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LM2733

SNVS209G – NOVEMBER 2002 – REVISED MAY 2019

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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))

(4)

Where the efficiency can be approximated from the curves provided.

8.2.2.9 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 tradeoffs in performance.

Larger inductors mean less inductor ripple current, which typically means less output voltage ripple (for a given

size of output capacitor). Larger inductors also mean more load power can be delivered because the energy

stored during each switching cycle is:

E =L/2 X (lp)

2

(5)

Where “lp” is the peak inductor current. An important point to observe is that the LM2733 device will limit its

switch current based on peak current. This means that since lp (maximum) 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. We will assume:

VIN = 5 V, VOUT = 12 V, VDIODE = 0.5 V, VSW = 0.5 V

Since the frequency is 1.6 MHz (nominal), the period is approximately 0.625 µs. The duty cycle will be 62.5%,

which means the ON time of the switch is 0.390 µs. It should be noted that when the switch is ON, the voltage

across the inductor is approximately 4.5 V.

Using the equation:

V = L (di/dt)

(6)

We can 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:

Figure 25. 10-µH Inductor Current,

5-V – 12-V Boost (LM2733X)

During the 0.390 µs ON time, the inductor current ramps up 0.176 A and ramps down an equal amount during

the OFF time. This is defined as the inductor “ripple current”. It can also be seen that if the load current drops to

about 33 mA, the inductor current will begin touching the zero axis which means it will be in discontinuous mode.

A similar analysis can be performed on any boost converter, to make sure the ripple current is reasonable and

continuous operation will be maintained at the typical load current values.