參數(shù)資料
型號: LM1572
廠商: National Semiconductor Corporation
英文描述: 1.5A, 500kHz Step-down Voltage Regulator
中文描述: 1.5A的,500kHz降壓型穩(wěn)壓器
文件頁數(shù): 10/17頁
文件大?。?/td> 584K
代理商: LM1572
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
V
and V
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
D
CCM
= 0.344
I
CL
= 2A (min value), so the required minimum inductance is
L
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
I
CLIM
= I
CL
{(m
C
T)
(D 0.5)} A
I
CLIM
= 2 {(0.42
2)
(0.65 0.5)} A
I
CLIM
= 1.87A
So the minimum inductance at this point is
L
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
P2
.
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
L
OPT
= 12μH
Step 5:
Conclusions
Several ’minimum inductances’ were calculated: 7.2, 5.2 and
6.2μHs. The optimum value is 12μH.Astandard 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
L
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