ILOAD(max) = (1 - DC) x (ISW(max) - DC (VIN - VSW))

2fL

LM27313, LM27313-Q1

SNVS487E – DECEMBER 2006 – REVISED JANUARY 2015

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Typical Applications (continued)

Combining all terms, we can develop an expression which allows the maximum available load current to be

calculated:

(12)

The equation shown to calculate maximum load current takes into account the losses in the inductor or turn-OFF

switching losses of the FET and diode. For actual load current in typical applications, we took bench data for

various input and output voltages and displayed the maximum load current available for a typical device in graph

form:

Figure 16. Max. Load Current vs VIN

8.2.1.2.11

Design Parameters VSW and ISW

The value of the FET ON voltage (referred to as VSW 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 VIN values below 5V, because the internal N-FET has less gate voltage in this

input voltage range (see Typical Characteristics). Above VIN = 5 V, 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 (ISW) is ensured to be at least 800 mA at duty cycles below 50%. For higher duty cycles, see

Typical Characteristics.

8.2.1.2.12

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:

TON(max) = 1/1.15M = 0.870 µs

(13)

Assume: VIN = 5 V, VOUT = 12 V, VSW = 0.2 V, and VDIODE = 0.3 V. The duty cycle is:

Duty Cycle = ((12 V + 0.3 V - 5 V) / (12 V + 0.3 V - 0.2 V)) = 60.3%

(14)

Therefore, the maximum switch ON time is:

(60.3% x 0.870 µs) = 0.524 µs

(15)

14

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