Reply to thread

Message: <blockquote data-quote="Sasika@Elakiri" data-source="post: 493396" data-attributes="member: 29714">Hema silva bro link eken meka hambawuna melo sasrak therenne ne ne <img src="/styles/default/xenforo/smilies/default/sad.gif"  class="smilie" loading="lazy" alt=":(" title="Sad    :(" data-shortname=":(" /><img src="/styles/default/xenforo/smilies/default/sad.gif"  class="smilie" loading="lazy" alt=":(" title="Sad    :(" data-shortname=":(" /><img src="http://file:///C:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/moz-screenshot.jpg" alt="" class="fr-fic fr-dii fr-draggable " style="" /><img src="http://file:///C:/DOCUME%7E1/ADMINI%7E1/LOCALS%7E1/Temp/moz-screenshot-1.jpg" alt="" class="fr-fic fr-dii fr-draggable " style="" />Power calculations <a href="http://en.wikipedia.org/wiki/AC_power" target="_blank">AC power</a>, which includes audio power, is best measured as an average power. This is the accurate method. It is based on this formula<a href="http://www.ac.wwu.edu/%7Evawter/PhysicsNet/Topics/ACCircuit/AC-AvePower.html" target="_blank">[2]</a>: <img src="http://upload.wikimedia.org/math/1/9/5/195cab4fdad8576b63419386ddfd3091.png" alt="" class="fr-fic fr-dii fr-draggable " style="" /> For a purely resistive load (not a speaker), a simpler equation can be used: <img src="http://upload.wikimedia.org/math/7/c/1/7c1c446258305f4c3c58251e5122d9ab.png" alt="" class="fr-fic fr-dii fr-draggable " style="" /> In the case of a steady sinusoidal tone (not music) into a purely resistive load, this can be calculated from the peak voltage and the resistance: <img src="http://upload.wikimedia.org/math/6/1/5/6151988bcdc773892c89c52b8419718a.png" alt="" class="fr-fic fr-dii fr-draggable " style="" /> Though a speaker is not purely resistive, these equations can be used to approximate power measurements for such a system, as follows. An ideal (100% efficient) <a href="http://en.wikipedia.org/wiki/Class_AB_amplifier" target="_blank">class AB amplifier</a> with a 12-volt <a href="http://en.wikipedia.org/wiki/Peak-to-peak" target="_blank">peak-to-peak</a> supply can drive a signal with a <a href="http://en.wikipedia.org/wiki/Peak_amplitude" target="_blank">peak amplitude</a> of 6 V. In an 8 <a href="http://en.wikipedia.org/wiki/Ohm_%28unit%29" target="_blank">ohm</a> (see <a href="http://en.wikipedia.org/wiki/Electrical_impedance" target="_blank">impedance</a>) loudspeaker this would deliver: Ppeak = (6 V)2 / 8 Ω = 4.5 watts peak instantaneous.<a href="http://www.google.com/search?q=%286+V%29%5E2+%2F+8+ohms" target="_blank">[3]</a> If this signal is <a href="http://en.wikipedia.org/wiki/Sinusoidal" target="_blank">sinusoidal</a>, its RMS value is 6 V × 0.707 = 4.242 V(RMS). This voltage into a speaker load of 8 Ω gives a power of: Pavg = (4.242 V)2 / 8 Ω = 2.25 watts average <a href="http://www.google.com/search?q=%284.242+V%29%5E2+%2F+8+ohms" target="_blank">[4]</a> Thus the output of an inexpensive car audio amplifier is limited by the voltage of the battery. In most actual car systems, the amplifiers are connected in a <a href="http://en.wikipedia.org/wiki/Bridge-tied_load" target="_blank">bridge-tied load</a> configuration, and speakers are no higher than 4Ω. High-power car amplifiers use a DC-to-DC converter to generate a higher supply voltage. The true power output of an amplifier can be estimated by examining the input current. Linear amplifiers tend to be about 60% efficient at best. A switch-mode amplifier (known as class D) can achieve much higher efficiency, sometimes as high as 95%. A linear car amplifier labeled "500 W PMPO" but fitted with a 5-amp fuse can, at most, deliver an average power of 5 A × 14.4 V × 60%, or about 43 watts.</blockquote>

Verification: Dawasata paya keeyak thibeda?

Top Bottom