ncp1570d ON Semiconductor, ncp1570d Datasheet - Page 12

• Manufacturer Part Number: ncp1570d
• Description: Low Voltage Synchronous Buck Controller
• Manufacturer: ON Semiconductor

provide the full rated switch current as inductor ripple
current, and will usually result in inefficient system
operation. The system will sink current away from the load
during some portion of the duty cycle unless load current is
greater than half of the rated switch current. Some value
larger than the minimum inductance must be used to ensure
the converter does not sink current. Choosing larger values
of inductor will reduce the ripple current, and inductor value
can be designed to accommodate a particular value of ripple
current by replacing I
I
response times to increase. The response times for both
increasing and decreasing current steps are shown below.
ripple voltage the system can tolerate. Output ripple voltage
is defined as the product of the output ripple current and the
output filter capacitor ESR.
V RIPPLE + ESR C I RIPPLE +
output inductors. Power dissipation is proportional to the
square of inductor current:
surrounding it is defined as the product of power dissipation
and thermal resistance to ambient:
approximately 45°C/W. The inductor temperature is given as:
V
the V
should be sufficient to ensure the controller IC does not operate
erratically due to injected noise.
Input Filter Capacitors
minimizes supply voltage variations due to changes in current
flowing through the switch FETs. These capacitors must be
chosen primarily for ripple current rating.
RIPPLE
CC
This equation identifies the value of inductor that will
However, reducing the ripple current will cause transient
Inductor value selection also depends on how much output
Thus, output ripple voltage can be calculated as:
Finally, we should consider power dissipation in the
The temperature rise of the inductor relative to the air
Ra for an inductor designed to conduct 20 A to 30 A is
A small RC filter should be added between module V
The input filter capacitors provide a charge reservoir that
Bypass Filtering
CC
T RESPONSE(INCREASING) +
T RESPONSE(DECREASING) +
:
L (RIPPLE) +
input to the IC. A 10 Ω resistor and a 0.1 μF capacitor
T(inductor) + DT(inductor) ) Tambient
DT(inductor) + (Ra)(P D )
P D + (I
SWITCH(MAX)
( f OSC )( I RIPPLE )( V IN(MIN) )
( V IN(MIN) * V OUT ) V OUT
2
L
) ( ESR L )
ESR C V IN * V OUT V OUT
with a desired value of
( V IN * V OUT )
f OSC L V IN
( L )( DI OUT )
( L )( DI OUT )
( V OUT )
http://onsemi.com
CC
and
12
current flowing in the input inductor L
output current is:
the switch FETs are off, and negative out of the capacitor
when the switch FETs are on. When the switches are off,
I
capacitor current is equal to the per−phase output current
minus I
to the output ripple current, we can approximate the input
capacitor current waveform as a square wave. We can then
calculate the RMS input capacitor ripple current:
I RMS(CIN) +
worst case input ripple current. This will require several
capacitors in parallel. In addition to the worst case current,
attention must be paid to the capacitor manufacturer’s
derating for operation over temperature.
5 V to 3.3 V conversion at 10 A at an ambient temperature
of 60°C. A droop voltage of 90 mV to 1.61 V and efficiency
of 80% is assumed. Average input current in the input filter
inductor is:
current rating for a 6.3 V, 1800 nF capacitor is 2000 mA at
100 kHz and 105°C. We determine the number of input
capacitors by dividing the ripple current by the
per−capacitor current rating:
I IN(RMS) +
IN(AVE)
Consider the schematic shown in Figure 24. The average
Input capacitor current is positive into the capacitor when
The input capacitance must be designed to conduct the
As an example, let us define the input capacitance for a
Input capacitor RMS ripple current is then
If we consider a Rubycon MBZ series capacitor, the ripple
V
IN
Number of capacitors + 4.74 A 2.0 A + 2.3
IN(AVE)
flows into the capacitor. When the switches are on,
I
I IN(AVE) + (10 A)(3.3 V 5 V) + 6.6 A
IN(AVE)
+ 4.74 A
I
RMS(CIN)
L
IN
. If we ignore the small current variation due
I IN(AVE) + I OUT
6.6 2 ) 3.3 V
I
2
IN(AVE)
[( 10 A * 6.6 A ) 2 * 6.6 A 2 ]
I OUT per phase * I IN(AVE) 2 * I
Figure 24.
)
C
CONTROL
5 V
IN
INPUT
V OUT
V IN
V OUT
L
V IN
OUT
IN
for any given
C
OUT
V
2
IN(AVE)
OUT