NCP1547
Short Circuit
When the V
FB
pin voltage drops below Foldback
Threshold, the regulator reduces the peak current limit by
40% and switching frequency to 1/4 of the nominal
frequency. These features are designed to protect the IC and
external components during over load or short circuit
conditions. In those conditions, peak switching current is
clamped to the current limit threshold. The reduced
switching frequency significantly increases the ripple
current, and thus lowers the DC current. The short circuit can
cause the minimum duty cycle to be limited by Minimum
Output Pulse Width. The foldback frequency reduces the
minimum duty cycle by extending the switching cycle. This
protects the IC from overheating, and also limits the power
that can be transferred to the output. The current limit
foldback effectively reduces the current stress on the
inductor and diode. When the output is shorted, the DC
current of the inductor and diode can approach the current
limit threshold. Therefore, reducing the current limit by 40%
can result in an equal percentage drop of the inductor and
diode current. The short circuit waveforms are captured in
Figure 9, and the benefit of the foldback frequency and
current limit is self−evident.
The pre−driver current is used to turn on/off the power
switch and is approximately equal to 12 mA 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 V
IN
pin when the switch is off. The pre−driver
current always returns to the V
SW
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 V
IN
+ V
O
when the switch is on, the power
dissipation due to pre−driver current can be calculated by
WDRV
+
12 mA
V 2
(VIN
*
VO
)
O )
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
V 2
WBASE
+
O
VIN
IS
60
where:
I
S
= 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
WSAT
+
VO
VIN
IS
VSAT
where:
V
SAT
= saturation voltage of the power switch which is
shown in Figure 5.
The switching loss occurs when the switch experiences
both high current and voltage during each switch transition.
This regulator has a 30 ns turn−off time and associated
power loss is equal to
I
WS
+
S
Figure 9. In Short Circuit, the Foldback Current and
Foldback Frequency Limit the Switching Current to
Protect the IC, Inductor and Catch Diode
Thermal Considerations
VIN
2
30 ns
fS
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
WIC
+
WQ
)
WDRV
)
WBASE
)
WSAT
)
WS
A calculation of the power dissipation of the IC is always
necessary prior to the adoption of the regulator. The current
drawn by the IC includes quiescent current, pre−driver
current, and power switch base current. The quiescent
current 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
current and generates power equal to
WQ
+
VIN
IQ
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,
TJ
+
WIC
R
qJA
)
TA
Minimum Load Requirement
where:
I
Q
= quiescent current.
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 V
O
divided by
12 mA should prevent any voltage overshoot at light load
conditions. Alternatively, the feedback resistors can be
valued properly to consume 12 mA current.
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