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This is pretty much right on. For normal solidstate amps you can figure the efficiency as 50% for figuring out current draw for a given power output. There are variables such as devices used, bias, and circuit losses but for the basis of deytermining current draw figure 50%. Just be aware of what type of power output is being quoted,average (rarely) or peak (often).

[QUOTE="Captain Kilowatt, post: 155584, member: 39"] This is pretty much right on. For normal solidstate amps you can figure the efficiency as 50% for figuring out current draw for a given power output. There are variables such as devices used, bias, and circuit losses but for the basis of deytermining current draw figure 50%. Just be aware of what type of power output is being quoted,average (rarely) or peak (often). [/QUOTE]

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