15A*120V*5 = 9KWatts. You're converter is only good up to 4KWatts. You can only run two of your 15A circuits.

or 5 ckts @800w? indeed, total not to exceed 4000??

 

I see no reason you couldn't do that.

 

now tcm has me thinkin "wall warts", there are some big chunky 120-12/15vac units out there, an his math in #13 and #14 sez bout 400va should work, might be an easy score on the cheap like from ebay or amazon, one on each branch ckt at the distribution panel.....

I'm thinking standard off the shelf automotive battery chargers from a pawn shop or garage sale. Dirt cheap and plenty of transformer to work with in the 20+amp size units.

 

or 5 ckts @800w? indeed, total not to exceed 4000??

Also, if you use a 12V transformer to make up the voltage difference, as proposed in post #4 and #6, then you'll only need

800/120*12 = 80 Watt transformer. (safety margin gives 100Watt transformer per leg) These should be cheap and easy to come by.

 

both sets of windings in series, transformer from batt chrgr. on the bench, line on 115vac side. load side(18vac). . . to where??

 

i can't envision the series hookup for boost, i tried a few sketches an kornfused me good... guess i don't have quite enough brain power

xfmr is two distinct windings with no continuity from side to side or side to core.

the bench test put 120 in got 18 out

 

both sets of windings in series, transformer from batt chrgr. on the bench, line on 115vac side. load side(18vac). . . to where??

The secondary goes in series with the primary on the hot line end. One way it will buck the input voltage down the other it will boost it up. 50/50 chance of getting it right the first time.

 

thanks for jumpin in there vu2nan, 4024k (@108) is the problem, do you have a 4024(@120) you might swap?

yes tcmtech, that matches one of my sketches, let me try a word picture here. horizontal line from line1 to load1, just right of line1 the line1 of the xfmr primary verticle down thru winding to line2. from that joint xfmr secondary winding is the path to load2.

recap, two horizontal parallel lines, top one straight from line1 to load1.
lower line contains the secondary winding squigel to load2 then the beginning of the secondary winding squigel is the point where the xfmr primary is drawn verticle-up from the lower line to the upper line?? yes? no? maybe-so?

 

and thus my confusion, "the secondary winding becomes the path to load2" all the amperes load2 consumes will go thru the secondary winding of the xfmr

what do you suppose would happen if i paralleled a conductor with the secondary winding to load2. the added conductor would carry most of the amperes, the winding would mostly be "boosting voltage", as in #14

 

yes? no? maybe-so?

See schematic in post two. Think about it and the parts you have and adapt from there.

what do you suppose would happen if i paralleled a conductor with the secondary winding to load2. the added conductor would carry most of the amperes, the winding would mostly be "boosting voltage", as in #14

Think about it for a while then tell us what you think will happen?

 

hi guys, thinking about it didnt do much good, its time to test theory, so i hooked it up with my smallest jumpers, and a flash of insight was revealed, well actually it was that 24 gauge jumper, the insight came after the smoke cleared. xfmr is fine, those dinky "sacrificial" jumpers are worthwhile on my bench...

its right here i want to pause and thank all ya'll for helping me cure another layer of my ignorance.

ok, lets try this again, hmmm, how to resemble that post 2 pic, might ought to hook up like this, that, an such, bench volts in 120, secondary volts 137 on the vom. an no smoke, woohoo.

the next attempt is actually like the right half of vu2nan's pic, so the 50/50 chance of buck or boost is about phasing, kewl. got lucky the first time, an "larnt" a bit more than i knew yesterday.

last question (for today) 400va/12=33.33amps, so, to guestimate the garage sale battery charger xfmr being re-purposed, the battery leads need to be at least awg 10 size or larger to assume the xfmr inside is large enough.

 

last question (for today) 400va/12=33.33amps, so, to guestimate the garage sale battery charger xfmr being re-purposed, the battery leads need to be at least awg 10 size or larger to assume the xfmr inside is large enough.

Not necessarily. For the transformers rough continuous VA rating just take its continuous rated amp output( not peak boost rating) times ~15 and call it close enough.

If your transformer came out of an older or larger model charger rated for 20 - 30 amps continuous charge with say 100 - 200 amp boost the odds are the transformer can take a pretty severe beating without problems. Even if the charger was a mid sized 15 - 20 amp rated unit the odds are unless you are trying to run your inverter for long periods of time at 3+ Kw outputs that too would not have any overloading issues either.

For the most part any decently built battery charger transformer is designed to run for 10 - 20 seconds in their peak boost mode which can be way over their normal continuous rating thus giving them a substantial short term overload capacity.