LM2635MX National Semiconductor, LM2635MX Datasheet - Page 11

• Manufacturer Part Number: LM2635MX
• Description: IC REG SYNCH BUCK 5-BIT 20-SOIC
• Manufacturer: National Semiconductor

Specifications of LM2635MX

• Applications: Power Supplies
• Current - Supply: 2.5mA
• Voltage - Supply: 4.5 V ~ 5.5 V
• Operating Temperature: 0°C ~ 125°C
• Mounting Type: Surface Mount
• Package / Case: 20-SOIC (7.5mm Width)
• Lead Free Status / RoHS Status: Lead free / RoHS Compliant
• Other names: *LM2635MX; LM2635MXTR

Applications Information
INDUCTOR SELECTION
The size of the output is determined by a number of param-
eters. Basically the larger the inductor, the smaller the output
ripple voltage, but the slower the converter’s response
speed during a load transient. On the other hand, a smaller
inductor requires higher switching frequency to maintain the
same level of output ripple, and probably results in a more
lossy converter, but has less inertia responding to load tran-
sient. In the case of Pentium II power supply, fast recovery of
the load voltage from transient window back to the steady
state window is considered important. This limits the highest
inductance value that can be used. The lowest inductance
value is limited by the highest switching frequency that can
be practically employed. As the switching frequency in-
creases, the switching loss in the MOSFETs tends to in-
crease, resulting in less converter efficiency and larger heat
sinks. A good switching frequency is probably a frequency
under which the MOSFET conduction loss is higher than the
switching loss because the cost of the MOSFET is directly
related to its R
the following equation:
where V
the switching frequency. For commonly used low R
MOSFETs, a reasonable switching frequency is 300 kHz.
Assume an output peak-peak ripple voltage of 18 mV is to be
guaranteed, the total output capacitor ESR is 9 mΩ, the input
voltage is 5V, and output voltage is 2.8V. The inductance
value according to the above equation will then be 2 µH. The
highest slew rate of the inductor current when the load
changes from no load to full load can be determined as
follows:
where D
around 0.9 for LM2635. For a load transient from 0A to 14A,
the highest current slew rate of the inductor, according to the
above equation, is 0.85A/µs, and therefore the shortest pos-
sible total recovery time is 14A/(0.85A/µs) = 16.5 µs. Notice
that the output voltage starts to recover whenever the induc-
tor starts to supply current.
The highest slew rate of the inductor current when the load
changes from full load to no load can be determined from the
same equation, but use D
Since the D
therefore −1.4A/µs. So the approximate total recovery time
will be 14A/(1.4A/µs) = 10 µs.
The input inductor is for limiting the input current slew rate
during a load transient. In the case that low ESR aluminum
electrolytic capacitors are used for the input capacitor bank,
voltage change due to capacitor charging/discharging is usu-
ally negligible for the first 20 µs. ESR is by far the dominant
factor in determining the amount of capacitor voltage
undershoot/overshoot due to load transient. So the worst
case is when the load changes between no load and full
load, under which condition the input inductor sees the high-
est voltage change across the input capacitors. Assume the
input capacitor bank is made up of three 16MV820GX, i.e.,
O_RIP
MAX
MIN
is the maximum allowed duty cycle, which is
DSON
is the peak-to-peak output ripple voltage, f is
of LM2635 at 300 kHz is 0%, the slew rate is
. The inductor size can be determined by
MIN
instead of D
MAX
(Continued)
.
DSON
11
the total ESR is 15 mΩ. Whenever there is a sudden load
current change, it has to initially be supported by the input
capacitor bank instead of the input inductor. So for a full load
swing between 0A and 14A, the voltage seen by the input
inductor is ∆V = 14A x 15 mΩ = 210 mV. Use the following
equation to determine the minimum inductance value:
where (di/dt)
slew rate, which is 0.1A/µs in the case of the Pentium II
power supply. So the input inductor size, according to the
above equation, should be 2.1 µH.
DYNAMIC POSITIONING OF LOAD VOLTAGE
Since the Intel VRM specifications have defined two operat-
ing windows for the MPU core voltage, one being the steady
state window and the other the transient window, it is a good
idea to dynamically position the steady state output voltage
in the steady state window with respect to load current level
so that the output voltage has more headroom for load
transient response. This requires information about the load
current. There are at least two simple ways to implement this
idea with LM2635. One is to utilize the output inductor DC
resistance, see Figure 7. The average voltage across the
output inductor is actually that across its DC resistance. That
average voltage is proportional to load current.
Since the switching node voltage V
input voltage and ground at the switching frequency, it is
impossible to choose point A as the feedback point, other-
wise the dynamic performance will suffer and the system
may have some noise problems. Using a low pass filter
network around the inductor, such as the one shown in the
figure, seems to be a good idea. The feedback point is C.
Since at the switching frequency the impedance of the 0.1
µF is much less than 5 kΩ, the bouncing voltage at point A
will be mainly applied across the resistor 5 kΩ, and point C
will be much quieter than A. However, V
majority of V
in steady state V
DC resistance. So at no load, output voltage is equal to V
and at full load, output voltage is I
further utilize the steady state window, a resistor can be
FIGURE 7. Dynamic Voltage Positioning by Utilizing
AB
MAX
Output Inductor DC Resistance
average, because of the resistor divider. So
C
is the maximum allowable input current
= I
O
x r
L
+ V
CORE
O
, where r
A
x r
bounces between the
CB
L
lower than V
average is still the
L
is the inductor
www.national.com
10011926
C
. To
C
,