= 0.8 × 69 µF (capacitance derated by 20% to account
for dc bias).
Therefore, from Equation 18,
R
COMP
= 30 kΩ.
Substituting R
COMP
in Equation 19 yields C
COMP
= 820 pF.
Table 10. Channel 1 Circuit Settings
Circuit Parameter Setting Value
Output Voltage, V
OUT
See Step 1 2.5 V
Reference Voltage, V
REF
Fixed, typical 0.6 V
Error Amplifier Transconductance, g
m
Fixed, typical 550 µS
Current-Sense Gain, G
CS
Fixed, typical
4 A/V
Switching Frequency, f
SW
See Step 2 600 kHz
Crossover Frequency, f
CROSS
1/12 f
SW
50 kHz
Zero Frequency, f
ZERO
1/8 f
CROSS
6.25 kHz
Output Inductor, L
OUT
See Step 3 3.3 µH
Output Capacitor, C
OUT
See Step 4 (47 + 22) µF
Compensation Resistor, R
COMP
See Equation 18 30 kΩ
Compensation Capacitor, C
COMP
See Equation 19 820 pF
CHANNEL 2 CONFIGURATION AND COMPONENTS
SELECTION
Complete the following steps to configure Channel 2:
1. For a target output voltage (V
OUT
) of 1.2 V, connect the
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
OUT2
.
2. Estimate the duty cycle (D) range. Ideally,
IN
OUT
V
V
D =
Therefore, for an output voltage of 1.2 V and a nominal
input voltage (V
IN
) of 5.0 V, the nominal duty cycle (D
NOM
)
is 0.24. Using the maximum input voltage (10% greater than
the nominal, or 5.5 V) results in the minimum duty cycle
(D
MIN
) of 0.22, whereas using the minimum input voltage
(10% less than the nominal, or 4.5 V) results in the maximum
duty cycle (D
MAX
) of 0.27.
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
SW
) of 600 kHz, which is chosen
based on the Channel 1 requirements, meets the duty cycle
ranges that were previously calculated. Therefore, this
switching frequency is acceptable.
3. Select the inductor by using the following equation:
IN
OUT
SW
L
OUT
IN
V
V
fI
VV
L ×
×
−
=
Δ
)(
In this equation, V
IN
= 5 V, V
OUT
= 1.2 V, ΔI
L
= 0.3 × I
L
= 0.9 A,
and f
SW
= 600 kHz, which results in L = 1.67 µH.
Therefore, when L = 2.2 µH (the closest standard value) in
Equation 5, ΔI
L
= 0.69 A.
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.
4. Select the output capacitor by using the following equations:
)-(8 ESRΔIΔVf
ΔI
C
LRIPPLE
SW
L
OUT_MIN
×××
≅








×
×≅
DROOPSW
OUT_STEPOUT_MIN
ΔVf
ΔIC
3
The first equation is based on the output ripple (ΔV
RIPPLE
),
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
L
= 0.69 A
f
SW
= 600 kHz
ΔV
RIPPLE
= 12 mV (1% of 1.2 V)
ESR = 3 mΩ (typical for ceramic capacitors)
ΔI
OUT_STEP
= 1.5 A
ΔV
DROOP
= 0.06 V (5% of 1.2 V)
The output ripple based calculation dictates that C
OUT
= 20 µF,
whereas the transient load based calculation dictates that
C
OUT
= 125 µF. To meet both requirements, use the latter to
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.