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<blockquote data-quote="vaiboy" data-source="post: 7792746" data-attributes="member: 570683">Well to begin to understand efficiency you need to understand current, it&#039;s the number of electrons that pass a point in a set amount of time. As each electron passes the point it gives up a fraction of it&#039;s energy in the form of heat. So the more electrons that pass the point the more heat that is generated in a set amount of time. So in the circuit there is useful energy that travels through the circuit and does work, and there is energy that escapes as heat.If we look at your example of a 1000W@4Ω amplifier the current through the output would be ~15.8 amps, we will forget about the supply side of the amp for a second. If the same amplifier were to be set up to output 1000W@1Ω the current would be ~31.6 amps. That&#039;s double the current at 4Ω, more useful energy is lost to heat but the same amount of work is being done (Work = Watts). Efficiency in amplifiers is a function of the supply to output, and the supply is doing more work to overcome the loss of useful energy to heat in the output. The amplifier would need to be set to output 250W@1Ω to match the current at 1000W@4Ω, but at 250W there would be less useful energy lost to heat on the supply side of the amplifier. So technically you could find an output wattage at 1Ω where the efficiency matches 1000W@4Ω, the output will be lower than 1000W and greater than 250W, but I don&#039;t know why you would want to.</blockquote>

[QUOTE="vaiboy, post: 7792746, member: 570683"] Well to begin to understand efficiency you need to understand current, it's the number of electrons that pass a point in a set amount of time. As each electron passes the point it gives up a fraction of it's energy in the form of heat. So the more electrons that pass the point the more heat that is generated in a set amount of time. So in the circuit there is useful energy that travels through the circuit and does work, and there is energy that escapes as heat. If we look at your example of a 1000W@4Ω amplifier the current through the output would be ~15.8 amps, we will forget about the supply side of the amp for a second. If the same amplifier were to be set up to output 1000W@1Ω the current would be ~31.6 amps. That's double the current at 4Ω, more useful energy is lost to heat but the same amount of work is being done (Work = Watts). Efficiency in amplifiers is a function of the supply to output, and the supply is doing more work to overcome the loss of useful energy to heat in the output. The amplifier would need to be set to output 250W@1Ω to match the current at 1000W@4Ω, but at 250W there would be less useful energy lost to heat on the supply side of the amplifier. So technically you could find an output wattage at 1Ω where the efficiency matches 1000W@4Ω, the output will be lower than 1000W and greater than 250W, but I don't know why you would want to. [/QUOTE]

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