Output Power of Audio Amplifier

Output power of audio amplifier

The output power is a important characteristic for audio amplifiers.

Introduction

Amplifiers have a specific power, which is indicated in Watts. The power is the product of current and voltage. In a loudspeaker the supplied electrical power is converted almost exclusively in heat. Only a fraction of the electrical power can be converted from loudspeakers into sound energy. This relationship describes the efficiency, which is indicated in per cent. Without proving it, e.g. a sound power of 30 Watts would cause an indescribable volume.

With hifi the amplifier power output is measured by alternating voltages. Measuring with DC voltage everything would be simpler. Measuring with DC voltages is not possible. Most audio amplifier does not transfer DC voltages. The background for it is that a woofer coil can be easily destroyed by DC voltage, therefore is in such a way designed the amplifiers that DC voltage is not amplified.

Meaning of the output power for an audio amplifier

How much output power needed, in addition a few questions:

How loud do you want to hear?
How large is s the efficiency of your loudspeaker?
How large is your room and how far far away sits you from the loudspeakers?
How does the distortion factor of your amplifier with increasing volume change with your loudspeaker?
How well can your amplifier dissipate the developing heat to the ambience?

Can you answer these questions clearly? Probably you do not know actually needed power output. I also not. It is perfectly indifferently how much power output an hifi amplifier has. The amp must supply only the desired volume in the desired quality with your loudspeakers. It is technically possible to build an amplifier with only 30 Watts that is better than a 200 Watt amplifier with same low volume level. With high loads, which demand more than rated output, to which "small one" naturally throws up the sponge, might be logical. The question of which amplifiers the better one is, that with 30 Watts or with 200 Watts, it is not a question of the power, but alone a question of the construction, specification and the electrical special equipment (e.g. protective circuits for amplifiers and loudspeakers).

What would you have rather?

A formula 1 engine with 800 HP or a modern marine diesel with 800 HP?

I rather take a modern marine diesel.

Why? The marine diesel has a torque plot, of it dreams the formula 1 engine at night. The marine diesel has a long service life and a full power firmness, of it dreams the formula 1 engine during whole race.

In the comparison, many amplifiers must be built after economical considerations. The placement robot in the assembly hangar, it must be constant in motion. In mass production the number of elements per equipment is to be small. Expensive elements are not applicable in many cases from cost reasons. The average customer's request required for payable achievement, understandably orients itself mass production to it. A good amplifier with high power and low manufacturing costs to build is a tightrope act.

In addition the elements must be often loaded to to their absolute maximum ratings. The manufacturer calculates by right into his development also that the user demands the maximum power only most rarely. In other words: many of the constructions are not appropriate for continuous stress under full load. Some cooling systems are not appropriate for the rated output. The transistors and thus the heat sinks, become very hot (try it). That heats the interior and other construction units e.g. electrolyte condensers, which lose on life span. How much per cent of the maximum power takes a music signal up at all, is entitled other question, whose answer extends the life of most amplifiers clearly

End of the discussion: the answer is that most amplifiers for a realistic medium volume music pleasure are sufficiently cooled, in the long term however only few are appropriate for the indicated rated output. The full power-firm well cooled 30 Watt amplifier achieves a similar size and still higher costs than the 200 Watts paper tiger.

The smaller need according to power made it possible to built better circuits. However only few customers would buy the small expensive exclusively developed amplifier. Therefore a majority of the industrie does not build this amplifier at all, only for enthusiasts. I want to never say thereby the industry build bad amplifiers, the industry build very good amplifiers in a very good price-power relationship. Somebody who is cocksure, nothing more leaves itself to improve, has only little knowledge.

Mathematics and Power

The mean power, the root mean square voltage and the root mean square current can be computed. T is the period duration. The RMS current from the temporally square average value, produces the same heat in an resistor as a equivalent large direct current. The effective power can be computed for Ohm's loads also in such a way:

These are valid the equations for the computation of the power of an amplifier, for Ohm's loads.

Der Widerstand "R" wird mit einer Leistung von 6,25 Watt beheizt. Dies entspricht genau dem linearen Mittelwert "meanpower" des zeitlichen Verlaufs der Leistung "power(t)". Von "power(t)" nochmal den Effektivwert zu bilden wäre falsch, da für Leistungen keine Effektivewerte existieren, sondern nur Mittelwerte. Die Berechnung eines Effektivwertes ergibt nur einen Sinn für Ströme und Spannungen, nicht für Leistungen. Die oft benutzte Bezeichnung "RMSpower" entspricht genau genommen dem zeitlich linearen Mittelwert der Leistung "meanpower"

The "R" is heated with an output of 6,25 Watts. This corresponds exactly to the linear mean value "meanpower" the course of "power(t)". To form the rms of "power(t)" again, would be wrong, since for power exists no effective values, but only for mean values . It's only wise to compute an rms for currents and voltages, not for power. The often used designation "RMSpower" corresponds exactly taken to the temporally linear mean value of the "meanpower".

The equation on the left side results in physically no sense. RMS can be formed only for currents or voltages, not for power.

Important equation (right) for sinusoidal voltages or currents, which simplifies the computations with Ohm's loads.

How the amplifiers power can be measured?

You must solve this equation it are generally valid for all wave forms of the input voltage and for complex loads.

Suitable measuring instruments and frequent measuring methods:

Power meter, special designed device, indicates directly RMS power.
Digital Storage Oscilloscope e.g. TDS 5054 with the possibility of algebraic operations. Two channels are used. Channel 1 measures the amplifiers output voltage by using a probe. Channel 2 measure the current with a sufficient fast current probe e.g. Tektronix P6042 , P6019 / 134, or AM503. Channel 1 is to be multiplied by channel 2 in the computer of the oscilloscope and for this product is during one period the integral to be formed. Modern digital memory of oscilloscopes controls this operation with few keypress. If no current probes are available, a shunt resistance can be used for the current measurement. The voltage drop at the shunt resistance in series with the load resistance can be measured.
Analogue Oscilloscope e.g.. Tektronix 475, 2465B or 2445A, require another measurement. As input signal for the amplifier a sinusoidal signal should be used and after possibility a purely Ohm's load. Measure the output voltage and note the amplitude. Divide it by root 2, thus get you the rms of the voltage. The temporally linear mean value of the achievement is then (rms of the voltage)²/ resistance. The current channel can be used also for the computation (p=i²*R).
Digital meter with "True RMS" e.g.. HP 3457A, HP 34401A. They measure in many cases either the True RMS voltage or the True RMS current. In the case of use applies for a Ohm's load: the temporally linear mean value of the power is then (rms of the voltage)²/ resistance. The current channel can be used also for the measurement (p=i²*R). A view into the operating instructions of the equipment is recommended, since they are not always suitable for all measuring frequencies. It recommends also here to use a sinusoidal testsignal. This is recommended in principle for all True RMS meters, even if these can determine the RMS of values still correctly with not sinusoidal voltages.
Digital meter without "True RMS" (often cheap small devices). They are calibrated in their AC ranges on sinusoidal voltages and currents. The use of sine wave voltages is compellingly necessary, in order to read off without correction the RMS voltage.
Moving iron movements adjust themselves with sufficient inertia instrument movements to the rms of the AC current. The attraction force of the metal parts hangs off of the square of the magnetic induction and depends thus on the square of the excited current.
Moving coil movements e.g. the Ballantine 3015A or HP 410C. They indicate the linear average value, as soon as they cannot follow the frequency mechanically any longer. For pure AC measurements is the average value zero. Therefore this instruments type e.g. internally an active full bridge rectifier becomes upstream. The instrument shows then the rectifying value. This stands with sinusoidal voltages however in firm relation to the rms. The scale of the measuring instrument scaled on sinusoidal sizes and is valid also only for these. For other signal forms e.g. conversion factors can triangle, rectangle etc. be computed.
Special Instruments with "high crest factor RMS to DC converter" e.g. HP 3400 or HP 3403C. Es sind konventionelle Drehspulmeßwerke oder digitale Anzeigen, denen ein besonderer Effektivwertumformer vorgeschaltet ist. There are conventional moving coils or digital meter, which are upstream a special rms to DC converter with high crest factor abilities.