NCV3020ADR2G ON Semiconductor, NCV3020ADR2G Datasheet - Page 17

• Manufacturer Part Number: NCV3020ADR2G
• Description: IC PWM CTLR SYNC 8SOIC
• Manufacturer: ON Semiconductor
• Series: -r

Specifications of NCV3020ADR2G

• Pwm Type: Voltage Mode
• Number Of Outputs: 1
• Frequency - Max: 360kHz
• Duty Cycle: 84%
• Voltage - Supply: 4.7 V ~ 28 V
• Buck: Yes
• Boost: No
• Flyback: No
• Inverting: No
• Doubler: No
• Divider: No
• Cuk: No
• Isolated: No
• Operating Temperature: -40°C ~ 125°C
• Package / Case: 8-SOIC (0.154", 3.90mm Width)
• Lead Free Status / Rohs Status: Lead free / RoHS Compliant

calculate the RMS current and peak current.
should support an rms current of 10.02 A and a peak current
of 11.2 A.
mechanical and electrical considerations. From a
mechanical perspective, smaller inductor values generally
correspond to smaller physical size. Since the inductor is
often one of the largest components in the regulation system,
a minimum inductor value is particularly important in
space−constrained applications.
perspective, the maximum current slew rate through the
output inductor for a buck regulator is given by Equation 10.
SlewRate
regulator’s ability to slew current through the output
inductor in response to output load transients. Consequently,
output capacitors must supply the load current until the
inductor current reaches the output load current level. This
results in larger values of output capacitance to maintain
tight output voltage regulation. In contrast, smaller values of
inductance increase the regulator’s maximum achievable
slew rate and decrease the necessary capacitance, at the
expense of higher ripple current. The peak−to−peak ripple
current for the NCP3020A is given by the following
equation:
I
PK
18
17
16
15
14
13
12
11
10
To keep within the bounds of the parts maximum rating,
An inductor for this example would be around 3.3 mH and
The final selection of an output inductor has both
This equation implies that larger inductor values limit the
9
8
7
6
5
4
3
2
1
0
10%
Figure 32. Ripple Current Ratio vs. Inductance
+ I
I
RMS
9 V
OUT
LOUT
18 V
+ 10 A @
15%
+ I
@ 1 )
+
OUT
V
15 V
IN
ra
@
2
20%
L
* V
OUT
³ 11.2 A + 10 A @ 1 )
12 V
1 ) ra
1 )
OUT
V
IN
12
(0.24)
25%
, (V)
³ 2.6
2
12
³ 10.02 A
2
From
ms
A
30%
+
12 V * 3.3 V
V
an
out
35%
3.3 mH
= 3.3 V
(0.24)
electrical
(eq. 10)
2
(eq. 8)
(eq. 9)
http://onsemi.com
40%
17
equation it is clear that the ripple current increases as L
decreases, emphasizing the trade−off between dynamic
response and ripple current.
copper and core losses. The copper losses can be further
categorized into dc losses and ac losses. A good first order
approximation of the inductor losses can be made using the
DC resistance as they usually contribute to 90% of the losses
of the inductor shown below:
geometry of the selected core, core material, and wire used.
Most vendors will provide the appropriate information to
make accurate calculations of the power dissipation then the
total inductor losses can be capture buy the equation below:
Input Capacitor Selection
produced during the on time of the upper MOSFET, so it
must have a low ESR to minimize the losses. The RMS value
of this ripple is:
D is the duty cycle, Iin
I
Loss in the input capacitors can be calculated with the
following equation:
is the effective series resistance of the input capacitance.
Due to large dI/dt through the input capacitors, electrolytic
or ceramics should be used. If a tantalum must be used, it
must by surge protected. Otherwise, capacitor failure could
occur.
Input Start−up Current
equation can be used.
output capacitance, V
t
the steady state input current during max load, then the input
fuse should be rated accordingly, if one is used.
SS
OUT
Ipp is the peak to peak current of the inductor. From this
The power dissipation of an inductor consists of both
The core losses and ac copper losses will depend on the
The input capacitor has to sustain the ripple current
The equation reaches its maximum value with D = 0.5.
P
To calculate the input startup current, the following
I
inrush
is the soft start interval. If the inrush current is higher than
CIN
is the load current.
is the power loss in the input capacitors and ESR
is the input current during startup, C
LP
Iin
RMS
tot
+ LP
P
+ I
CIN
I
INRUSH
I
LP
PP
OUT
+ ESR
CU_DC
OUT
CU
+
RMS
@ D @ ( 1 * D )
+
+ I
V
is the desired output voltage, and
L
) LP
OUT
OUT
CIN
C
is the input RMS current, and
RMS
OUT
( 1 * D )
@ I
@ F
CU_AC
t
2
SS
@ V
IN*RMS
@ DCR
SW
OUT
) LP
2
OUT
Core
is the total
(eq. 12)
(eq. 13)
(eq. 14)
(eq. 15)
(eq. 16)
(eq. 11)
OUT
CIN