LT3825EFE-PBF LINER [Linear Technology], LT3825EFE-PBF Datasheet - Page 16

• Manufacturer Part Number: LT3825EFE-PBF
• Description: Isolated No-Opto Synchronous Flyback Controller with Wide Input Supply Range
• Manufacturer: LINER [Linear Technology]

APPLICATIONS INFORMATION
Setting Feedback Resistive Divider
The expression for V
tion is rearranged to yield the following expression for the
feedback resistors:
Continuing the example, if ESR + R
3.32k, then:
choose 37.4k.
It is recommended that the Thevenin impedance of the
resistive divider (R1||R2) is roughly 3k for bias current
cancellation and other reasons.
Current Sense Resistor Considerations
The external current sense resistor is used to control peak
primary switch current, which controls a number of key
converter characteristics including maximum power and
external component ratings. Use a noninductive current
sense resistor (no wire-wound resistors). Mounting the
resistor directly above an unbroken ground plane con-
nected with wide and short traces keeps stray resistance
and inductance low.
The dual sense pins allow for a fully Kelvined connection.
Make sure that SENSE
nect close to the sense resistor to preserve this.
Peak current occurs at 98mV of sense voltage V
the nominal sense resistor is V
peak switch current of 10A requires a nominal sense resistor
of 0.010Ω. Note that the instantaneous peak power in the
sense resistor is 1W, and that it is rated accordingly. The
use of parallel resistors can help achieve low resistance,
low parasitic inductance and increased power capability.
Size R
V
assume that our worst-case conditions yield an I
LT3825
16
SENSE
R1= R2
R1= 3.4k
SENSE
and maximum V
using worst-case conditions, minimum L
V
5 + 8 • 0.008
OUT
1.232 • 1/ 3
+I
OUT
SEC
+
and SENSE
IN
V
developed in the Operation sec-
FB
. Continuing the example, let us
• ESR + R
(
• N
– 1 = 37.6k
SF
SENSE
–
DS(ON)
are isolated and con-
DS(ON)
/I
PK
. For example, a
= 8mΩ, R2 =
)
– 1
SENSE
PK
40%
. So
P
,
above nominal so I
on R
110% = 80mV/3.64A and nominal R
to the nearest available lower value.
Selecting the Load Compensation Resistor
The expression for R
section as:
Continuing the example:
This value for R
cal methods are required for producing the best results.
This is because several of the required input variables are
diffi cult to estimate precisely. For instance, the ESR term
above includes that of the transformer secondary, but its
effective ESR value depends on high frequency behavior,
not simply DC winding resistance. Similarly, K1 appears
as a simple ratio of V
fi ciency, but theoretically estimating effi ciency is not a
simple calculation.
The suggested empirical method is as follows:
1. Build a prototype of the desired supply including the
2. Temporarily ground the C
3. Calculate a value for the K1 constant based on V
If ESR + R
actual secondary components.
compensation function. Measure output voltage while
sweeping output current over the expected range.
Approximate the voltage variation as a straight line,
ΔV
and the measured effi ciency.
R
K1=
R
CMP
CMP
SENSE
OUT
/ΔI
V
= 1.96k
=
= 0.116 •
IN
V
DS(ON)
and minimum V
K
OUT
OUT
1
• Eff
•
R
= R
CMP
ESR R
SENSE
= 8m
PK
=
S(OUT)
20m
is a good starting point, but empiri-
= 3.64A . If there is a 10% tolerance
48 • 90%
CMP
+
IN
• –
to V
.
(
DS ON
5
was derived in the Operation
1
SENSE
• 1– 0.455
8m
(
CMP
(
OUT
DC
)
= 0.116
)
pin to disable the load
= 80mV, then R
times (differential) ef-
•
SENSE
R
1
•
)
= 20mΩ. Round
N
• 37.4k
SF
=
R
IN
S S OUT
SENSE
(
, V
3525fa
OUT
)
•