LT1316CS8 Linear Technology, LT1316CS8 Datasheet - Page 8

• Manufacturer Part Number: LT1316CS8
• Description: IC DC/DC CONVERTER STEP-UP 8SOIC
• Manufacturer: Linear Technology
• Type: Step-Up (Boost)r

Specifications of LT1316CS8

• Internal Switch(s): Yes
• Synchronous Rectifier: No
• Number Of Outputs: 1
• Voltage - Output: 1.23 ~ 30 V
• Current - Output: 500mA
• Voltage - Input: 1.5 ~ 12 V
• Operating Temperature: 0°C ~ 70°C
• Mounting Type: Surface Mount
• Package / Case: 8-SOIC (3.9mm Width)
• Lead Free Status / RoHS Status: Contains lead / RoHS non-compliant
• Power - Output: -
• Frequency - Switching: -

APPLICATIONS
LT1316
Discontinuous Mode Operation
A boost converter with a high V
a high duty cycle in continuous mode. For duty cycles
exceeding the LT1316’s guaranteed minimum specifica-
tion of 0.73, the circuit will need to be designed for
discontinuous operation. Additionally, very low peak cur-
rent limiting below 50mA may necessitate operating in this
mode unless high inductance values are acceptable. When
operating in discontinuous mode, a different equation
governs available output power. For each switch cycle, the
inductor current ramps down to zero, completely releas-
ing the stored energy. Energy stored in the inductor at any
time is equal to 1/2 LI
each cycle, the equation for maximum power out is:
When designing for very low peak currents (< 50mA), the
inductor size needs to be large enough so that on-time is
a least 1 s. On-time can be calculated by the equation:
where V
Also, at these low current levels, current overshoot due to
power transistor turn-off delay will be a significant portion
of peak current. Increasing inductor size will keep this to
a minimum.
Design Example 1
Requirements: V
1. Find duty cycle
8
Because duty cycle is less than the LT1316 minimum
specification (0.73), the circuit can be designed for
continuous operation.
P
Where f =
On-Time =
DC =
OUT(MAX)
SAT
V
V
OUT
OUT
= 0.2V.
= 1/2L(I
V
– V
– V
I
IN
(V
IN
PEAK
I
IN
= 2V, V
SAT
U
– V
PEAK
IN
– V
(L)
2
+ V
PEAK
SAT
. Because this energy is released
+ V
1
INFORMATION
• L
SAT
U
D
OUT
D
+ t
2
)f
)
=
OUT
OFF
= 5V and I
5 – 0.2 + 0.4
5 – 2 + 0.4
:V
W
IN
ratio operates with
LOAD
= 10mA.
U
= 0.654
2. I
3. Find L
4. Find R
Design Example 2
Requirements: V
1. Find duty cycle:
2. Find P
3. Set the on-time to the data sheet minimum of 3.4 s and
Find R
R
Because duty cycle exceeds LT1316 minimum specifi-
cation of 73%, the circuit must be designed for discon-
tinuous operation.
Multiply P
P
find L
Overshoot =
L =
DC =
L =
PEAK
SET
OUT(MAX)
=
= 293 H
=
2(0.196W)(3.4 s + 2 s)
2P
V
5 – 2 + 0.4
0.4(58mA)
=
OUT(MAX)
SET
SET
(t
(3.4 s
V
OUT(MAX)
OUT
47k
V
ON
OUT
2(I
1 – DC
0.4(I
OUT
OUT
resistor
from Figure 3 for 58mA – 1.8mA = 56.2mA
= P
2
OUT
=
– V
)(V
– V
– V
2
PEAK
IN
OUT
by 1.4 to give a safe operating margin
)(3.3 – 0.2)
330 H
V
IN
)
IN
L
SAT
= 3.3V, V
=
IN
(t
IN
(1.4) = (5mA)(28V)(1.4) = 0.196W
– V
+ V
2
ON
1 – 0.654
)
2(10mA)
+ V
2 s
300ns
+ V
SAT
D
+ t
D
D
= 1.8mA
OFF
t
)
OUT
OFF
2
=
2
)
= 58mA
28 – 3.3 + 0.4
28 – 0.2 + 0.4
= 28V and I
= 52 H
LOAD
= 0.89
= 5mA.