ADP2116ACPZ-R7 datasheet(30/36 Pages) AD

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ADP2116

Rev. 0 | Page 30 of 36

Calculate the compensation component values of the

feedback loop by using the following equation:

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

REF

OUT

CROSS

COMP

)

(

where:

gm = 550 μS.

GCS = 4 A/V.

VREF = 0.6 V.

VOUT = 2.5 V.

COUT = 0.8 × 69 μF (capacitance derated by 20% to account

for dc bias).

Therefore, from Equation 18,

RCOMP = 30 kΩ.

Substituting RCOMP in Equation 19 yields CCOMP = 820 pF.

Table 10. Channel 1 Circuit Settings

Circuit Parameter

Setting

Value

Output Voltage, VOUT

See Step 1

2.5 V

Reference Voltage, VREF

Fixed, typical

0.6 V

Error Amplifier Transconductance, gm

Fixed, typical

550 μS

Current-Sense Gain, GCS

Fixed, typical

4 A/V

Switching Frequency, fSW

See Step 2

600 kHz

Crossover Frequency, fCROSS

1/12 fSW

50 kHz

Zero Frequency, fZERO

1/8 fCROSS

6.25 kHz

Output Inductor, LOUT

See Step 3

3.3 μH

Output Capacitor, COUT

See Step 4

(47 + 22) μF

Compensation Resistor, RCOMP

See Equation 18

30 kΩ

Compensation Capacitor, CCOMP

See Equation 19

820 pF

CHANNEL 2 CONFIGURATION AND COMPONENTS

SELECTION

Complete the following steps to configure Channel 2:

For a target output voltage (VOUT) 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, VOUT2.

Estimate the duty cycle (D) range. Ideally,

OUT

D =

Therefore, for an output voltage of 1.2 V and a nominal

input voltage (VIN) of 5.0 V, the nominal duty cycle (DNOM)

is 0.24. Using the maximum input voltage (10% greater than

the nominal, or 5.5 V) results in the minimum duty cycle

(DMIN) of 0.22, whereas using the minimum input voltage

(10% less than the nominal, or 4.5 V) results in the maximum

duty cycle (DMAX) 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 (fSW) 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.

Select the inductor by using the following equation:

OUT

−

)

(

In this equation, VIN = 5 V, VOUT = 1.2 V, ΔIL = 0.3 × IL = 0.9 A,

and fSW = 600 kHz, which results in L = 1.67 μH.

Therefore, when L = 2.2 μH (the closest standard value) in

Equation 5, ΔIL = 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.

Select the output capacitor by using the following equations:

)

(

ESR

ΔI

ΔV

ΔI

RIPPLE

OUT_MIN

≅

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

≅

DROOP

OUT_STEP

OUT_MIN

ΔV

ΔI

The first equation is based on the output ripple (ΔVRIPPLE),

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:

ΔIL = 0.69 A

fSW = 600 kHz

ΔVRIPPLE = 12 mV (1% of 1.2 V)

ESR = 3 mΩ (typical for ceramic capacitors)

ΔIOUT_STEP = 1.5 A

ΔVDROOP = 0.06 V (5% of 1.2 V)

The output ripple based calculation dictates that COUT = 20 μF,

whereas the transient load based calculation dictates that

COUT = 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.

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