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working in the computer field all my life, there was a decade or two when equipment like modems and external drives had a usable life of only a few years.  during this time when equipment was tossed, i routinely threw the AC adapter into a draw that today still contains dozens of small modular power supplies comprising many different voltages and current capacities.  when i hit upon a good operating voltage for an accessory off a standard variable transformer, i can usually find a comparable or close enough fixed modular supply to act as a dedicated supply.

alternately here in town we have two very large electronic surplus stores and if i ever need an oddball supply, i can usually pick something up for a buck or two.

cheers...gary

@stan2004 posted:

A discussion about AC voltage-dropping is not complete without a cameo appearance of the diode method.  Many OGR threads on this.  Typical implementation would look something like this and cost less than $5.  You'd get multiple, simultaneously use-able outputs.  Not as convenient as a multiple output transformer as it does require component interconnections, wiring, etc.



AC voltage dropping using bridge rectifiers



Not sure if I am doing something wrong, but I can't seem to get the a voltage drop.  I can get a .7 drop.  To get 1.4 drop, do I have to connect the "-" and "+" together and have the output be the wire under the "C" on the diode, or do the "-" and "+" not come into play when you want a 1.4 drop.

I have the KBU 1010 and am using AC.

Thanks

Tony

Last edited by Tony H

Yes.  "-" and "+" MUST be connected together.  If feeding the "hot" into the terminal under the letter "A" then you should get about -0.7V drop at the "-/+" junction, and about -1.4V drop at the terminal under the letter "C".  Then to get more drop, take the "C" terminal from the first bridge and feed the "A" terminal of the next bridge.  And so on down the line.

Note that if measuring the drop using JUST a voltmeter with no load from an accessory, light bulbs, or whatever, the drop will be substantially less.   It might be only -0.5V (or less).  The 2nd output would be only -1V (or less). And so on.

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