Datasheet

LT3837
16
3837fd
APPLICATIONS INFORMATION
Setting Feedback Resistive Divider
Use the equation developed in the Operation section for
the feedback divider.
It is recommended that the Thevenin impedance of the
resistors on the FB Pin is roughly 3k for bias current
cancellation and other reasons.
For the example using primary winding sensing if
ESR = 0.002 and R
DS(ON)
= 0.004 then:
R1=
3k
1.237
•
3.3+10• 0.002+0.004
(
( )
1/ 3
( )






– 0.7








= 22.75k
So, choose 22.1k.
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
+
and SENSE
–
are isolated and con-
nect close to the sense resistor to preserve this.
Peak current occurs at 98mV of sense voltage V
SENSE
. So
the nominal sense resistor is V
SENSE
/I
PK
. For example, a
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
SENSE
using worst-case conditions, minimum L
P
,
V
SENSE
and maximum V
IN
. Continuing the example, let us
assume that our worst-case conditions yield an I
PK
10%
above nominal so I
PK
= 10.41A . If there is a 5% tolerance
on R
SENSE
and minimum V
SENSE
= 80mV, then R
SENSE
•
105% = 88mV/10.41A and nominal R
SENSE
= 8.05mΩ.
Round to the nearest available lower value 8.0mΩ.
Selecting the Load Compensation Resistor
The expression for R
CMP
was derived in the Operation
section for primary winding sensing as:
R
CMP
=K1•
R
SENSE
• 1–DC
( )
ESR+R
DS(ON)
•R1 • N
SP
=R
S(OUT)
Continuing the example:
K1=
V
OUT
V
IN
• Eff






=
3.3
9 88%
( )
= 0.417
If ESR = 0.002Ω and R
DS(ON)
= 0.004Ω
R
CMP
= 0.417 •
8.0mΩ • 1– 0.52
( )
0.002Ω + 0.004Ω
• 22.1kΩ•0.33
=
1.93k
Ω
This value for R
CMP
is a good starting point, but empiri-
cal methods are required for producing the best results.
This is because several of the required input variables are
difficult 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
IN
to V
OUT
times (differential) ef-
ficiency, but theoretically estimating efficiency is not a
simple calculation.
The suggested empirical method is as follows:
1. Build a prototype of the desired supply including the
actual secondary components.
2. Temporarily ground the C
CMP
pin to disable the load
compensation function. Measure output voltage while
sweeping output current over the expected range.
Approximate the voltage variation as a straight line,
∆V
OUT
/∆I
OUT
= R
S(OUT)
.
3. Calculate a value for the K1 constant based on V
IN
, V
OUT
and the measured (differential) efficiency.
4. Compute:
R
CMP
=K1•
R
SENSE
R
S(OUT)
•R1•N
SP
orN
SF