Technical support

The problem of maximum power consumption of power amplifier

In the project, the customer needs us to provide the maximum power consumption of the power load, and then compare the maximum power consumption with the power load power reserved for the audio part of the project, or require Party A to provide the power load power.

Currently circulating the proposed method of measurement, that is, the two-channel 8 ohm power superposition divided by the efficiency of the amplification type, this method is often inaccurate, because there are two important parts for the efficiency of the amplifier, the amplification part and the power supply part, which work together to form a power factor (working efficiency). And the same amplifier type due to the power supply design and amplification design differences, resulting in completely different efficiency, such as the maximum power consumption calculated by this method and the actual maximum power consumption difference is very large. It seems to be a very simple problem, but how to accurately estimate the maximum power consumption of the amplifier? It should be a problem for most engineers, although the audio source and signal processing part is marked by a fixed power consumption (relative to the load and working state of the power supply part is stable, the conversion efficiency and loss are stable), but the high-power conversion equipment such as the power amplifier does not have a fixed nominal maximum power consumption (the driving load is not fixed, the signal component is always changing state, and the power supply is stable. And the magnification is also unstable), many reasons lead us to the maximum power consumption of the power amplifier part is not clear, then there is a way to assist us to carry out a relatively accurate assessment of the maximum power consumption of each amplifier? I can tell you that the following methods can be simple and effective to measure the maximum power of any amplifier products in practical applications.

First, a power strip is modified, disconnect the live wire end, connect it to a high-precision digital multimeter in series, and use a 10A or 20A current file for measurement (remember that the current file does not measure voltage, otherwise the consequences will be responsible), if the power amplifier and other equipment are connected, the current reading of the multimeter is the total current of the current device. The total power (i.e. apparent power) can be obtained by calculating the voltage and current.

The first thing to know: The measured total power (apparent power) includes active power and reactive power, active power is the actual loss of power in the amplification and conversion process of power amplifier equipment, reactive power is the power of inductor or capacitor components in the power amplifier power circuit and 220V AC power supply reciprocating exchange (in the AC circuit, all the inductive or capacitive components, After the power is applied, the magnetic field of the inductor or the electric field between the capacitor plates will be established, so that during the first half of the "instantaneous power positive" time of each alternating current cycle, they will absorb energy from the power supply to create a magnetic field or electric field; The second half of the "instantaneous power is negative" time, its established magnetic or electric field energy is returned to the power supply. Therefore, in the whole cycle, the average value of this power is equal to zero, so the energy of the power supply and the magnetic field energy or electric field energy in a reversible energy conversion, without consuming power), reactive power is the basis of active power work, reactive power is equivalent to the bucket, active power is equivalent to water, no bucket can not transport water, And the total weight in the transportation process is the total weight of water and water, so the maximum total power consumption (apparent power) we measured is the power that the site needs to provide us. This is the biggest difference in power loss between linear and nonlinear loads in AC and DC circuits.

The second point to understand: What is the Power factor (PF)? The power factor parameter is the result obtained by dividing the active power rate (actual loss power) by the apparent power of the line. The closer the active power is to the apparent power, the higher the utilization rate of the power supply is, that is, the higher the PF power factor is. When the power factor is up to 1, the active power and apparent power are equal. This situation exists in the direct current circuit of linear load, but it is difficult to achieve in the alternating current circuit, and the circuit needs to be compensated, which fully explains why the PFC power factor of the power amplifier is a very key parameter for the efficiency of the power amplifier. Continuing with the analogy of the relationship between water and the bucket, the higher the PFC power factor, the lighter the bucket, and the less force is required in the process of transporting water. On the contrary, the lower the PFC power factor, the heavier the bucket, the greater the force required in the process of transporting water, and the more wasted work.

The third knowledge point that needs to be understood: In which power output and load state the power consumption of the amplifier is the largest? The normal power amplifier is based on 1KHz sine wave signal input, and the maximum nominal output power of one-fourth of the maximum nominal output power under 4 ohMs load is the maximum power consumption of the power amplifier. After demonstration and measurement, In the case of 20Hz-20kHz pink noise signal input, the power consumption of 4 ohm at full load (the panel peak lamp is in a shining state) is equal to one quarter of the nominal maximum output power of 4 Ohm load in the case of 1kHz sine wave signal input, so the maximum power consumption of many manufacturers is defined in this way. The maximum active power consumption of POWERSOFT amplifiers is labeled in this way.

By understanding the above three knowledge points, I believe that you can not only accurately measure the maximum power consumption in the practical application of any type of power amplifier. It can also be argued to understand why POWERSOFT's power factor is 0.95, and what is good for the user?

• Demonstration 1. Take POWERSOFT K3 power amplifier as an example to measure the maximum apparent power in the application environment.

  Answer: Input signal 1KHz sine wave signal, drive dual-channel 4 euro linear load, running in a quarter of 4 euro maximum output power (650W that is 51V) state, the current in the measured power supply line is 7.163A, Apparent power = total current X voltage =7.163A X 218.6V =1565.83VA, this value is basically consistent with the specification specification, the error part is the voltage difference, the specification takes 230V voltage as the standard, the actual measurement of the ambient voltage is 218.6V, so slightly produce a small error. The apparent power of 1565.83VA is the power rate of a single power amplifier in daily application, remember that the unit of this parameter is VA is not W (we can think of it as power W to facilitate communication with Party A), the power includes reactive power and active power, multiple power amplifiers can superposition the apparent power of a single power supply. The power value after stacking is increased by about 30% and provided to the user to facilitate future equipment additions and safe operation needs.

• Demonstration 2, POWERSOFT K3 power amplifier in the application environment at the maximum apparent power of the active power is what?

  Answer: POWERSOFT K3 Specification nominal (PFC) power factor is 0.95, so active power = apparent power X PFC (power factor) =1565.83VA X 0.95=1487.5W, It is calculated that the actual active power consumed by POWERSOFT K3 amplifier in the process of energy conversion is 1487.5W.

• Demonstration 3. What is the apparent power and active power of the Class 3 H amplifier of the analog power supply under the same working state?

  Answer: The same 4 OHM dual-channel linear load, the input signal is 1KHz sine wave signal, the output is 650W (51V) state, the current in the measured power line is 14.24A, the apparent power = total current X voltage =14.24A X 215.7V =3071.57VA, The same load and output power, but for the power requirements of the field power supply, the power demand of the class 3 H power amplifier of the analog power supply is twice that of the POWERSOFT K3 power amplifier, which fully shows that if the use of POWERSOFT K3 power amplifier not only reduces the volume, but also reduces the volume of the power amplifier. And for the site power load is half and twice the work.

Measured by the instrument, the power factor of the third-level H-type power amplifier of the analog power supply is 0.773. It can be calculated that the active power of the power amplifier = apparent power X PFC (power factor) =3071.57VA X 0.773=2374.32W. It is calculated that the actual active power consumed in the process of energy conversion is 2374.32W.

This means that under the same load and output state, POWERSOFT K3 is twice as small as the level 3 class H power amplifier of the analog power supply for the power supply of the environment, saving about 1 degree of power per hour (about 1 yuan) for a single power amplifier, and multiple power amplifiers can save a lot of electricity every year.

Through the above knowledge points, first, the technical team can fully understand how to obtain the power supply demand of the use environment according to the maximum power consumption of the actual power amplifier equipment, so as to avoid power waste and increase the cost of Party A's infrastructure. Second, the business team can fully understand the role of PFC and if the user understands the advantages of POWERSOFT digital Class D power amplifier in the application through technical means, it is conducive to the promotion of power amplifier applications.