Depending on a lot of things. Here is my rule of thump.
5kW effektive = P * 0.78 => P = 6410,25W = U * U / 4 Ohm => 25641 (V*V) = U * U => U = 160,12815380508713348325366976769, easy said 160. Voltage lost in power output devices => U = 165. Vtrafo = U * 1.4142 = 233,3452377915606830522786394946V, easy said 234V.
5kW effektive = P * 0.78 => P = 6410,25W = I * I * 4 Ohm => 1603 (A*A) = I * I => I = 40,032038451271783370813417441922, easy said 40. Current lost in power output devices = 0 => I = 40. Itrafo = I * 1.4142 = 56,613851707229787530538735429226, easy said 57A.
So in small telling for a standard non switched setup and a 5kW amp You will need 2 secondary windings each at approx 234V and 57 A Peak or rated 165V 40A as trafo ratings are RMS as well.......
Some might say they could do better than 78% efficientcy, but i doubt it. Why ? I have worked with amps since 82 and have only seen a small handfull being better.
Oh and this is minimum for only 1 amp
5kW effektive = P * 0.78 => P = 6410,25W = U * U / 4 Ohm => 25641 (V*V) = U * U => U = 160,12815380508713348325366976769, easy said 160. Voltage lost in power output devices => U = 165. Vtrafo = U * 1.4142 = 233,3452377915606830522786394946V, easy said 234V.
5kW effektive = P * 0.78 => P = 6410,25W = I * I * 4 Ohm => 1603 (A*A) = I * I => I = 40,032038451271783370813417441922, easy said 40. Current lost in power output devices = 0 => I = 40. Itrafo = I * 1.4142 = 56,613851707229787530538735429226, easy said 57A.
So in small telling for a standard non switched setup and a 5kW amp You will need 2 secondary windings each at approx 234V and 57 A Peak or rated 165V 40A as trafo ratings are RMS as well.......
Some might say they could do better than 78% efficientcy, but i doubt it. Why ? I have worked with amps since 82 and have only seen a small handfull being better.
Oh and this is minimum for only 1 amp
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