LT1976EFE Linear Technology, LT1976EFE Datasheet - Page 15

• Manufacturer Part Number: LT1976EFE
• Description: IC REG SW STEP DWN 1.5A 16-TSSOP
• Manufacturer: Linear Technology
• Type: Step-Down (Buck)r

Specifications of LT1976EFE

• Internal Switch(s): Yes
• Synchronous Rectifier: No
• Number Of Outputs: 1
• Voltage - Output: 1.2 ~ 54 V
• Current - Output: 1.5A
• Frequency - Switching: 200kHz
• Voltage - Input: 3.3 ~ 60 V
• Operating Temperature: -40°C ~ 125°C
• Mounting Type: Surface Mount
• Package / Case: 16-TSSOP Exposed Pad, 16-eTSSOP, 16-HTSSOP
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• Power - Output: -

APPLICATIO S I FOR ATIO
Example: with V
0.08Ω, ESL = 10nH:
MAXIMUM OUTPUT LOAD CURRENT
Maximum load current for a buck converter is limited by
the maximum switch current rating (I
rating for the LT1976 is 1.5A. Unlike most current mode
converters, the LT1976 maximum switch current limit
does not fall off at high duty cycles. Most current mode
converters suffer a drop off of peak switch current for duty
cycles above 50%. This is due to the effects of slope
compensation required to prevent subharmonic oscilla-
tions in current mode converters. (For detailed analysis,
see Application Note 19.)
The LT1976 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 (I
Discontinuous operation occurs when:
V
I
I
I
dt
P-P
di
OUT MAX
OUT DIS
RIPPLE
=
(
(
=
3 3
( )
.
12 33
)
= (0.362A)(0.08) + (10e – 9)(363e3)
= 0.0289 + 0.003 = 32mV
12
e
)
≤
( )(
–
=
3 3 12 3 3
(
V
.
I
5
PK
OUT
IN
e
=
2
( )( )(
= 12V, V
−
–
U
3 63 5
L f V
(
.
V
6 200 3
(
IN
– .
V
)(
OUT
e
–
2
U
( )( )( )
V
IN
L f V
)
)(
P-P
OUT
OUT
e
)
V
IN
/2) is less than I
)
= 3.3V, L = 33μH, ESR =
)
=
–
IN
0 362
W
V
.
OUT
P-P
PK
)
A
=
). The current
I
PK
U
OUT
–
I
P
.
2
-P
For V
Note that there is less load current available at the higher
input voltage because inductor ripple current increases. At
V
conditions:
To calculate actual peak switch current in continuous
mode with a given set of conditions, use:
If a small inductor is chosen which results in discontinous
mode operation over the entire load range, the maximum
load current is equal to:
CHOOSING THE INDUCTOR
For most applications the output inductor will fall in the
range of 15μH to 100μH. 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 LT1976 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.
IN
I
I
I
I
OUT MAX
OUT MAX
SW PK
OUT MAX
= 15V, duty cycle is 33% and for the same set of
OUT
(
(
(
(
)
= 5V, V
=
)
)
)
=
=
I
=
=
=
OUT
1 5 0 24 1 26
1 5 0 42 1 08
1 5
1 5
2
. – .
. – .
. –
. –
(
IN
I
V
+
PK
OUT
= 8V and L = 20μH:
V
2
2 20
2 20
OUT
2
(
(
)(
( )( )( )
2
f L V
LT1976/LT1976B
( )( )( )
V
L f V
(
e
=
e
=
IN
V
( )(
( )(
–
–
IN
5 8 5
5 15 5
–
.
.
6 200 3 8
6 200 3 15
–
V
)(
)(
IN
OUT
V
IN
A
A
–
OUT
–
)
)
e
e
)
)
)( )
)( )
15
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