功放放大倍数电压= w=2⼀8,这个怎么算啊 w=500,v=63.25,怎么得出来的?
推荐回答（1个）
主题：从电工的角度看音响，论音箱是如何被烧的？功放和音箱功率比2:1
已知：音箱额定功率：500W，阻抗8 峰值功率：2000W
功放的额定功率1000W*2 8 输入灵敏度电压0.775V
1500W*2 4
证明：1、音箱的额定输入电压 P=U2/R U＝ = =63.25V
2、音箱的额定电流I=U/R=63.25/8=7.9A
3、功放的额定输出电压P=U2/R U= = =89.44V
4、功放的放大倍数：89.44/0.775＝115倍
功放输入灵敏度电压换算成dBu＝20LG（0.775V/0.775V）=0dBu
现在就把实际工作当中的连接图画出来：
file:///C:/Users/ADMINI~1/AppData/Local/Temp/msohtml1/01/clip_image011.gif file:///C:/Users/ADMINI~1/AppData/Local/Temp/msohtml1/01/clip_image012.gif file:///C:/Users/ADMINI~1/AppData/Local/Temp/msohtml1/01/clip_image014.gif file:///C:/Users/ADMINI~1/AppData/Local/Temp/msohtml1/01/clip_image015.gif file:///C:/Users/ADMINI~1/AppData/Local/Temp/msohtml1/01/clip_image016.gif 首先我们会把调音台的PFL按下，拧GAIN按钮，直到电平表显示在“0dBu”停下，增益就调好了，接着把PFL再按下，让它弹起来，我们把方推子推到0dB位置。再把主推子推到0dB位置，主电平就显示“0dBu”电平给下级处理器，我们再把处理器的输入，输出增益设置到0dB，处理器就输出“0dB”的电平功放，再把功放的增益拧到0dB，就把0dB的电平给功放内部的放大电路（放大倍数＝89.44/6.775＝115倍）放大，功放就输出89.44V的电压给音箱（音箱的额定电压63.25V ，额定电流为7.9A但此时功放输出铃音箱的电压为89.44V.此时音箱音圈流过的电流为89.44/8=11A），由上面已知音箱的功率和阻抗求得音箱的额定电压为63.25V，额定电流为7.9A，那说明此音箱的音圈长期允许通过的电流为7.9A，音箱音圈耐压63.25V而不致于损坏。要想限制功放输出电压为63.25V，那就要限制功放的输入电压了，输入电压＝63.25V/115倍＝0.55V。把0.55V换算成dBu＝20LG（0.55V/0.775V）＝-3dBu。要想把前线的输出电压限制到0.55V功放输入，把调音台的分推子，主推子，处理器的输入、输出增益，功放的增益旋钮其中一个打到-3DB位置就可以保证音箱长期工作了（功放的输出电压不大于63.25V，流过音箱音圈的电流不大于7.9A）.
音箱是如何被烧的？
如果把调音台方推子，主推子推到0dB，处理器输入，输出增益设置在0dB，功放增益旋钮扭到0dB，功放就输出0.775*115=89.44V 的电压给音箱，使音箱音圈过电压工作，这个电压使音箱的音圈工作在11A的过电流状态下，当然在这样危险的状态下短时间工作不会烧音箱音圈，但长时间工作会让音箱音圈损坏留下伏笔，因为长时间让音圈工作在过电压的状态，让音圈的绝缘损坏，同时长时间的过电流会使音圈发热，这样就会使在涂在音圈上的绝缘漆慢慢损坏从而导致音圈的匝与匝之间短路，从而使音箱音圈的阻抗慢慢就小，而流过音圈的电流慢慢加大，直到音箱不响了，也标致着音圈被烧断了，音箱也就宣布死亡了。要想音箱不被烧怎么办？
加限幅器
第一个限幅器就是音响师本人把调音台的主推子和方推子其中之一推到（-3dB）的位置上，让调音台输出给下级的电平值为-3 dBu，下线所有增益都打到0dB。最后到功放输入电平值为-3dBu。
第二个限幅器就是处理器里面的限幅器，启动电平设置为-3dBu。
启动电平＝（输入灵敏度-20LG（功放额定输出功率/音箱额定功率）再把启动时间，恢复时间设置为最快，压缩比设置为无限大：1，这样的话不管前面输入到限幅器的电平值有多大都限制到-3 dBu的电平给功放，从而使音箱不过压，不过流而被烧毁
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