SBOA022

APPLICATION BULLETIN
®
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POWER AMPLIFIER STRESS
AND POWER HANDLING LIMITATIONS
BY BRUCE TRUMP (602) 746-7347
To achieve reliable power amplifier designs you must consider the stress on the amplifier compared to its power
RI
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 Q1 and Q2 provide positive and negative
output current to the load. IOUT is shown flowing out of the
amplifier, so Q1 is supplying the output current. For positive
output current, Q2 is “off” and can be ignored.
V+
RF
Q1
+
VCE
–
VCE = (V+) –VO
IOUT
VIN
The stress on Q1 under load is related to the output current
and the voltage across Q1 (its collector-to-emitter voltage,
VCE). The product of these quantities, IOUT • VCE, is the power
dissipation of Q1. This power dissipation is one important
consideration, but the “safe operating area” provides a more
complete description of the amplifier’s limits.
Q2
VO
RL
V–
FIGURE 1. Simplified Power Op Amp Circuit.
As VCE 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 dissipation. VCE • IO is a constant 120W (at 25°C) in Figure 2. All
points on this region of the curve produce the same maximum junction temperature. Exceeding the safe output current in this region may damage the transistor junction.
SAFE OPERATING AREA
The power handling ability of a power transistor is characterized by its Safe Operating Area (SOA), Figure 2. The
SOA curve shows permissible voltage, (VCE) and current,
(IOUT). The maximum safe current is a function of VCE. The
characteristic shape of this curve has four distinct regions.
At low VCE, 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.
Max Current
SAFE OPERATING AREA
Thermal
Limits
10
ms
0.5
s
1m
TC = +85°C
t=
TC = +125°C
5m
2.0
s
IO (A)
t=
t=
TC = +25°C
5.0
Second
Breakdown
Region
1.0
0.5
Thermal Limitation
(TJ = 200°C)
0.2
Voltage
Breakdown
Second Breakdown
Limited
0.1
1
2
5
10
20
50
100
VCE |VS – VOUT| (V)
FIGURE 2. Safe Operating Area (SOA)— OPA502. (Figure 2 in PDS-1166)
©
SBOA022
1993 Burr-Brown Corporation
AB-039
Printed in U.S.A. April, 1993
As VCE 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 VCE.
Exceeding the safe output current in the second breakdown
region can produce a localized thermal runaway, destroying
the transistor.
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 calculations.)
If the op amp must survive a short-circuit to one of the power
supplies, for instance, the maximum VCE would be the total
of both supplies—a very demanding case.
Not all applications must (or can be) designed for shortcircuit 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.
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 accounts for the affect of case temperature on junction temperature. Additional lines may show the maximum safe
current for pulses of various durations according to the
thermal time constants of a device.
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 VCE is large—exactly the condition
that exists with a short-circuit. While useful in some applications, the foldover limiter can produce unusual behavior—
especially with reactive loads. See the OPA502 data sheet
for details.
The SOA curve should be interpreted as an absolute maximum rating. Operation at any point on the thermal limit
portion of the curve produces the maximum allowable junction 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).
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.
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.
At maximum output voltage, the voltage across the conducting transistor, VCE, is at a minimum and the power dissipation 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 VCE 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 = I2R), 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.
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?
Peak amplifier dissipation occurs at an output voltage of
(V+)/2, or 50% output. At this point, VCE is (V+)/2 and IO is
(V+)/(2RL). The amplifier dissipation at this worst-case
point is the product of VCE and IO, or (V+)2/(4RL). 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.
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 dissipate 120W during short-circuit. It would require an
“infinite” or ideal heat sink to maintain the case temperature at 25°C in normal room ambient conditions.
2
1.0
PS
Power from
Power Supply
0.9
0.6
100
60
0.5
PL
Power Delivered 0.4
to Load
0.3
40
0.2
80
Worst
Case
20
0.1
0
(V+)2
0.7
120
)
0.8
140
Power • RL
160
(
180
Normalized Output
Power for VS = ±40V, RL = 8Ω (Watts)
POWER—CONTINUOUS dc
200
Power
Supply
Power
Amplifier
Dissipation
PD
=
PS
Power
Delivered
to Load
–
PL
PD
Power Dissipation
of Amplifier
0
0
10
20
30
40
50
60
70
80
90
100
DC Output Voltage (% of V+ Supply)
FIGURE 3. DC Power Dissipation, Resistive Load.
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.
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 30Ω load connected to ground. What is the worst
case power dissipation and SOA requirement?
Answer—
The worst case occurs at half output, where VO = 35V,
and VCE = 35V. The output current at this point would
be 35V/30Ω = 1.17A which is within the SOA. Power
dissipation would be 35V • 1.17A = 41W.
Unusual current waveforms can be estimated with an approximation 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 VCE. Applications 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.
Other points to consider: The maximum output voltage
would be approximately 65V, and 65V/30Ω = 2.17A.
At this point, VCE = 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, VCE = 70V
where the maximum safe current is 0.4A.
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.
Supply Voltage
V+
Output Voltage
Waveform
VCE = (V+)/2
Equivalent
Rectangular Pulse
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 transistors, lowering the effective thermal resistance of the
package.
0
Equivalent Duration
If the instantaneous peak dissipation point is within the SOA
of the amplifier, the primary concern is providing a sufficient heat sink to prevent overheating. Since this peak
FIGURE 4. Pulsed Loads.
3
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.
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 8Ω 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, PD, is the
difference of the first two curves. The shape of the PD 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.
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, PS, 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 approximately three times the worst-case amplifier dissipation with
a resistive load!
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 instantaneous output voltage hovers near the crucial half-supplyvoltage value for a large portion of the ac cycle.
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.
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/RL.
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.
AC applications rarely must endure continuous operation at
the maximum dissipation point of Figure 5. An audio amplifier, 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.
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 waveform is asymmetrical, measure and calculate the positive
POWER—AVERAGE ac, RESISTIVE LOAD
0.7
PL
Power Delivered
to Load
0.6
80
0.4
Worst
Case
60
0.3
40
0.2
PD
Power Dissipation
of Amplifier
20
0
0
10
20
30
40
50
60
70
80
90
0.1
0
100
AC Peak Output Voltage (% of V+ Supply)
FIGURE 5. AC Power Dissipation, Resistive Load.
4
Power
Supply
Power
)
(V+)2
0.5
PS
Power from
Power Supply
Power • RL
100
(
120
Normalized Output
Power (W) VS = ±40V, RL = 8Ω
140
Amplifier
Dissipation
PD
=
PS
Power
Delivered
to Load
–
PL
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.
V+ Supply
Load
Voltage
VCE
UNUSUAL LOADS
Usually an op amp sources current to the load (Q1 conducting, 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 (Q2
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–.
Load
Current
V– Supply
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.
At this instant, VO = 0, but IO at maximum.
FIGURE 6. Voltage and Current Waveforms for Inductive
Load.
MOTOR LOADS
Motor loads can be tricky to evaluate. They are like a
reactive load since stored energy (mechanical) can be delivered back to the amplifier. Motor and load inertia can cause
the amplifier to dissipate very high power when speed is
changed.
and negative supplies separately and add the powers. If the
waveform is symmetrical, you can measure one and multiply 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.
Electro-mechanical systems can be modeled with electric
circuits. This is a science in itself—beyond the scope of this
discussion.
For sinusoids, finding the load power is easy—
PLOAD = (IO rms) • (VO rms) • cos(θ)
Where θ is the phase angle between load voltage and
current. (See Figure 8 for measurement methods.)
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,
(VCE), not the amplifier output voltage. The most stressful
conditions may occur with moderate current, but low load
voltage.
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
V+
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.
PS = (V+) • I1 + |V–| • I2
PD = PS – PL
I1
PD
Load
I2
V–
NOTE: I1 and I2 are readings from D'Arsonval-type average-responding
meters. RMS-responding meters will not provide accurate results.
FIGURE 7. Measuring Power Supply Power.
5
A) DUAL TRACE TIME SWEEP
B) X-Y OSCILLOSCOPE DISPLAY
CONTINUOUS AC SIGNALS
Resistive
Load
1
0
Oscilloscope Y-Input
IOUT (A)
Oscilloscope
Channel B
Load Voltage
Oscilloscope
Channel A
Load Current
MOTOR LOAD
–1
Highest
Stress
20
0
1
0
–1
Complex Reactive
Load
–20
Time Sweep
–20
0
Purely Reactive
Load (max current
at 0V)
20
Oscilloscope X-Input
VOUT (V)
VIN
Power
Amp
VOUT
Load ZL
Current Sense
Make RS << ZL
FIGURE 8. Voltage/Current Waveforms, Unusual Loads.
6
RS
I OUT = V
RS
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