Duty Cycle =

VOUT + VDIODE - VIN

VOUT + VDIODE - VSW

LM27313, LM27313-Q1

SNVS487E – DECEMBER 2006 – REVISED JANUARY 2015

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Typical Applications (continued)

(3)

This applies for continuous mode operation.

The equation shown for calculating duty cycle incorporates terms for the FET switch voltage and diode forward

voltage. The actual duty cycle measured in operation will also be affected slightly by other power losses in the

circuit such as wire losses in the inductor, switching losses, and capacitor ripple current losses from self-heating.

Therefore, the actual (effective) duty cycle measured may be slightly higher than calculated to compensate for

these power losses. A good approximation for effective duty cycle is:

DC (eff) = (1 - Efficiency x (VIN / VOUT))

where

•

the efficiency can be approximated from the curves provided.

(4)

8.2.1.2.8

Inductance Value

The first question we are usually asked is: “How small can I make the inductor?” (because they are the largest

sized component and usually the most costly). The answer is not simple and involves trade-offs in performance.

More inductance means less inductor ripple current and less output voltage ripple (for a given size of output

capacitor). More inductance also means more load power can be delivered because the energy stored during

each switching cycle is:

E = L/2 x (lp)

2

where

•

lp is the peak inductor current.

(5)

An important point to observe is that the LM27313 will limit its switch current based on peak current. This means

that because lp(max) is fixed, increasing L will increase the maximum amount of power available to the load.

Conversely, using too little inductance may limit the amount of load current which can be drawn from the output.

Best performance is usually obtained when the converter is operated in “continuous” mode at the load current

range of interest, typically giving better load regulation and less output ripple. Continuous operation is defined as

not allowing the inductor current to drop to zero during the cycle. It should be noted that all boost converters shift

over to discontinuous operation as the output load is reduced far enough, but a larger inductor stays “continuous”

over a wider load current range.

To better understand these tradeoffs, a typical application circuit (5V to 12V boost with a 10 µH inductor) will be

analyzed.

Because the LM27313 typical switching frequency is 1.6 MHz, the typical period is equal to 1/fSW(TYP), or

approximately 0.625 µs.

We will assume: VIN = 5 V, VOUT = 12 V, VDIODE = 0.5 V, VSW = 0.5 V. The duty cycle is:

Duty Cycle = ((12 V + 0.5 V - 5 V) / (12 V + 0.5 V - 0.5 V)) = 62.5%

(6)

The typical ON time of the switch is:

(62.5% x 0.625 µs) = 0.390 µs

(7)

It should be noted that when the switch is ON, the voltage across the inductor is approximately 4.5 V.

Use the equation:

V = L (di/dt)

(8)

Then, calculate the di/dt rate of the inductor which is found to be 0.45 A/µs during the ON time. Using these

facts, we can then show what the inductor current will look like during operation:

12

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