LT3085
17
3085fb
Without series resistor R
S
, power dissipation in the LT3085
equals:
P
TO TAL
= 5V – 3.3V
()
•
0.5A
60
+ 5V – 3.3V
()
• 0.5A
= 0.86W
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
0.5A
= 2.4Ω
Power dissipation in the LT3085 now equals:
P
TOTAL
= 5V – 3.3V
()
•
0.5A
60
+ 0.5V
()
• 0.5A = 0.26W
The LT3085’s power dissipation is now only 30% compared
to no series resistor. R
S
dissipates 0.6W of power. Choose
appropriate wattage resistors to handle and dissipate the
power properly.
The second technique for reducing power dissipation,
shown in Figure 9, uses a resistor in parallel with the
LT3085. This resistor provides a parallel path for current
fl ow, reducing the current fl owing through the LT3085.
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)
= 0.5A and
I
OUT(MIN)
= 0.35A. 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
315mA
= 7.30Ω
(5% Standard value = 7.Ω)
The maximum total power dissipation is (5.5V – 3.2V) •
0.5A = 1.2W. However the LT3085 supplies only:
0.5A –
5.5V – 3.2V
7.5Ω
= 0.193A
Therefore, the LT3085’s power dissipation is only:
P
DIS
= (5.5V – 3.2V) • 0.193A = 0.44W
R
P
dissipates 0.71W of power. As with the fi rst technique,
choose appropriate wattage resistors to handle and dis-
sipate the power properly. With this confi guration, the
LT3085 supplies only 0.36A. Therefore, load current can
increase by 0.3A to 0.143A while keeping the LT3085 in
its normal operating range.
Figure 9. Reducing Power Dissipation Using a Parallel Resistor
+
–
LT3085
IN
V
CONTROL
OUT
V
OUT
V
IN
C2
3085 F09
SET
R
SET
R
P
C1
APPLICATIONS INFORMATION
