LM1572 datasheet(10/17 Pages) NSC

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Application Information (Continued)

harmonic instability is not of concern if any one or more of

the above conditions are not true. And in that case, the

inductor selection considerations become identical to those

for voltage mode control. Note that the worst case condition

to check for subharmonic instability, and/or to choose an

inductance large enough to prevent these oscillations, is at

the lowest desired input voltage.

Designing for discontinuous conduction mode is clearly an

attractive option for some experienced designers. One rea-

son for this is that subharmonic instability is then of no

concern. However discontinuous mode is possible only if the

maximum load is less than half the current limit. Further, the

design procedure is rather complicated and iterative too.

Therefore it is considered out of the scope of this section.

The required equation (for Continuous Conduction Mode,

’CCM’) is

where the duty cycle is

SW and VD are the forward drops across the switch and the

diode respectively. A sample calculation follows.

Example: The input voltage range is 8.5-16V. The output is

5V @ 1.5A. It can be assumed that both the switch forward

drop and diode drop are 0.5V, by default.

Step 1: Current Limit at maximum input

The duty cycle at 16V input is

CCM = 0.344

CL = 2A (min value), so the required minimum inductance is

MIN_CL_CCM = 7.2µH

Step 2: Current Limit at maximum input

The calculation is repeated at the lowest desired input volt-

age of 8.5V. The duty cycle at this point is

The current limit at this duty cycle is

CLIM =ICL − {(mC• T)•(D − 0.5)} A

CLIM = 2 − {(0.42• 2)•(0.65 − 0.5)} A

CLIM = 1.87A

So the minimum inductance at this point is

MIN_CL_CCM = 5.2µH

There is one more possible influence on the value of mini-

mum inductance, which is now discussed.

Step 3: Subharmonic Instability

The LM1572 is current mode controlled. If the minimum input

voltage is less than roughly twice the output voltage, the duty

cycle is close to or greater than 0.5. That makes two of the

three conditions required for subharmonic instability. There-

fore also assuming continuous conduction mode, the value

of inductance must be large enough to prevent these oscil-

lations (occurring at f/2). The relevant equation is

On the left is the minimum inductance required. The right

side contains ’Q, which is the quality factor of the half fre-

quency peaking prior to the outbreak of subharmonic oscil-

lations. Q should typically not be greater than 2, or subhar-

monic oscillations become increasingly likely. Further, to

avoid voltage mode control type of response (excessive

slope compensation), neither must Q be less than about 0.2.

Very large values of L lead to smaller and smaller Q. So

acceptable values of Q usually lie within the range 0.2 to 2.

This calculation should always be done at the worst case:

i.e. the minimum input voltage.

LMIN_f/2 = 6.2µH

But what is the optimum value? This is examined next.

Step 4: Optimum Inductance

The designer should be clear that choosing smaller induc-

tance values may not necessarily lead to smaller sized in-

ductors. The relationship between inductance and inductor

size is not always intuitive. The size of an inductor is deter-

mined by its energy handling requirement which is 1⁄2*L*I

So very small inductors may lead to excessively high peak

currents, which can also increase the size of the input and

output capacitors. The reader is referred to AN-1197 for

further details. There it is shown that ’r’ , the current ripple

ratio, should be around 0.4. Using this as the yardstick, the

’optimum’ value of inductance is

This calculation must always be done at the maximum input

voltage, which is 16V for this example

OPT = 12µH

Step 5: Conclusions

Several ’minimum inductances’ were calculated: 7.2, 5.2 and

6.2µHs. The optimum value is 12µH. A standard value higher

than all the ’minimums can be picked. Here, a standard

8.2µH/1.5A was chosen so as to provide the smallest sized

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