Datasheet
LTC3863
23
3863f
For more information www.linear.com/3863
APPLICATIONS INFORMATION
the error amplifier output drive current on ITH of 100µA.
The effect causes ITH to appear clamped in response
to a transient load current step which causes excessive
output droop.
An R
ITH
greater than 20k allows ITH to swing 1.5V with
margin for temperature and part to part variation and
should never have this issue. In applications with less
severe transient load step requirements, R
ITH
can safely
be set as low as 10k. We do not recommend less than
10k in any application. If R
ITH
is too small then either
the operating frequency will need to be increased or the
output capacitor increased to increase the R
ITH
required
to stabilize the system. We strongly recommend that any
system with an R
ITH
less than 20k be experimentally veri-
fied with worst-case load steps.
Design Example
Consider an inverting converter with the following speci-
fications:
V
IN
= 4.5V to 55V, V
OUT
= –5V, I
OUT(MAX)
= 1.8A, and
f = 320kHz (Figure 7).
The output voltage is programmed according to:
V
OUT
= –0.8V •
R
FB2
R
FB1
If R
FB2
is chosen to be 188k, then R
FB1
needs to be 30.1k.
The FREQ pin is tied to signal ground in order to program
the switching frequency to 350kHz. The on-time required
to generate –5V output from 55V V
IN
in continuous mode
can be calculated as:
t
ON(CCM)
=
5V + 0.5V
320kHz • 55V + 5V + 0.5V
( )
= 260ns
This on-time, t
ON
, is larger than LTC3863’s minimum on-
time with sufficient margin to prevent cycle skipping. Use
a lower frequency if a larger on-time margin is needed to
account for variations from minimum on-time and switch-
ing frequency. As load current decreases, the converter
will eventually start cycle skipping.
Next, set the inductor value such that the inductor ripple
current is 60% of the average inductor current at maximum
V
IN
= 55V and full load = 1.8A:
L =
55V
2
• 5V + 0.5V
( )
0.6 •1.8A • 320kHz • 55V + 5V + 0.5V
( )
2
≈ 13.1µH
Select a standard value of 12μH.
The resulting ripple current at minimum V
IN
of 4.5V is:
∆I
L
=
5V • 5V + 0.5V
( )
12µH• 320kHz • 5V + 5V + 0.5V
( )
= 0.644A
Figure 7. Design Example (4.5V to 55V Input, –5V, 1.8A at 320kHz)
320kHz
16mΩ
L1
12µH
Q1
D1
CAP
0.47µF
PGND
LTC3863
3863 F07
SS
ITH
FREQ
SGND
RUN
V
IN
PLLIN/MODE
SENSE
GATE
V
FBN
V
FB
10k
52.3k
187k
C
OUT3
100µF
20V
V
OUT
–5V
1.8A
30.1k
C
IN1
: CDE AFK686M63G24T-F
C
IN2
: TDK CGA6M3X7S2A475K
C
OUT1
: TDK C4532X7R1C336M
C
OUT3
: PANASONIC 20SVP100M
D1: VISHAY SS8PH9-M3/87A
L1: MSS1278-123ML
Q1: VISHAY Si7469DP
V
IN
4.5V TO 55V
C
IN1
68µF
63V
C
IN2
4.7µF
100V
×2
15nF
220pF
0.1µF
C
OUT1
33µF
16V
×2
12pF
+
+