APPLICATION BULLETIN
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POWER AMPLIFIER STRESS
AND POWER HANDLING LIMITATIONS
BY BRUCE TRUMP (602) 746-7347
FIGURE 2. Safe Operating Area (SOA)-- OPA502. (Figure 2 in PDS-1166)
SAFE OPERATING AREA
10
V
CE
|V
S
V
OUT
| (V)
I
O
(A)
10
5.0
2.0
1.0
0.5
0.2
0.1
5
2
1
20
50
100
T
C
= +25C
T
C
= +85C
T
C
= +125C
t = 0.5ms
t = 1ms
t = 5ms
Thermal Limitation
(T
J
= 200C)
Second Breakdown
Limited
Max Current
Thermal
Limits
Second
Breakdown
Region
Voltage
Breakdown
To achieve reliable power amplifier designs you must con-
sider the stress on the amplifier compared to its power
handling limitations. Power handling limits are specified by
the Safe Operating Area (SOA) curves of the power amp.
Stress on the amplifier depends on amplifier load and signal
conditions which can be evaluated with straightforward
techniques.
Consider the simplified power op amp shown in Figure 1.
Output transistors Q
1
and Q
2
provide positive and negative
output current to the load. I
OUT
is shown flowing out of the
amplifier, so Q
1
is supplying the output current. For positive
output current, Q
2
is "off" and can be ignored.
The stress on Q
1
under load is related to the output current
and the voltage across Q
1
(its collector-to-emitter voltage,
V
CE
). The product of these quantities, I
OUT
V
CE
, is the power
dissipation of Q
1
. This power dissipation is one important
consideration, but the "safe operating area" provides a more
complete description of the amplifier's limits.
SAFE OPERATING AREA
The power handling ability of a power transistor is charac-
terized by its Safe Operating Area (SOA), Figure 2. The
SOA curve shows permissible voltage, (V
CE
) and current,
(I
OUT
). The maximum safe current is a function of V
CE
. The
characteristic shape of this curve has four distinct regions.
At low V
CE
, maximum output current can be safely delivered
to the load. Exceeding the maximum current in this region
can overstress wire bonds or metallization on the die and
destroy the device.
FIGURE 1. Simplified Power Op Amp Circuit.
As V
CE
is increased, the power dissipation of the transistor
increases until self-heating raises the junction temperature to
its maximum safe value. All points along this thermally
limited region (dotted lines) produce the same power dissi-
pation. V
CE
I
O
is a constant 120W (at 25
C) in Figure 2. All
points on this region of the curve produce the same maxi-
mum junction temperature. Exceeding the safe output cur-
rent in this region may damage the transistor junction.
V
IN
V+
Q
1
Q
2
R
L
I
OUT
V
O
V
CE
= (V+) V
O
V
V
CE
+
R
F
R
I
1993 Burr-Brown Corporation
AB-039
Printed in U.S.A. April, 1993
2
As V
CE
is further increased, beyond the thermally limited
region, the safe output current decreases more rapidly. This
so-called second breakdown region is a characteristic of
bipolar output transistors. It is caused by the tendency of
bipolar transistors to produce "hot spots"--points on the
transistor where current flow concentrates at high V
CE
.
Exceeding the safe output current in the second breakdown
region can produce a localized thermal runaway, destroying
the transistor.
The final limit is the breakdown voltage of the transistor.
This maximum power supply voltage cannot be exceeded.
Often, an SOA curve provides information showing how the
safe output current varies with case temperature. This ac-
counts for the affect of case temperature on junction tem-
perature. Additional lines may show the maximum safe
current for pulses of various durations according to the
thermal time constants of a device.
The SOA curve should be interpreted as an absolute maxi-
mum rating. Operation at any point on the thermal limit
portion of the curve produces the maximum allowable junc-
tion temperature--a condition not advised for long-term
operation. Although operation on the second-breakdown
portion of the curve produces lower temperature, this line is
still an absolute maximum. Operation below this limit will
provide better reliability (i.e.--better MTTF).
HEAT SINKING
In addition to assuring that an application does not exceed
the safe operating area of the power amplifier, you must also
assure that the amplifier does not overheat. To provide an
adequate heat sink, you must determine the maximum power
dissipation. The following discussions detail methods and
considerations that affect SOA requirements and power
dissipation and heat sink requirements.
SHORT-CIRCUITS
Some amplifier applications must be designed to survive a
short-circuit to ground. This forces the full power supply
voltage (either V+ or V) across the conducting output
transistor. The amplifier will immediately go into current
limit. To survive this condition a power op amp with
adjustable current limit must be set to limit at a safe level.
Example 1
What is the maximum current limit value which would
protect against short-circuit to ground when OPA502
(Figure 2) power supplies are
40V?
Answer--
If the case temperature could be held to 25
C, the
current limit could be set to 3A, maximum. This would
be unlikely, however, since the amplifier would dissi-
pate 120W during short-circuit. It would require an
"infinite" or ideal heat sink to maintain the case tem-
perature at 25
C in normal room ambient conditions.
If the case temperature were held to 85
C, a 2A current
limit would be safe. Power dissipation would be 80W,
requiring a heat sink of 0.75
C/W--a large heat sink.
(See Application Bulletin AB-038 for heat sink calcula-
tions.)
If the op amp must survive a short-circuit to one of the power
supplies, for instance, the maximum V
CE
would be the total
of both supplies--a very demanding case.
Not all applications must (or can be) designed for short-
circuit protection. It is a severe condition for a power
amplifier. Additional measures such as fuses or circuitry to
sense a fault condition can limit the time the amplifier must
endure a short-circuit. This can greatly reduce the heat sink
requirement.
An additional feature of the OPA502 and OPA512 power
amplifiers, the optional fold-over circuit, can be connected
on the current limit circuit. This can be set to reduce the
current limit value when V
CE
is large--exactly the condition
that exists with a short-circuit. While useful in some appli-
cations, the foldover limiter can produce unusual behavior--
especially with reactive loads. See the OPA502 data sheet
for details.
RESISTIVE LOADS--DC OPERATION
Consider a power amplifier driving a resistive load. It is
tempting to check for safe operation only at maximum
output voltage and current. But this condition is not usually
the most stressful.
At maximum output voltage, the voltage across the conduct-
ing transistor, V
CE
, is at a minimum and the power dissipa-
tion is low. In fact, if the amplifier output could swing all the
way to the power supply rail, the current output would be
high, but the amplifier power dissipation would be zero
because V
CE
would be zero.
Figure 3 plots power from the power supply, load power,
and amplifier power dissipation as a function of output
voltage delivered to a resistive load. The power delivered to
the load increases with the square of the output voltage
(P = I
2
R), while the power from the power supply increases
linearly. The amplifier dissipation (equal to the difference of
the first two curves) follows a parabola. If the amplifier
output could swing all the way to the power supply rail
(dotted portion of lines), all the power from the supply
would be delivered to the load and the amplifier dissipation
would be zero.
Peak amplifier dissipation occurs at an output voltage of
(V+)/2, or 50% output. At this point, V
CE
is (V+)/2 and I
O
is
(V+)/(2R
L
). The amplifier dissipation at this worst-case
point is the product of V
CE
and I
O
, or (V+)
2
/(4R
L
). Check this
condition to assure that it is within the SOA of the amplifier.
Also be sure that you have sufficient heat sinking for the
calculated power dissipation to prevent overheating.
3
Example 2
An unbalanced power supply is often used with power
amplifiers to allow a large unipolar output voltage. A
+70V/5V power supply is used with the OPA502 to
drive a 30
load connected to ground. What is the worst
case power dissipation and SOA requirement?
Answer--
The worst case occurs at half output, where V
O
= 35V,
and V
CE
= 35V. The output current at this point would
be 35V/30
= 1.17A which is within the SOA. Power
dissipation would be 35V 1.17A = 41W.
Other points to consider: The maximum output voltage
would be approximately 65V, and 65V/30
= 2.17A.
At this point, V
CE
= 5V, a safe value on the SOA.
If the current limit were set to accommodate the full
output of 2.17A, it would not be safe for short-circuits
to ground. With a short-circuit to ground, V
CE
= 70V
where the maximum safe current is 0.4A.
PULSED OPERATION
Some applications must handle pulses of current or varying
current waveforms with a low duty-cycle. The SOA plot
sometimes shows an ability to supply larger currents for
short duration pulses. In Figure 2, the SOA limits are labeled
for 5ms, 1ms and 0.5ms pulses. The duty-cycle must be low
(approximately 5% or less), so that heating in the output
transistor is given time to dissipate.
Unusual current waveforms can be estimated with an ap-
proximation to a rectangular pulse as shown in Figure 4.
With a resistive load, the most stressful condition is when
the output voltage is approximately half the supply voltage
as shown. For other types of loads, evaluate any condition
that produces significant load current and high V
CE
. Appli-
cations which pulse currents beyond the dc SOA of the
amplifier should be evaluated very carefully since they are
pushing the limits of the device. Good reliability is achieved
by taking a conservative approach to SOA limits.
AC SIGNALS
Imagine a time-varying signal that rapidly transverses the
curves in Figure 3. The point of maximum dissipation is
passed only briefly. If the signal changes rapidly enough
(above 50Hz), the thermal time constant of the device causes
the junction temperature to be determined by the average
power dissipation. So, ac applications are generally less
demanding than dc applications of the same peak voltage
and current requirements.
If the signal is bipolar, such as a sine wave centered around
zero, each output transistor "rests" for a half-cycle. The total
amplifier dissipation is shared between the two output tran-
sistors, lowering the effective thermal resistance of the
package.
If the instantaneous peak dissipation point is within the SOA
of the amplifier, the primary concern is providing a suffi-
cient heat sink to prevent overheating. Since this peak
FIGURE 3. DC Power Dissipation, Resistive Load.
Power
Power
Amplifier
Supply
Delivered
Dissipation
Power
to Load
P
D
=
P
S
P
L
FIGURE 4. Pulsed Loads.
Supply Voltage
V
CE
= (V+)/2
Output Voltage
Waveform
Equivalent
Rectangular Pulse
Equivalent Duration
V+
0
POWER--CONTINUOUS dc
200
180
160
140
120
100
80
60
40
20
0
DC Output Voltage (% of V+ Supply)
Power for V
S
= 40V, R
L
= 8
(Watts)
0
10
20
30
40
50
60
70
80
90
100
Normalized Output
Power R
L
(V+)
2
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
P
S
Power from
Power Supply
P
L
Power Delivered
to Load
P
D
Power Dissipation
of Amplifier
Worst
Case
( )
4
condition is passed only briefly during an ac cycle, ac
applications operate reliably, closer to the SOA limit.
Figure 5 shows the power curves for a power amplifier with
40V supplies and an 8
resistive load. Again, powers are
plotted with respect to the percentage of maximum voltage
output. As with dc, the power delivered from the power
supply increases linearly with output voltage and the power
delivered to the load increases with the square of the output
voltage. The power dissipated by the amplifier, P
D
, is the
difference of the first two curves. The shape of the P
D
curve
is similar to the dc signal case, but does not approach zero
at 100% output voltage. This is because at full ac output
voltage, the output is rapidly transversing the whole curve (0
to 100%) of Figure 4. Figure 5 shows the average dissipation
of this dynamic condition.
Amplifier dissipation reaches a maximum when the peaks of
the ac output waveform are approximately 63% of the power
supply voltage. For this sine wave amplitude, the instanta-
neous output voltage hovers near the crucial half-supply-
voltage value for a large portion of the ac cycle.
The normalized values read from the right side of the curve
in Figure 5 can be scaled to any supply voltage and load
resistance. To find your amplifier dissipation at a given
signal level, multiply the reading taken from the right-side
scale by (V+)
2
/R
L
.
AC applications rarely must endure continuous operation at
the maximum dissipation point of Figure 5. An audio ampli-
fier, for instance, with voice or music typically dissipates
much less than this worst-case value, regardless of the signal
amplitude. Yet, since a continuous sine wave signal of any
amplitude is conceivable, this worst-case condition is a
useful benchmark. Depending on the application, you might
want to design for this condition.
REACTIVE LOADS--AC SIGNALS
Figure 6 shows the relationship of voltage and current in
purely inductive load. The current lags the load voltage by
90
. At peak current, the load voltage is zero. This means
that the amplifier must deliver peak current with the full V+
across the conducting transistor (V for negative half-cycle
peak current). The situation is equally severe for a capacitive
load. Check for this condition of voltage and current on the
SOA curve.
Once again, consider the curve in Figure 5. Power amplifier
dissipation is equal to the power from the power supply
minus the power delivered to the load. The power from the
power supply, P
S
, is the same whether the load impedance is
resistive or reactive. But if the load is completely reactive
(inductive or capacitive), the power delivered to the load is
zero. So the power dissipated by the amplifier is equal to the
power from the power supply. At full output this is approxi-
mately three times the worst-case amplifier dissipation with
a resistive load!
A reactive load is a very demanding case, requiring a large
heat sink compared to a resistive load. Fortunately, purely
reactive loads are rare. An ac motor, for instance, could not
be purely inductive, or it would be incapable of performing
any mechanical work.
FINDING POWER DISSIPATION
Unusual loads and signals can be challenging to evaluate.
Use the principle that amplifier power dissipation is equal to
the power from the supplies minus the load power.
Power delivered from the power supplies can be measured
as shown in Figure 7. The power from each supply is equal
to the average current times its voltage. If the output wave-
form is asymmetrical, measure and calculate the positive
FIGURE 5. AC Power Dissipation, Resistive Load.
Power
Power
Amplifier
Supply
Delivered
Dissipation
Power
to Load
P
D
=
P
S
P
L
POWER--AVERAGE ac, RESISTIVE LOAD
140
120
100
80
60
40
20
0
AC Peak Output Voltage (% of V+ Supply)
Power (W) V
S
= 40V, R
L
= 8
0
10
20
30
40
50
60
70
80
90
100
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
P
S
Power from
Power Supply
P
D
Power Dissipation
of Amplifier
P
L
Power Delivered
to Load
Worst
Case
Normalized Output
Power R
L
(V+)
2
( )
5
and negative supplies separately and add the powers. If the
waveform is symmetrical, you can measure one and multi-
ply by two. Use an average-responding meter to measure the
current. A simple D'Arsonval type meter movement with a
current shunt works well. Do not use an rms-responding
meter.
For sinusoids, finding the load power is easy--
P
LOAD
= (I
O
rms) (V
O
rms) cos(
)
Where
is the phase angle between load voltage and
current. (See Figure 8 for measurement methods.)
For complex waveforms, the load power is more difficult to
measure. You may know something about your load which
allows you to determine load power. If not, you can build a
circuit that measures load power using a multiplier IC to
FIGURE 6. Voltage and Current Waveforms for Inductive
Load.
continuously multiply load voltage and current. The average
dc output of the multiplier is proportional to the average load
power. See the MPY100 data sheet for a circuit to measure
power with a multiplier.
UNUSUAL LOADS
Usually an op amp sources current to the load (Q
1
conduct-
ing, Figure 1), when the output voltage is positive. But
depending on the type of load and the voltage to which it is
referenced, an op amp might have to sink current (Q
2
conducting) with positive output voltage. Or, it could be
required to source current with negative output voltage. In
these cases, the voltage across the conducting transistor is
larger than V+ or V.
An example of this situation is a power op amp connected as
a current source. The output of a current source might be
connected to any voltage potential within its compliance
range. Sourcing high current to a negative potential node
would produce high dissipation and require good SOA.
MOTOR LOADS
Motor loads can be tricky to evaluate. They are like a
reactive load since stored energy (mechanical) can be deliv-
ered back to the amplifier. Motor and load inertia can cause
the amplifier to dissipate very high power when speed is
changed.
Electro-mechanical systems can be modeled with electric
circuits. This is a science in itself--beyond the scope of this
discussion.
You can, however, measure the V-I demand of a motor (or
any other load) under actual load conditions. Figure 8 shows
a current sense resistor placed in series with the load. With
load voltage and current displayed on separate oscilloscope
traces, you can find the conditions of maximum stress. Be
sure to consider the voltage across the conducting transistor,
(V
CE
), not the amplifier output voltage. The most stressful
conditions may occur with moderate current, but low load
voltage.
An X-Y type display of voltage and current (Figure 8B) may
also help identify troublesome conditions. More demanding
combinations of voltage and current are those that deviate
from a straight-line resistive load.
V+ Supply
V
CE
At this instant, V
O
= 0, but I
O
at maximum.
Load
Voltage
Load
Current
V Supply
FIGURE 7. Measuring Power Supply Power.
Load
V
V+
P
D
I
1
I
2
NOTE: I
1
and I
2
are readings from D'Arsonval-type average-responding
meters. RMS-responding meters will not provide accurate results.
P
S
= (V+) I
1
+ |V| I
2
P
D
= P
S
P
L
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