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LT1977 ๋ฐ์ดํ„ฐ์‹œํŠธ(HTML) 13 Page - Linear Technology

๋ถ€ํ’ˆ๋ช… LT1977
์ƒ์„ธ๋‚ด์šฉ  High Voltage 1.5A, 500kHz Step-Down Switching Regulator with 100uA Quiescent Current
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LT1977 ๋ฐ์ดํ„ฐ์‹œํŠธ(HTML) 13 Page - Linear Technology

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LT1977
13
1977f
APPLICATIO S I FOR ATIO
The LT1977 is able to maintain peak switch current limit
over the full duty cycle range by using patented circuitry to
cancel the effects of slope compensation on peak switch
current without affecting the frequency compensation it
provides.
Maximum load current would be equal to maximum
switch current for an infinitely large inductor, but with
finite inductor size, maximum load current is reduced by
one-half peak-to-peak inductor current. The following
formula assumes continuous mode operation, implying
that the term on the right (IP-P/2) is less than IOUT.
II
VV
V
Lf V
I
I
OUT MAX
PK
OUT
IN
OUT
IN
PK
P
()
โ€“
โ€“
โ€“
=
()(
)
()( )( )
=
2
-P
2
Discontinuous operation occurs when:
I
VV
V
Lf V
OUT DIS
OUT
IN
OUT
IN
()
โ€“
()( )(
)
โ‰ค
()
2
For VOUT = 5V, VIN = 8V and L = 15ยตH:
I
ee
A
OUT MAX
()
.โ€“
โ€“
โ€“
.โ€“ .
.
=
()(
)
()(
)( )
==
15
58 5
2 15
6 500 3 8
1 5 0 125
1 375
Note that there is less load current available at the higher
input voltage because inductor ripple current increases. At
VIN = 15V, duty cycle is 33% and for the same set of
conditions:
I
ee
A
OUT MAX
()
.โ€“
โ€“
โ€“
.โ€“ .
.
=
()(
)
()(
)( )
==
15
515 5
2 15
6 500 3 15
15 022
1 28
To calculate actual peak switch current in continuous
mode with a given set of conditions, use:
II
VV
V
Lf V
SW PK
OUT
OUT
IN
OUT
IN
()
โ€“
=+
()
()( )( )
2
If a small inductor is chosen which results in discontinous
mode operation over the entire load range, the maximum
load current is equal to:
I
If L V
VV
V
OUT MAX
PK
IN
OUT
IN
OUT
()
โ€“
=
()( )( )
()(
)
22
2
CHOOSING THE INDUCTOR
For most applications the output inductor will fall in the
range of 5ยตH to 33ยตH. Lower values are chosen to reduce
physical size of the inductor. Higher values allow more
output current because they reduce peak current seen by
the LT1977 switch, which has a 1.5A limit. Higher values
also reduce output ripple voltage and reduce core loss.
When choosing an inductor you might have to consider
maximum load current, core and copper losses, allow-
able component height, output voltage ripple, EMI, fault
current in the inductor, saturation and of course cost.
The following procedure is suggested as a way of han-
dling these somewhat complicated and conflicting
requirements.
1. Choose a value in microhenries such that the maximum
load current plus half the ripple current is less than the
minimum peak switch current (IPK). Choosing a small
inductor with lighter loads may result in discontinuous
mode of operation, but the LT1977 is designed to work
well in either mode.
Assume that the average inductor current is equal to
load current and decide whether or not the inductor
must withstand continuous fault conditions. If maxi-
mum load current is 0.5A, for instance, a 0.5A inductor
may not survive a continuous 2A overload condition.
For applications with a duty cycle above 50%, the
inductor value should be chosen to obtain an inductor
ripple current of less than 40% of the peak switch
current.
2. Calculate peak inductor current at full load current to
ensure that the inductor will not saturate. Peak current
can be significantly higher than output current, especially
with smaller inductors and lighter loads, so donโ€™t omit
this step. Powdered iron cores are forgiving because they
saturate softly, whereas ferrite cores saturate abruptly.
Other core materials fall somewhere in between. The
following formula assumes continuous mode of opera-
tion, but it errs only slightly on the high side for discon-
tinuous mode, so it can be used for all conditions.


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