Datasheet

LT3083
17
3083fa
+
–
LT3083
IN
V
CONTROL
OUT
V
OUT
V
IN
C2
3083 F08
SET
R
SET
R
P
C1
Figure 8. Reducing Power Dissipation Using a Parallel Resistor
The second technique for reducing power dissipation,
shown in Figure 8, uses a resistor in parallel with the
LT3083. This resistor provides a parallel path for current
flow, reducing the current flowing through the LT3083.
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.
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)
= 2A 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) •
2A = 4.6W. However, the LT3083 supplies only:
2A −
5.5V
−
3.2V
3.6
Ω
= 1.36A
Therefore, the LT3083’s power dissipation is only:
P
DISS
= (5.5V – 3.2V) • 1.36A = 3.13W
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
LT3083 supplies only 1.36A. Therefore, load current can
increase by 1.64A to a total output current of 3.64A while
keeping the LT3083 in its normal operating range.
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