Datasheet

LM27313
www.ti.com SNVS487D –DECEMBER 2006–REVISED APRIL 2013
Figure 16. Max. Load Current vs V
IN
DESIGN PARAMETERS V
SW
AND I
SW
The value of the FET "ON" voltage (referred to as V
SW
in the equations) is dependent on load current. A good
approximation can be obtained by multiplying the "ON Resistance" of the FET times the average inductor
current.
FET on resistance increases at V
IN
values below 5V, since the internal N-FET has less gate voltage in this input
voltage range (see Typical Performance Characteristics curves). Above V
IN
= 5V, the FET gate voltage is
internally clamped to 5V.
The maximum peak switch current the device can deliver is dependent on duty cycle. The minimum switch
current value (I
SW
) is ensured to be at least 800 mA at duty cycles below 50%. For higher duty cycles, see
Typical Performance Characteristics curves.
THERMAL CONSIDERATIONS
At higher duty cycles, the increased ON time of the FET means the maximum output current will be determined
by power dissipation within the LM27313 FET switch. The switch power dissipation from ON-state conduction is
calculated by:
P
SW
= DC x I
IND(AVG)
2
x R
DS(ON)
(13)
There will be some switching losses as well, so some derating needs to be applied when calculating IC power
dissipation.
MINIMUM INDUCTANCE
In some applications where the maximum load current is relatively small, it may be advantageous to use the
smallest possible inductance value for cost and size savings. The converter will operate in discontinuous mode in
such a case.
The minimum inductance should be selected such that the inductor (switch) current peak on each cycle does not
reach the 800 mA current limit maximum. To understand how to do this, an example will be presented.
In this example, the LM27313 nominal switching frequency is 1.6 MHz, and the minimum switching frequency is
1.15 MHz. This means the maximum cycle period is the reciprocal of the minimum frequency:
T
ON(max)
= 1/1.15M = 0.870 µs (14)
We will assume: V
IN
= 5V, V
OUT
= 12V, V
SW
= 0.2V, and V
DIODE
= 0.3V. The duty cycle is:
Duty Cycle = ((12V + 0.3V - 5V) / (12V + 0.3V - 0.2V)) = 60.3% (15)
Therefore, the maximum switch ON time is:
(60.3% x 0.870 µs) = 0.524 µs (16)
An inductor should be selected with enough inductance to prevent the switch current from reaching 800 mA in
the 0.524 µs ON time interval (see Figure 17):
Copyright © 2006–2013, Texas Instruments Incorporated Submit Documentation Feedback 11
Product Folder Links: LM27313