MSK5450 M. S Kennedy, MSK5450 Datasheet - Page 3

• Manufacturer Part Number: MSK5450
• Description: High Current, Very Low Output Surface Mount Voltage Regulators
• Manufacturer: M. S Kennedy

The maximum resistor value can be calculated from the
following:
R1 max =
Where:
REGULATOR PROTECTION:
MINIMIZING POWER DISSIPATION:
To maximize the performance and reduce power dissipation of
the MSK 5450 series devices, Vin should be maintained as
close to dropout or at Vin minimum when possible. See Input
Supply Voltage requirements. A series resistor can be used to
lower Vin close to the dropout specification, lowering the input
to output voltage differential. In turn, this will decrease the
power that the device is required to dissipate. Knowing peak
current requirements and worst case voltages, a resistor can
be selected that will drop a portion of the excess voltage and
help to distribute the heating. The circuit below illustrates this
method.
INPUT CAPACITOR:
The MSK 5450 series are high performance linear regulators
for high current, low voltage applications requiring fast tran-
sient response. The devices are fully protected from damage
due to fault conditions, offering constant current limiting and
thermal shutdown. The thermal shutdown junction tempera-
ture is typically 150°C.
If the device is to be located more than 4 inches from the bulk
supply capacitance, a minimum 1uF capacitor should be placed
as close to the input pin as possible for proper bypassing. A
smaller value capacitor such as 0.01uF should be placed in
parallel with the larger value capacitor. Larger input capacitor
values will help to improve ripple rejection.
OUTPUT CAPACITOR:
The MSK 5450 series devices require a minimum of external
components to maintain stability. A minimum of output ca-
pacitance is necessary for stable operation. Due to the wide
bandwidth design, the device will operate with a wide range of
capacitance and ESR values. For most applications, a 10uF
ceramic capacitor will suffice. Ideally, this should be an X7R
ceramic capacitor or a tantalum capacitor due to their thermal
performance. There is no upper limit to the amount of output
capacitance that may be used.
Vin min=Minimum input voltage
Vout max=Maximum output voltage across the full
Vdrop=Worst case dropout voltage (Typically 500mV)
Iout peak=Maximum load current
Quiescent Current=Max. quiescent current at Iout peak
APPLICATION NOTES
Iout peak + Quiescent Current
Vin min - (Vout max + Vdrop)
temperature range
3
THERMAL SHUTDOWN:
The MSK 5450 series of devices is equipped with a thermal shut-
down circuit that will turn off the device when the junction tem-
perature reaches approximately 150°C. It is important for the user
to be aware that high temperature operation will limit the current
capability of the device due to this thermal shutdown protection.
In cases of maximum input voltage and high case temperature, the
output current available may be less than 3 Amps. See curve be-
low for clarification.
PACKAGE CONNECTIONS:
The MSK 5450 series are highly thermally conductive devices and
the thermal path from the package heat sink to the internal junc-
tions is very short. Standard surface mount soldering techniques
should be used when mounting the device. Some applications may
require additional heat sinking of the device.
HEAT SINK SELECTION:
To select a heat sink for the MSK 5450, the following for-
mula for convective heat flow may be used:
First, the power dissipation must be calculated as follows:
Power Dissipation = (Vin - Vout) x Iout + Vin x Quiescent Current
Next, the user must select a maximum junction temperature.
The equation may now be arranged to solve for the required
heat sink to ambient thermal resistance (R sa).
EXAMPLE:
An MSK 5450-1.3 is configured for Vin=+1.8V and Vout=+1.3V.
Iout is a continuous 5Amp DC level. Under these conditions the
maximum ground current would be 150mA. The ambient tempera-
ture is +100°C and the maximum junction temperature is 125°C.
R jc = 2.0°C/W and R cs = 0.5°C/W typically.
Power Dissipation = (1.8V - 1.3V) x 5A + (1.8 x 150mA)
Solve for R sa given P
In this example, a heat sink with a thermal resistance of no
more than 6.5°C/W must be used to maintain a junction tem-
perature of no more than 125°C.
R sa = 125°C - 100°C
= 6.53°C/W
[
2.77W
D
: = 2.77 Watts
PRELIMINARY
]
- 2.0°C/W - 0.5°C/W
Rev. - 5/07