AD8531ART Analog Devices, AD8531ART Datasheet - Page 9

• Manufacturer Part Number: AD8531ART
• Description: Low Cost, 250 ma Output Single-supply Amplifiers
• Manufacturer: Analog Devices

REV. B
Power Dissipation
Although the AD8531/AD8532/AD8534 is capable of providing
load currents to 250 mA, the usable output load current drive
capability will be limited to the maximum power dissipation
allowed by the device package used. In any application, the
absolute maximum junction temperature for the AD8531/
AD8532/AD8534 is 150 C, and should never be exceeded for
the device could suffer premature failure. Accurately measuring
power dissipation of an integrated circuit is not always a
straightforward exercise, so Figure 34 has been provided as a
design aid for either setting a safe output current drive level or
in selecting a heatsink for the three package options available on
the AD8531/AD8532/AD8534.
These thermal resistance curves were determined using the
AD8531/AD8532/AD8534 thermal resistance data for each
package and a maximum junction temperature of 150 C. The
following formula can be used to calculate the internal junction
temperature of the AD8531/AD8532/AD8534 for any application:
where T
To calculate the power dissipated by the AD8531/AD8532/
AD8534, the following equation can be used:
where I
The quantity within the parentheses is the maximum voltage
developed across either output transistor. As an additional
design aid in calculating available load current from the
AD8531/AD8532/AD8534, Figure 1 illustrates the AD8531/
AD8532/AD8534 output voltage as a function of load resistance.
Figure 34. Maximum Power Dissipation vs. Ambient
Temperature
1.5
0.5
P
junction-to-case; and
T
V
V
LOAD
1
0
JA
J
DISS
A
S
OUT
0
= junction temperature;
TSSOP
= is supply voltage; and
PDIP
SOIC
= Ambient temperature of the circuit.
= package thermal resistance,
JA
JA
JA
= power dissipation;
= is output voltage.
= 158 C/W
= 240 C/W
= is output load current;
= 103 C/W
P
DISS
25
T
J
= I
= P
TEMPERATURE – C
LOAD
DISS
50
(V
JA
S
–V
+ T
OUT
A
T
FREE AIR
NO HEATSINK
75
)
J
MAX = 150 C
85
100
–9–
Power Calculations for Varying or Unknown Loads
Often, calculating power dissipated by an integrated circuit to
determine if the device is being operated in a safe range is not
as simple as it might seem. In many cases power cannot be
directly measured. This may be the result of irregular output
waveforms or varying loads; indirect methods of measuring
power are required.
There are two methods to calculate power dissipated by an
integrated circuit. The first can be done by measuring the pack-
age temperature and the board temperature. The other is to
directly measure the circuit’s supply current.
Calculating Power by Measuring Ambient and Case
Temperature
Given the two equations for calculating junction temperature:
where T
ture.
where T
data sheet.
The two equations can be solved for P (power):
Once power has been determined it is necessary to go back and
calculate the junction temperature to assure that it has not been
exceeded.
The temperature measurements should be directly on the pack-
age and on a spot on the board that is near the package but
definitely not touching it. Measuring the package could be diffi-
cult. A very small bimetallic junction glued to the package could
be used or it could be done using an infrared sensing
device if the spot size is small enough.
Calculating Power by Measuring Supply Current
Power can be calculated directly knowing the supply voltage and
current. However, supply current may have a dc component
with a pulse into a capacitive load. This could make rms current
very difficult to calculate. It can be overcome by lifting the sup-
ply pin and inserting an rms current meter into the circuit. For
this to work you must be sure all of the current is being deliv-
ered by the supply pin you are measuring. This is usually a good
method in a single supply system; however, if the system uses
dual supplies, both supplies may need to be monitored.
Input Overvoltage Protection
As with any semiconductor device, whenever the condition
exists for the input to exceed either supply voltage, the device’s
input overvoltage characteristic must be considered. When an
overvoltage occurs, the amplifier could be damaged depending
on the magnitude of the applied voltage and the magnitude of
the fault current. Although not shown here, when the input
voltage exceeds either supply by more than 0.6 V, pn-junctions
internal to the AD8531/AD8532/AD8534 energize allowing
current to flow from the input to the supplies. As illustrated in
the simplified equivalent input circuit (Figure 32), the AD8531/
AD8532/AD8534 does not have any internal current limiting
resistors, so fault currents can quickly rise to damaging levels.
JA
J
C
is the junction to ambient thermal resistance.
is junction temperature, and T
is case temperature and
P = (T
T
AD8531/AD8532/AD8534
A
+ P
T
T
A
J
J
= T
= T
– T
JA
C
A
C
= T
)/ (
+ P
+ P
JA
C
+ P
JC
and
JA
JC
–
A
is ambient tempera-
JC
JA
JC
)
are given in the