ncp5214 ON Semiconductor, ncp5214 Datasheet - Page 25

• Manufacturer Part Number: ncp5214
• Description: 2-in-1 Notebook Ddr Power Controller
• Manufacturer: ON Semiconductor

rated DC current and 3.5 mW DCR is chosen.
ripple voltage is considered. Suppose the output ripple
voltage is 2% of the nominal output voltage.
should meet both undershoot and overshoot requirements.
Sometimes, it may take several times of iterations between
the process of selecting inductance of the inductor and ESR
and capacitance of the output capacitor.
estimated by:
is used.
capacitor is considered:
larger should be selected. Two capacitors each with 1.5 A
rated ripple current can be connected in parallel to provide
a total of 3.0 A rated rms ripple current.
capacitance of 220 mF, ESR of 15 mW, rated voltage of
2.5 V, and rated rms ripple current of 1.5 A are used.
calculated as:
Thus, inductor with 1.8 mH inductance, 14 A maximum
c. Calculate ESR and capacitance of output filter
First, the ESR required to achieve the desired output
To ensure that undershoot requirement of less than 100 mV is achieved, the capacitance must be:
To make sure that overshoot requirement of less than 100 mV is fulfilled, capacitance must be:
Therefore, output capacitor with capacitance of 440 mF
Then, the voltage rating of the output capacitor is
Thus, output capacitor with 2.5 V or larger rated voltage
Finally, the rated rms ripple current of the output
Thus, capacitor with rated rms ripple current of 3.0 A or
Therefore, two same capacitors in parallel each with
Since the L = 1.8 mH, C
ESR v
I COUT(rms) w
capacitor:
+ 15.8 mW
(0.02
V rated w 1.25
(20 V−1.836 V)
1.8 mH
1.8 V)
(20 V−1.8 V)
R 3 +
OUT
1.8 mH
400 kHz
C OUT w
1.836 V + 2.3 V
C OUT w
2p
= 440 mF, and the target close loop bandwidth is 100 kHz, the value of R
100 kHz
1.836 V
1.8 V
400 kHz
100 mV−7 A
20 V
1.8 mH
(100 mV ) 1.836 V) 2 − (1.836 V) 2
+ 2.3 A
1.925 V
7 A
20 V
(eq. 48)
(eq. 52)
(eq. 53)
7 A )
http://onsemi.com
7.5 mW
NCP5214
20 V
2 1.8 mH 400 kHz 20 V
(20 V−1.836 V) 1.836 V
4.3 kW
25
undershoot requirement is considered, such that:
value of 7.5 mW is selected for more design margin and
better performance. Then, two same SP−Caps or POSCAPs
each with 15 mW ESR in parallel having a resultant ESR
of 7.5 mW should be good enough to meet the
requirements.
capacitance
requirements.
load condition, such that:
MOSFET data sheet, the maximum R
selected for RL1.
range between 2.0 kW and 5.0 kW.
the close loop compensation should be considered at
maximum V
as below:
Second, the ESR required to meet the transient load
Therefore, the suitable ESR is 12 mW or smaller, and the
Then, check that whether the previously supposed
d. Calculate the resistance value of OCP current limit
First, the OCP current limit is estimated at maximum
Thus, I
Then, the value of RL1 is calculated by:
Therefore, the resistor with standard value of 4.7 kW is
e. Calculate the RC values of the compensation network:
First, 4.3 kW is chosen as the value of R
Since the worst case of stability is at the maximum V
I LIMIT u 8 A ) 2 A )
1−
V RAMP + 1.25 V ) 0.045
setting resistor:
400 kHz
1.8 mH
1.8 V−36 mV
+ 11.16 A
LIMIT
20 V
RL1 + 11.5 A
IN
2
meets
. Then the ramp amplitude can be calculated
+ 317.6 mF
ESR v 100 mV
is set to 11.5 A. Suppose from the high−side
440 mF
+ 335.9 mF
the
+ 7.3 kW
26 mA
7 A
2
(20 V−1.836 V)
undershoot
10 mW
1.8 mH
+ 14.3 mW
(20 V−5 V) + 1.925 V
+ 4.4 kW
DS(on)
400 kHz
and
1
which is in the
is 10 mW.
1.836 V
overshoot
3
(eq. 50)
(eq. 51)
(eq. 57)
(eq. 49)
(eq. 54)
(eq. 55)
(eq. 56)
20 V
can be
IN
,