Application Information
VO
VIN
WSAT
=
× IS × VSAT
where VSAT = saturation voltage of the power switch which
is shown in Figure 3.
The switching loss occurs when the switch experiences
both high current and voltage during each switch transi-
tion. This regulator has a 30ns turn-off time and associated
power loss is equal to
IS × VIN
WS =
× 20ns × fS
2
The turn-on time is much shorter and thus turn-on loss is
not considered here.
The total power dissipated by the IC is sum of all the above
W
IC = WQ + WDRV + WBASE + WSAT + WS
Figure 7: In short circuit, the foldback current and foldback frequency
limit the switching current to protect the IC, inductor and catch diode.
The IC junction temperature can be calculated from the
ambient temperature, IC power dissipation and thermal
resistance of the package. The equation is shown as follows,
Thermal Considerations
A calculation of the power dissipation of the IC is always
necessary prior to the adoption of the regulator. The cur-
rent drawn by the IC includes quiescent current, pre-driver
current, and power switch base current. The quiescent cur-
rent drives the low power circuits in the IC, which include
comparators, error amplifier and other logic blocks.
Therefore, this current is independent of the switching cur-
rent and generates power equal to
TJ = WIC × RθJA + TA
The maximum IC junction temperature shall not exceed
125°C to guarantee proper operation and avoid any dam-
ages to the IC.
Minimum Load Requirement
As pointed out in the previous section, a minimum load is
required for this regulator due to the pre-driver current
feeding the output. Placing a resistor equal to VO divided
by 12mA should prevent any voltage overshoot at light
load conditions. Alternatively, the feedback resistors can be
valued properly to consume 12mA current.
W
Q = VIN × IQ
where IQ= quiescent current.
The pre-driver current is used to turn on/ off the power
switch and is approximately equal to 12mA in worst case.
During steady state operation, the IC draws this current
from the Boost pin when the power switch is on and then
receives it from the VIN pin when the switch is off. The pre-
driver current always returns to the VSW pin. Since the pre-
driver current goes out to the regulator’s output even when
the power switch is turned off, a minimum load is required
to prevent overvoltage in light load conditions. If the Boost
pin voltage is equal to VIN + VO when the switch is on, the
power dissipation due to pre-driver current can be calculat-
ed by
Component Selection
Input Capacitor
In a buck converter, the input capacitor witnesses pulsed
current with an amplitude equal to the load current. This
pulsed current and the ESR of the input capacitors deter-
mine the VIN ripple voltage, which is shown in Figure 8.
For VIN ripple, low ESR is a critical requirement for the
input capacitor selection. The pulsed input current possess-
es a significant AC component, which is absorbed by the
input capacitors. The RMS current of the input capacitor
can be calculated using:
2
VO
W
DRV = 12mA × ( VIN – VO +
)
VIN
The base current of a bipolar transistor is equal to collector
current divided by beta of the device. Beta of 60 is used
here to estimate the base current. The Boost pin provides
the base current when the transistor needs to be on. The
power dissipated by the IC due to this current is
I
RMS = IO D(1 – D)
where D = switching duty cycle which is equal to VO/ VIN.
IO = load current.
2
VO
IS
60
WBASE
=
×
VIN
where IS = DC switching current.
When the power switch turns on, the saturation voltage
and conduction current contribute to the power loss of a
non-ideal switch. The power loss can be quantified as
8
CHERRY [ CHERRY SEMICONDUCTOR CORPORATION ]
