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In Reply to: Another SMPS Question posted by Triode_Kingdom on May 10, 2017 at 19:15:29:
Following an exhaustive search, I can say that data sheets for the ICs used as converters simply don't show this. Representative schematics only show input voltage, not current, and none of the design calculations for selecting support components refer to DC input current. However, I think the answer lies in the efficiency ratings. If a buck converter required 12.6V/4A input to produce 6.3V/4A output, it would be 50% efficient. That's no better than a linear regulator. The efficiency rating for these is typically much higher than that, closer to 90%. Using that number, 25.2W at the output (6.3V/4A) should require 28W input. If the input voltage is 12.6, 28W equals an input current of 2.2A. So, it is in fact very much like a conventional transformer in terms of voltage and current conversion. This means it should be feasible to connect a small, inexpensive 6.3V buck converter to a 12.6V SMPS and have both voltages available in an amplifier for filaments.
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Follow Ups
- Answered My Own Q - Triode_Kingdom 06:59:46 05/11/17 (1)
- RE: Answered My Own Q - cpotl 12:10:36 05/11/17 (0)
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