Guangzhou Yun Electronic Technology Co., Ltd. sound
Tel: 18122256892
Fax: 18122256892
Mobile: 18122256892 Manager Chen
E-mail: 3119837087@qq.com
Address: Panyu District, Guangzhou City Agricultural Science Institute South Street on the 8th floor, No. 1, third floor, two
The speaker power is not the bigger the better, the best is applicable, for ordinary home users about 20 square meters of room, the real sense of the 60W power (referring to the speaker's effective output power of 30W x 2) is sufficient, but Amplifier reserves the power the better, preferably the actual output power of more than 2 times.
First, the sound pressure level calculation formula
The following formula is the sound field, in the r distance, a point of the sound pressure level is more convenient formula:
SPL = PWL + 10log (Q / (4 * 3.14 * r * r) + 4 / R),
Where, SPL: sound pressure level at r distance point, unit (dB), sound pressure level reference is 0dB = 20μPa.
PWL: sound power level of the sound source, unit (dB), sound power level reference 0dB = 10-12W. Sound power refers to the sound power output on the speaker, not the amplifier output power to the speaker. The sound power level is the sound power expressed in decibels. Therefore, the sound power 1W is the sound power level 120dB. Q: The directivity of the sound source.
① When the sound source in the middle of the room, surrounded by the edge, the sound energy to the sphere of radiation, pointing coefficient is 1.
② sound source on the ground to the hemisphere surface radiation, pointing coefficient is 2.
③ placed in the two walls to cross the rib to 1/4 spherical radiation, the directional coefficient of 4.
④ placed in the corner of the room to 1/8 spherical radiation, the directional coefficient of 8. r: distance from the sound source in m, R: room constant in m2, S: indoor surface area in m2, a: average sound absorption in the room, without unit.
R represents the room constant sound processing capacity of the room, and the wall area and sound absorption capacity of the room, the larger ＆#118alue of this room. Is the average sound absorption rate, a constant less than one. As the classroom so-called sound more active room, about 0.25; and as bedroom lack of reverberation room, good sound absorption, it is about 0.35. In the formula, Q / 4πr2 on the left side of parentheses indicates the sound pressure of the direct sound, which is inversely proportional to the square of the distance. The 4 / R on the right is the sound pressure of the reflected sound associated with the room. The sum of these two terms is the sound pressure generated by the sound source at that point, which is usually expressed in decibels.
Since the sound source is usually set at 0 dB, all other points are negative. As long as you know the amount of the above formula, can be substituted into this type of operation.
Second, the sound pressure level attenuation curve solution
When the A point is Q = 4, R = 100, the distance of 3m at the sound pressure level attenuation of the solution process:
Find the horizontal line with Q = 4 at the pointing coefficient, rightward to the intersection with the 3m diagonal, and then vertically up to find the intersection with the curve of the room constant R = 100. This is the A point, and then horizontally to the left along the extension of the coordinates of the intersection point, you can read the sound pressure level attenuation of 11dB.
A more practical example is to use the back of the formula and the curve two ways to solve. There is a long 5.4m, 3.6m wide, 2.5m high room, the average wall absorption rate = 0.3, if the sound on the front of the wall in the room, with a speaker stand, close to the back wall, the audience away from the speaker 3m. Ask the sound pressure level at this time attenuation.
First of all, no matter which solution must first calculate the room constant R. The six surface area of ＆#8203;＆#8203;the room:
First look at the functional solution process: where the left one is Q / 4πr2 = 2 / 4π32 = 0.0177, the other on the right 4 / R = 4/36 = 0.125. Therefore, the attenuation of sound pressure level is 10log (0.0177 + 0.125) = -8.5dB. The graphic process is as follows: Estimate the position of R = 36 in the space of R = 20 ~ 50 from Q = 2 line → r = 3 line, mark a bit "B" and then read the attenuation on the ordinate Yes, it's more than -8dB.
Third, the speaker's output power estimation
Above the calculated attenuation also need to add the sound power level PWL can be the point of the sound pressure level. So you also have to know how to calculate the sound power. Sound power is from the speaker output port count, so first of all to understand the speaker sensitivity or [output sound pressure level] concept.
The sensitivity of the loudspeaker is defined as 1W of electrical power input measured at 1m at the sound pressure level in dB / W / m. This is an inherent characteristic of the speaker, and when this ＆#118alue is 92 dB / W / m, the electroacoustic conversion efficiency of the speaker is 1%. That is, the amplifier needs to output 100W electric power to get 1W sound power. If the sensitivity is less than 92 dB / W / m, the 100 W electric power will also change less than 1 W sound power, and the difference will be doubled by 3 dB. For example, a speaker with 89 dB / W / m sensitivity requires 200W of electrical power to get 1W of acoustic power, while a 86 dB / W / m of speaker requires up to 400W of electrical power to achieve 1W of acoustic power.
Using these relationships, the electrical power can be converted to sound power PWL, PWL = 10 log (sound power output W / 10-12 W), sound power 1 W = 120 dB sound power level. For speakers with a sensitivity of 92dB / W / m, the PWL when driven with 100W electric power is 120dB. In fact, in actual use, there is no need to consider the concept of acoustic power, can be directly input to the speaker's power and speaker sensitivity to calculate PWL.
For example, there is a single 2A3 stereo amplifier, 3.5W per channel, totaling 7W, speaker sensitivity of 92dB / W / m, then PWL how big? 3 meters away from the sound pressure level is how much? Consider, with a sensitivity of 92 speakers, the total sound power of the two speakers only 7/100 = 0.07W, expressed in decibels PWL is 120 +10 log0.07 = 120-11.55 = 108.45dB, or directly to calculate the more simple power PWL = 10log (7 / 10-12) = 108.45dB, of course, this result is also for the specific speaker sensitivity of 92 dB / W / m. Listening to 3m of that room, plus the above-calculated sound pressure level attenuation of -8.5dB, the actual sound pressure level SPL = 108.45-8.5≈100dB is obtained.
If the sensitivity of the speaker is not 92 dB / W / m, just add the difference between the actual speaker sensitivity and 92 dB / W / m. With multiples or the actual level to be very troublesome, as long as decibel plus plus minus can be. For example, when the speaker sensitivity is only 88 dB / W / m, then the sound power level is 108.45 + (88-92) = 104.45 dB, and if the speaker sensitivity is 94 dB / W / m, the PWL is 108.45 + The sound pressure level at (94-92) = 110.45.3 m accordingly becomes 96 dB and 102 dB.
The sound pressure level can also be measured by a noise meter or a sound pressure gauge, but the measured ＆#118alue is the average level of the music, which is 10 db smaller than the peak level of the music. That is instructed by the instrument when the music level is 88dB, the actual level of the music peak level 98dB. In other words, you hear the right sound at the music scene, measured with the instrument as 88dB, then the audio playback system according to the audience in the 98dB sound pressure level to design.
So far, we know how to calculate the sound pressure level at some point in the light of the obvious conditions. It is unclear how certain parameters are obtained or how they are set. In addition, for audio and home theater enthusiasts it is even more important how to design the power amplifier's power, the choice of the speaker's sensitivity and so on according to the required sound pressure level.
S = 2 × (5.4 × 3.6 + 5.4 × 2.5 + 3.6 × 2.5) = 83.88≈84m2, R = 084 × 0.3 / (1-0.3) = 36. Because there is a speaker shelf, so the speaker is not close to the wall edge of the wall, only a wall desirable Q = 2, if the speaker shelf is low or no speaker shelf can take a ＆#118alue in the middle of 2 to 4, the problem in 2 dollars.
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