LT1977
W U U
APPLICATIO S I FOR ATIO
U
IPK22 f L V
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.
( )( )(
)
IN
IOUT(MAX)
=
2 V
OUT)(
V – V
IN OUT
(
)
CHOOSING THE INDUCTOR
For most applications the output inductor will fall in the
range of 5µH to 33µH. 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.
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
.
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.
V
V – V
IN OUT
(
OUT)(
)
IP-P
2
I
OUT(MAX) = IPK
–
= IPK –
2 L f V
( )( )(
)
IN
Discontinuous operation occurs when:
VOUT V – V
(
)
IN
OUT
IOUT(DIS)
≤
2(L)(f)(V )
IN
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.
For VOUT = 5V, VIN = 8V and L = 15µH:
5 8 – 5
( )(
)
IOUT(MAX) = 1.5 –
2 15e – 6 500e3 8
( )( )( )
= 1.5 – 0.125 = 1.375A
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.
Note that there is less load current available at the higher
inputvoltagebecauseinductorripplecurrentincreases.At
VIN = 15V, duty cycle is 33% and for the same set of
conditions:
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.
5 15 – 5
( )(
)
I
OUT(MAX) = 1.5 –
2 15e – 6 500e3 15
)( )( )
(
= 1.5 – 0.22 = 1.28A
2. Calculate peak inductor current at full load current to
ensure that the inductor will not saturate. Peak current
canbesignificantlyhigherthanoutputcurrent,especially
with smaller inductors and lighter loads, so don’t omit
thisstep.Powderedironcoresareforgivingbecausethey
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-
To calculate actual peak switch current in continuous
mode with a given set of conditions, use:
VOUT V – VOUT
(
IN
)
ISW(PK) = IOUT
+
2 L f V
( )( )( IN
)
If a small inductor is chosen which results in discontinous
mode operation over the entire load range, the maximum
load current is equal to:
tinuous mode, so it can be used for all conditions.
1977f
13
Linear [ Linear ]
