The relationship between the output power of power amplifier and the maximum electricity consumption

How can the maximum power consumption of this amplifier be calculated roughly by the output power of a power amplifier?

First, what is the purpose of understanding the relationship between power output and maximum power consumption? It is only to determine the maximum power consumption of the site, and know how thick the cable should be, so that the whole sound system will always work in a safe state of electricity. How should we determine the total power consumption according to the power of power amplifier?

To understand this problem, we must first understand the composition of the total power consumption of the amplifier. The total power consumption of the power amplifier is composed of 2 parts: the output power of the amplifier and the heat loss power generated during the output of the power amplifier. According to the law of conservation of energy, we can see that the total power consumption of power amplifier = power output power + power dissipation power.

From the formula, it can be seen that when the power amplifier's total power is determined, the greater the output power of the power amplifier, the smaller the power loss power of the power amplifier, the higher the efficiency of the amplifier, and the lower the efficiency of the amplifier.

The efficiency of the power amplifier can be expressed in the following formula:

Power amplifier efficiency = power output power / power amplifier total power *100%

After making clear the composition of the total power of the amplifier, we should know the 3 power states of the amplifier.

1/8 Power, 1/3 Power, Maximum Power.

Here is the explanation for them:

1/8 Power: a typical procedure for accidental limiting. This rating can be used in most cases.

1/3 Power: a severe procedure that represents a serious limit.

Maximum Power: (sine) is a continuous sine wave drive with 1% limiter.

The power value of the above three states will be marked in the instructions of the imported power amplifier. If the 1/8 Power and 1/3 Power are not marked in the instructions of some amplifiers, only the power values under the corresponding impedance are removed to 8 and 3, the output power of 1/8 Power and 1/3 Power under the impedance can be obtained.

Next, according to the manual of LAB FP6400 power amplifier, I will explain in detail how much electricity is needed to ensure that the amplifier operates stably. The following table is contained in the manual of LAB FP6400 power amplifier.

FP6400 Power 1/3 Power 1/8 Power Idle

8 ohms 2 x 1300W 1230W 575W 105W

4 ohms 2 x 2300W 1975W 900W 105W

2 ohms 2 x 3200W 2950W 1290W 105W

We use the power amplifier, in general, the maximum use of the 2 Channel 4 ohms this one gear power, in this case, the operation of high efficiency amplifier, and relatively stable operation. So I chose this power amplifier's 2 Channel 4 ohms power as a specific example to analyze.

The above table gives the 2 Channel 4 ohms, and the 1/3 Power is 1975W, which is the power used, and the system can be multiplied to 2 for peak power when the system is distributed. So for this amplifier, the 4000 watts distribution is the best solution.

In front of the law of conservation of energy, it is known that the power amplifier's total power = power amplifier output power + power amplifier heat loss power, and then the output power, heat loss power and efficiency of the power amplifier in the 2 Channel 4 ohms, 1/3 Power.

1/3 Power (output power) =2300/3 x 2=1533W

The table shows that the 2 Channel 4 ohms, 1/3 Power (power consumption) is 1975W.

Heat loss power = power consumption output power =1975-1533=442W

Power amplifier efficiency = power output power / power amplifier total power consumption *100%=1533/1975 * 100%=77.6%

For others, "AB class, power utilization rate of about 50%, H class of about 60%-70%, D class of more than 80%" argument, personally think is not necessarily correct. The reason is that although all H class power amplifiers are available, there is still a big difference in efficiency between manufacturers. The efficiency of the LAB FP6400 listed above can reach 77.6%, but most manufacturers can not do this efficiently.

Next, take the CREST AUDIO (peak) power amplifier, which is also the world's top brand, as an example. The following table is related to some parameters of the top PRO9001 of CREST AUDIO power amplifier:

4 ohms Stereo Power:2 x 2050W

Class:H

1/3 Power Curr Draw:30A (120V, 4 ohms)

Next, let's take a look at the output power, thermal loss power and efficiency of PRO9001 in the 2 Channel 4 ohms, 1/3 Power.

1/3 Power (output power) =2050/3 x 2=1367W

1/3 Power (electric power) =30 x 120=3600W

Heat loss power = power consumption output power =3600-1376=2233W

Power amplifier efficiency = power output power / power amplifier total power consumption *100%=1376/3600 * 100%=38%

Similarly, the efficiency of H class power amplifier LAB FP6400 can reach 77%, while the efficiency of CREST AUDIO PRO9001 is only 38%. Let's take a look at the efficiency of CREST AUDIO AB power amplifier.

The following table is related to some parameters of the top PRO3301 of CREST AUDIO power amplifier:

4 ohms Stereo Power:2 x 330W

Class:AB

1/3 Power Curr Draw:9A (120V, 4 ohms)

Calculate the output power, thermal loss power and efficiency of PRO3301 in the 2 Channel 4 ohms, 1/3 Power.

1/3 Power (output power) =330/3 x 2=220W

1/3 Power (electric power) =9 x 120=1080W

Heat loss power = power consumption output power =1080-220=860W

Power amplifier efficiency = power output power / power amplifier total power consumption *100%=220/1080 * 100%=20%

Do you see the difference between the "AB class, the power utilization rate of about 50%, about 60%-70% of the H class, and more than 80% of the D class" proposed by the "trumpet killer"?

Conclusion: the method of determining the maximum power consumption at the site is based on the 1/3 Power (power consumption) of the amplifier and multiplied by 2. The next thing to do is to select the corresponding cable according to the numerical value.