LT1766 Linear Technology, LT1766 Datasheet - Page 18

• Manufacturer Part Number: LT1766
• Description: 5.5V to 60V 1.5A/ 200kHz Step-Down Switching Regulator
• Manufacturer: Linear Technology

LT1766/LT1766-5
APPLICATIO S I FOR ATIO
switch off spike will also cause the SW node to go below
ground. The LT1766 has special circuitry inside which
mitigates this problem, but negative voltages over 0.8V
lasting longer than 10ns should be avoided. Note that
100MHz oscilloscopes are barely fast enough to see the
details of the falling edge overshoot in Figure 7.
A second, much lower frequency ringing is seen during
switch off time if load current is low enough to allow the
inductor current to fall to zero during part of the switch off
time (see Figure 8). Switch and diode capacitance reso-
nate with the inductor to form damped ringing at 1MHz to
10 MHz. This ringing is not harmful to the regulator and it
has not been shown to contribute significantly to EMI. Any
attempt to damp it with a resistive snubber will degrade
efficiency.
THERMAL CALCULATIONS
Power dissipation in the LT1766 chip comes from four
sources: switch DC loss, switch AC loss, boost circuit
current, and input quiescent current. The following formu-
las show how to calculate each of these losses. These
formulas assume continuous mode operation, so they
should not be used for calculating efficiency at light load
currents.
Switch loss:
18
P
SW
2V/DIV
R
SW OUT
Figure 7. Switch Node Resonance
I
V
SW RISE
U
IN
2
V
OUT
50ns/DIV
U
SW FALL
t
EFF
W
( / )
1 2
1766 F07
I
OUT
U
V
IN
f
Boost current loss:
Quiescent current loss:
Example: with V
Total power dissipation in the IC is given by:
P
P
P
SW
Q
BOOST
R
t
t
t
t
f = Switch frequency
P
P
P
EFF
r
f
Ir
Q
BOOST
TOT
SW
= (V
= (V
= t
40 0 0015
0.2A/DIV
= Effective switch current/voltage overlap time
10V/DIV
= (t
= Switch resistance ( 0.3) hot
0 04 0 388 0 43
If
= P
= 0.43W + 0.02W + 0.08W = 0.53W
V
( .
IN
0 3 1 5
IN
.
IN
= (I
.
r
/1.7)ns
/1.2)ns
SW
Figure 8. Discontinuous Mode Ringing
+ t
5
0 0015
V
V
L = 47 H
40
OUT
V
IN
OUT
.
2
f
OUT
+ P
= 40V
40
2
+ t
IN
1 36
= 5V
.
/0.05)ns
)
/
2
BOOST
Ir
= 40V, V
V
5 0 003
I
+ t
OUT
IN
( .
If
V
97 10
)
OUT
+ P
0 02
/
.
1 s/DIV
36
•
.
OUT
)
Q
W
0 003
W
.
9
= 5V and I
0 08
.
1 2 1 40 200 10
/
W
1766 F08
OUT
SWITCH NODE
VOLTAGE
INDUCTOR
CURRENT
AT I
= 1A:
OUT
= 0.1A
•
1766fa
3