Datasheet

LTC1871-7
20
18717fd
applicaTions inForMaTion
Checking Transient Response
The regulator loop response can be verified by looking at
the load transient response at minimum and maximum
V
IN
. Switching regulators generally take several cycles to
respond to an instantaneous step in resistive load current.
When the load step occurs, V
O
immediately shifts by an
amount equal to (∆I
LOAD
)(ESR), and then C
O
begins to
charge or discharge (depending on the direction of the load
step) as shown in Figure 14. The regulator feedback loop
acts on the resulting error amp output signal to return V
O
to its steady-state value. During this recovery time, V
O
can
be monitored for overshoot or ringing that would indicate
a stability problem.
A second, more severe transient can occur when con-
necting loads with large (>1µF) supply bypass capacitors.
The discharged bypass capacitors are effectively put in
parallel with C
O
, causing a nearly instantaneous drop in
V
O
. No regulator can deliver enough current to prevent
this problem if the load switch resistance is low and it is
driven quickly. The only solution is to limit the rise time
of the switch drive in order to limit the inrush current di/
dt to the load.
Boost Converter Design Example
The design example given here will be for the circuit shown
in Figure 9. The input voltage is 8V to 28V, and the output
is 42V at a maximum load current of 1.5A.
1. The maximum duty cycle is:
D =
V
O
+ V
D
– V
IN
V
O
+ V
D






=
42 + 0.4 – 8
42 + 0.4
= 81.1%
2. Pulse-skip operation is chosen so the MODE/SYNC pin
is shorted to INTV
CC
.
3. The operating frequency is chosen to be 250kHz to
reduce the size of the inductor. From Figure 5, the
resistor from the FREQ pin to ground is 100k.
4. An inductor ripple current of 40% of the maximum load
current is chosen, so the peak input current (which is
also the minimum saturation current) is:
I
IN(PEAK)
= 1+
χ
2






•
I
O(MAX)
1– D
MAX
= 1.2 •
1.5
1– 0.81
= 9.47A
The inductor ripple current is:
∆I
L
= χ •
I
O(MAX)
1– D
MAX
= 0.4 •
1.5
1– 0.81
= 3.2A
And so the inductor value is:
L =
V
IN(MIN)
∆I
L
• f
•D
MAX
=
8
3.2 • 250k
• 0.81= 8.1µH
Figure 14a. Load Transient Response for the Circuit in Figure 9
Figure 14b. Load Transient Response for the Circuit in Figure 9
V
OUT
500mV/DIV
I
OUT
0.5A/DIV
0.5A
250µs/DIV
18717 F14a
1.5A
V
IN
= 8V
V
OUT
500mV/DIV
I
OUT
0.5A/DIV
0.5A
250µs/DIV
18717 F14b
1.5A
V
IN
= 28V