Datasheet

LT3080-1
15
30801fb
Figure 8. Reducing Power Dissipation Using a Series Resistor
+
–
LT3080-1
IN
V
CONTROL
OUT
V
OUT
V
IN
′
V
IN
C2
30801 F08
SET
R
SET
25mΩ
R
S
C1
applicaTions inFormaTion
The second technique for reducing power dissipation,
shown in Figure 9, uses a resistor in parallel with the
LT3080-1. This resistor provides a parallel path for current
flow, reducing the current flowing through the LT3080-1.
This technique works well if input voltage is reasonably
constant and output load current changes are small. This
technique also increases the maximum available output
current at the expense of minimum load requirements.
Reducing Power Dissipation
In some applications it may be necessary to reduce
the power dissipation in the LT3080-1 package without
sacrificing output current capability. Two techniques are
available. The first technique, illustrated in Figure 8, em-
ploys a resistor in series with the regulator’s input. The
voltage drop across R
S
decreases the LT3080-1’s IN-to-
OUT differential voltage and correspondingly decreases
the LT3080-1’s power dissipation.
As an example, assume: V
IN
= V
CONTROL
= 5V, V
OUT
= 3.3V
and I
OUT(MAX)
= 1A. Use the formulas from the Calculat-
ing Junction Temperature section previously discussed.
Without series resistor R
S
, power dissipation in the
LT3080-1 equals:
P
TOTAL
= 5V – 3.3V
( )
•
1A
60






+ 5V – 3.3V
( )
• 1A = 1.73
W
If the voltage differential (V
DIFF
) across the NPN pass
transistor is chosen as 0.5V, then R
S
equals:
R
S
=
5V – 3.3V − 0.5V
1A
= 1.2Ω
Power dissipation in the LT3080-1 now equals:
P
TOTAL
= 5V – 3.3V
( )
•
1A
60






+ 0.5V
( )
• 1A = 0.53W
The LT3080-1’s power dissipation is now only 30% com-
pared to no series resistor. R
S
dissipates 1.2W of power.
Choose appropriate wattage resistors to handle and dis-
sipate the power properly.
Figure 9. Reducing Power Dissipation Using a Parallel Resistor
+
–
IN
V
CONTROL
OUT
V
OUT
V
IN
C2
30801 F09
SET
R
SET
R
P
C1
LT3080-1
25mΩ
As an example, assume: V
IN
= V
CONTROL
= 5V, V
IN(MAX)
=
5.5V, V
OUT
= 3.3V, V
OUT(MIN)
= 3.2V, I
OUT(MAX)
= 1A and
I
OUT(MIN)
= 0.7A. Also, assuming that R
P
carries no more
than 90% of I
OUT(MIN)
= 630mA.
Calculating R
P
yields:
R
P
=
5.5V – 3.2V
0.63A
= 3.65Ω
(5% Standard value = 3.6Ω)
The maximum total power dissipation is (5.5V – 3.2V) •
1A = 2.3W. However, the LT3080-1 supplies only:
1A –
5.5V – 3.2V
3.6Ω
= 0.36A
Therefore, the LT3080-1’s power dissipation is only:
P
DIS
= (5.5V – 3.2V) • 0.36A = 0.83W
R
P
dissipates 1.47W of power. As with the first technique,
choose appropriate wattage resistors to handle and dis-
sipate the power properly. With this configuration, the
LT3080-1 supplies only 0.36A. Therefore, load current can
increase by 0.64A to 1.64A while keeping the LT3080-1 in
its normal operating range.