Datasheet

LTC1871-1
20
18711fb
APPLICATIONS INFORMATION
Figure 13. Load Transient Response for a 3.3V Input,
5V Output Boost Converter Application, 0.7A to 7A Step
dissipated in this resistor would be 514mW at maxi-
mum output current. Assuming an effi ciency of 90%,
this sense resistor power dissipation represents 1.3%
of the overall input power. In other words, for this ap-
plication, the use of V
DS
sensing would increase the
effi ciency by approximately 1.3%.
For more details regarding the various terms in these
equations, please refer to the section Boost Converter:
Power MOSFET Selection.
3. The losses in the inductor are simply the DC input cur-
rent squared times the winding resistance. Expressing
this loss as a function of the output current yields:
P
R(WINDING)
=
I
O(MAX)
1–D
MAX






2
•R
W
4. Losses in the boost diode. The power dissipation in the
boost diode is:
P
DIODE
= I
O(MAX)
• V
D
The boost diode can be a major source of power loss
in a boost converter. For the 3.3V input, 5V output at
7A example given above, a Schottky diode with a 0.4V
forward voltage would dissipate 2.8W, which represents
7% of the input power. Diode losses can become signifi -
cant at low output voltages where the forward voltage
is a signifi cant percentage of the output voltage.
5. Other losses, including C
IN
and C
O
ESR dissipation and
inductor core losses, generally account for less than
2% of the total additional loss.
Checking Transient Response
The regulator loop response can be verifi ed by looking at
the load transient response. 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 (ΔI
LOAD
)(ESR),
and then C
O
begins to charge or discharge (depending on
the direction of the load step) as shown in Figure 13. 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.
I
OUT
2V/DIV
V
OUT
(AC)
100mV/DIV
100µs/DIV
18711 F13
V
IN
= 3.3V
V
OUT
= 5V
MODE/SYNC = INTV
CC
(PULSE-SKIP MODE)
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 1. The input voltage is 3.3V, and the output is 5V
at a maximum load current of 7A (10A peak).
1. The duty cycle is:
D=
V
O
+ V
D
–V
IN
V
O
+ V
D






=
5 + 0.4 – 3.3
5 +
0.4
= 38.9%
2. Pulse-skip operation is chosen so the MODE/SYNC pin
is shorted to INTV
CC
.
3. The operating frequency is chosen to be 300kHz to
reduce the size of the inductor. From Figure 5, the
resistor from the FREQ pin to ground is 80k.
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+

2






•
I
O(MAX)
1–D
MAX
= 1.2 •
7
1– 0.39
= 13.8A