LT3085 LINER [Linear Technology], LT3085 Datasheet - Page 18

• Manufacturer Part Number: LT3085
• Description: Adjustable 500mA Single Resistor Low Dropout Regulator
• Manufacturer: LINER [Linear Technology]

LT3085
APPLICATIONS INFORMATION
Without series resistor R
equals:
If the voltage differential (V
transistor is chosen as 0.5V, then R
Power dissipation in the LT3085 now equals:
The LT3085’s power dissipation is now only 30% compared
to no series resistor. R
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
5.5V, V
I
than 90% of I
18
OUT(MIN)
P
R
P
TOTAL
TOTAL
S
=
OUT
5V – 3.3V − 0.5V
= 0.35A. Also, assuming that R
= 0.86W
= 5V – 3.3V
= 5V – 3.3V
= 3.3V, V
(
(
OUT(MIN)
0.5A
OUT(MIN)
S
)
= 630mA.
dissipates 0.6W of power. Choose
)
•
S
•
⎛
⎝ ⎜
, power dissipation in the LT3085
⎛
⎝ ⎜
IN
= 2.4Ω
0.5A
0.5A
60
= 3.2V, I
= V
60
DIFF
⎞
⎠ ⎟
CONTROL
) across the NPN pass
⎞
⎠ ⎟
+ 0.5V
+ 5V – 3.3V
(
S
(
OUT(MAX)
equals:
P
)
= 5V, V
carries no more
• 0.5A = 0.26W
= 0.5A and
)
IN(MAX)
• 0.5A
=
Calculating R
The maximum total power dissipation is (5.5V – 3.2V) •
0.5A = 1.2W. However the LT3085 supplies only:
Therefore, the LT3085’s power dissipation is only:
R
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
P
(5% Standard value = 7.Ω)
P
R
0.5A –
dissipates 0.71W of power. As with the fi rst technique,
DIS
P
=
= (5.5V – 3.2V) • 0.193A = 0.44W
5.5V – 3.2V
5.5V – 3.2V
315mA
P
7.5Ω
C1
yields:
V
SET
R
CONTROL
= 7.30Ω
SET
LT3085
= 0.193A
+
–
OUT
IN
R
P
C2
3085 F09
V
V
IN
OUT
3085fa