The limits of power amplifier use

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The limits of power amplifier use [Copy link]

In order to achieve the reliability design of the power amplifier, the amplifier's tolerance must be considered. The power range limit is determined by the safe operating area (SOA) curve of the power amplifier. The amplifier's tolerance depends on the amplifier's load and signal status.

Figure 1 shows a simplified power operational amplifier, with output transistors Q1 _and Q2 _providing positive and negative output currents to the load. _IOUT represents the current flowing out of the amplifier, so Q1 _supplies the output current. For positive output current, Q2 _is off and can be omitted.

When Q ₁ is loaded, its handling is a function of the output current and the voltage across Q ₁ (its collector-emitter voltage, V _{CE ). The product of these two quantities, I}_OUT · V _{CE ,} is the power dissipated in Q _1. This power dissipation is an important consideration, but the “ safe operating area ” provides a more complete description of an amplifier’s limitations.

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Safe working area

The power application range of a power transistor is determined by its safe operating area (SOA) (see Figure 2). The SOA curve shows the allowable voltage (V _CE ) and current (I _OUT ), and the maximum safe current is a function of V _{CE . When V}_CE is low, a larger output current can be delivered to the load. In this area, if the maximum current is exceeded, the chip may be overloaded and the device may be damaged. As V _CE increases, the power consumption of the transistor also increases until the junction temperature rises to its maximum safe value. All points along this thermally limited area (dashed line) produce the same power consumption. In Figure 2, V _CE · I _O is a constant of 120W (at 25 ℃ ), and all points on this curve in this area produce the same maximum junction temperature. Exceeding the safe current in this area may damage the transistor.

When V _CE is further increased , the safe output current drops faster beyond the thermal limit region. This so-called second breakdown region is a characteristic of bipolar transistors. It is caused by " local overheating " in bipolar transistors . In the second breakdown region, exceeding the safe output current will produce local thermal runaway, thereby damaging the transistor.

The ultimate limit is the breakdown voltage of the transistor, which cannot exceed this maximum supply voltage. Usually the SOA curve is a curve that shows how the safe output current changes with the case temperature, which shows that the case temperature has an effect on the junction temperature. Other curves show the maximum safe current for pulses of various durations based on the thermal time constant of the device. The SOA curve should be understood as the absolute maximum range. Operating at any point in the thermally limited section of the curve will produce the maximum allowable junction temperature (a state that is not recommended for long-term operation).

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Although operating above the secondary breakdown region of the curve only produces lower temperatures, this line is still an absolute maximum and operating below this line will provide better reliability (i.e. better mean time to failure - MTTF).

Heat dissipation

In addition to ensuring that you do not exceed the safe operating area of the power amplifier, you must also ensure that the amplifier does not overheat. In order to provide an adequate heat sink, you must determine the maximum power dissipation. The following details the methods and issues that affect the SOA power consumption and heat sink requirements.

Short Circuit

Some amplifier applications must be designed to withstand a short circuit to ground. This forces the full supply voltage (either V ₊ or V- ₎ to be applied across the conducting output transistor, and the amplifier immediately enters a current cutoff state. To withstand this condition, a power op amp with adjustable current limit must be controlled to a safe level.

When the power supply of the OPA502 (Figure 2) is ± 40V, what should be the maximum current limit value to protect against short-circuit to ground?

If the case temperature is maintained at 25 °C , the current limit should be a maximum of 3A. If the case temperature is maintained at 85 °C , a current limit of 2A will be safe, the power dissipation will be 80W, and a 0.75 °C /W heat sink can be used. For example, if the op amp must withstand a short circuit to one supply, the maximum VCE will be the sum of the two supplies.

It is generally considered that it is not necessary to design for short-circuit protection for all applications, but for power amplifiers, this is a critical condition. Auxiliary means such as fuses or circuits that sense fault conditions can guarantee the time the amplifier must withstand a short circuit. This can greatly reduce the requirements for the heat sink.

Resistive load

When examining a power amplifier driving a resistive load, people only check safety at maximum output voltage and current, but this state is not always its maximum tolerance.

_{At maximum output voltage, the voltage V CE} across the conducting transistor is at its minimum value, and power consumption is at its lowest. In fact, if the amplifier output can follow the power curve, the output current can become very large, but the amplifier power consumption will be zero because V _CE is zero.

Figure 3 depicts the power from the source, the power to the load, and the amplifier power dissipation as a function of the output voltage with a resistive load. The power delivered to the load increases with the square of the output voltage (*P=V _O² /R), while the power from the source increases linearly and the amplifier losses vary along a parabola. If the amplifier output could always follow the trajectory of the source (dashed line), the full power of the source would be delivered to the load, and the amplifier power would be zero.

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The peak power dissipation of the amplifier occurs at an output voltage of V ₊ / ² or 50% of the output, at which point V _CE is V ₊ / ² and I _O is V ₊ / 2R _L . The power dissipation of the amplifier at this worst-case point is the product of V _CE and I _O , or (V ₊ ) ² / (4R _L ). Check this condition to ensure that it is within the safe operating area (SOA) of the amplifier. Also ensure that there is adequate heat sinking for the calculated power dissipation to prevent overheating.

Pulse application

Some applications must handle current pulses or changing current waveforms with low duty cycles. SOA curves sometimes show the ability to deliver high currents for short duration pulses. SOA limits for 5ms, 1ms, and 0.5ms pulses are shown in Figure 2. The duty cycle must be low (about 5% or less) to give time for the heat on the output transistors to dissipate.

Use an approximation to a rectangular pulse to estimate abnormal current waveforms, as shown in Figure 4. For resistive loads, the condition with maximum load is when the output voltage is approximately half of the supply voltage shown. For other types of loads, evaluate any condition that produces significant load current and high V _CE . Evaluate applications where the pulse current exceeds the amplifier's DC SOA range with particular care because they are close to the device's limiting values. Good reliability is achieved by choosing a constant value close to the SOA limit.

AC signal

Imagine a time-varying signal that rapidly crosses the curve in Figure 3, only briefly passing through the point of maximum power dissipation. If the signal changes quickly enough (more than 50 Hz), the junction temperature caused by the device's thermal time constant is dominated by the average power dissipation. Therefore, an AC application will generally require less power than a DC application of the same peak voltage and current.

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If the signal is bidirectional, such as a sine wave centered about zero, each output transistor "rests" for half a cycle, and the total amplifier power dissipation is evenly divided between the two output transistors, while reducing the effective package thermal resistance.

If the instantaneous peak loss point is within the amplifier's SOA, the first concern is to provide a large enough heat sink to prevent overheating. Since this peak condition only occurs briefly during an AC cycle, AC applications can operate reliably closer to the SOA limit.

Figure 5 shows the power curve for a power amplifier with ± 40V supplies and an 8Ω resistive load. In addition, the power is plotted relative to the percentage of the maximum voltage output. As in the DC case, the power supplied by the supply increases linearly with the output voltage, and the power supplied to the load increases with the square of the output voltage. The power consumed by the amplifier, PD _, is the difference between the first two curves. The shape of the _PD curve is similar to that of the DC signal, but it cannot approach zero at 100% output voltage. This is because at full AC output voltage, the output quickly sweeps across the entire curve of Figure 4 (0 to 100%). Figure 5 shows the average loss in this dynamic state.

The amplifier's losses are at their maximum when the peak of the AC output waveform is approximately 63% of the supply voltage. For this sine wave amplitude, the instantaneous output voltage is at a critical value close to half the supply voltage for most of the AC cycle.

For any supply voltage and load resistance, you can use the normalized values indicated on the right side of the curve in Figure 5. To find the loss of your amplifier at a given signal level, multiply the reading from the right scale by (V+) 2 ^/ R _L.

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AC applications rarely have to experience continuous operation at the maximum loss point of Figure 5. For example, an audio amplifier with speech or music will generally have losses much less than this worst-case value, regardless of the signal amplitude. Since a continuous sine wave signal of arbitrary amplitude is possible, this worst-case condition is a useful benchmark. Depending on the application, you may need to design for this condition.

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Reactive load-AC signal

Figure 6 shows the relationship between voltage and current in a purely inductive load. The current lags the load voltage by 90 ° , and the load voltage is zero when the current is at its peak. This means that the amplifier must provide the peak current when the voltage across the turn-on transistor is full scale V _{+ (V}_- for the negative half cycle of the peak current ). This is also severe for capacitive loads, check the voltage and current on the SOA curve in this state.

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Revisiting the curves in Figure 5, the power amplifier losses equal the power drawn from the supply minus the power delivered to the load. The power drawn from the supply, PS, is the same whether the load impedance is resistive or reactive. However, if the load is purely reactive (inductive or capacitive), the power delivered to the load is zero. Therefore, the power consumed by the amplifier is equal to the power drawn from the supply, which is about three times the worst-case losses of an amplifier with a resistive load at full output.

Reactive loads are very lossy and require a large heat sink compared to resistive loads. Fortunately, purely reactive loads are rare. For example, an AC motor cannot be purely inductive or it would not do any mechanical work.

Power loss

Evaluating unique loads and signals can be complicated. Using the principle that the power dissipated by an amplifier is equal to the power drawn by the supply minus the power drawn by the load, the power delivered by the 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 and negative supplies separately and add the two powers. If the waveform is symmetrical, you can measure once and multiply by 2. Use an average-responding meter to measure the current. A simple D'Arsonval-type meter with a current shunt arrangement works well. Do not use an rms-responding meter.

For a sinusoidal signal, it is easy to find the power in the load:

PLOAD＝(I _o rms) · (V _o rms) · cos( θ )

Where θ is the phase angle between the load voltage and current (see measurement method Figure 8).

For complex waveforms, load power is more difficult to measure. You may know something about the load to determine the load power. Otherwise, you can use a multiplier integrated circuit to build a circuit to measure load power by multiplying the voltage and current in sequence. The average DC output of the multiplier is proportional to the average load power.

Unique Loads

Normally, when the op amp's output is positive, it sources current to the load (Q1 _is on, Figure 1). Depending on the loads and voltages involved, the op amp may have to sink current (Q2 is on) for a positive output _, or it may be required to source current for a negative output voltage. In these cases, the voltage across the turn-on transistor is greater than either V ₊ or _V- .

An example of this is a power operational amplifier used as a current source. Within the compliance range of the current source, its output can be connected to any voltage potential. When a large current flows to a negative potential node, large losses may occur, requiring a good SOA.

Motor load

Motor loads can be tricky to evaluate because they can return stored energy (mechanical energy) to the amplifier, so they act much like a resistive load. The inertia of the motor and load can cause the amplifier to dissipate significant power as speed changes.

Mechatronic systems can be modeled using circuits, which is a discipline in itself (and beyond the scope of this article).

However, you can measure the VI consumption of a motor (or any other load) under significant load conditions. Figure 8 shows a current sense resistor connected in series with the load. Using the load voltage and current displayed on the oscilloscope sweeps, you can find the maximum load condition. Be sure to look at the voltage across the pass transistor (VCE), not the amplifier output voltage. Maximum load conditions may occur at moderate currents, but with lower load voltages.

An XY display of voltage and current (Figure 8B) can also help identify troublesome conditions. The largest power dissipations of the voltage and current combination are those that deviate from a linear resistive load.