Determining the amperage of an unknown transformer
in [Design]
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From: Xfmr Guy on 11 Aug 2010 00:00 On Tue, 10 Aug 2010 12:43:38 -0700 (PDT), Nunya <jack_shephard(a)cox.net> wrote: <SNIP> > > Not only that, but using weight as the rule is not going to be >very accurate either. It's just about as accurate as an estimate based on length and/or weight measurements can be. The power handling capability of a transformer is directly proportional to the WaAc product; see page 4.3 of: http://www.diegm.uniud.it/mattavelli/elettronica_industriale/Magnetica/CoreDesign.pdf Since the WaAc product has dimensionality (length)^4, and weight has dimensionality (length)^3, the power handling capability of a transformer should be proportional to (weight)^(4/3). If weight vs. power handling is plotted on loglog paper, you will get a scatter plot along a straight line with a different slope than if you had plotted WaAc vs power handling. A plot of weight vs power handling is not quite as good as a plot of WaAc vs power handling for a number of reasons. The ratio of WaAc to core volume is the same for all transformers with the same aspect ratio; that is, for all EI core transformers with a "square stack", that ratio will be the same no matter how big or small the transformer. If the core weight vs power handling for a bunch of transformers having the same aspect ratio is plotted, the plot should look the same as a plot of WaAc vs power handling, just with a different slope. But, transformers having an "oversquare" or "undersquare" stack will have a different ratio of WaAc to volume, and the plot of WaAc vs power handling will not map to a plot of weight vs power handling in the same way as for a "square" stack. Another source of error with a weight vs power handling plot is that the measured weight of a finished transformer isn't just the weight of the core; the weight of the copper, insulation, and mounting hardware, is also included. Nevertheless, a plot of transformer weight vs power handling is a good start for determining the power rating of a transformer. To refine the result, a temperature rise test should be performed. A fairly extensive discussion of this problem, with plots of a large number of commercial transformers is found at: http://forum.allaboutcircuits.com/showthread.php?t=38273
From: Xfmr Guy on 11 Aug 2010 00:45 On Wed, 11 Aug 2010 10:25:37 +1000, Grant <omg(a)grrr.id.au> wrote: <SNIP> > >I been wondering lately how to tell the power split between two >windings on a nice looking transformer in the 'junk box'. > A well designed transformer should have the same current density in all the windings. It's not too hard to show that given a couple (or three) windings, with one winding carrying a current I1 and with an output voltage V1, and a DC resistance R1, and similarly, a second winding with I2, R2 and V2, the current density in winding 1 is proportional to (I1*R1)/V1, and in winding 2, it's proportional to (I2*R2)/V2. Then the currents in the two windings (for equal current densities) are related like this: I1/I2 = (R2*V1)/(R1*V2)
From: Robert Baer on 11 Aug 2010 01:05 Scott wrote: > I have a toroidal power transformer that I'm considering using for a > power supply project. It has two secondaries -- 18v-0-18v and 0-48v. > Part number is 7-40-0011 (already tried googling it, no luck). I'm > pretty sure I purchased it from one of the online surplus places, > maybe All Electronics, probably about a decade ago. > > What's the best way to determine how many amps the secondary is rated > for? Can I just throw a resistive load on it an measure the current? > or does this risk damage? As in any "unknown" transformer, start by determining the approximate power rating from the cross-section; all else then falls in place.
From: Robert Baer on 11 Aug 2010 01:08 Jan Panteltje wrote: > On a sunny day (Tue, 10 Aug 2010 08:32:06 -0700 (PDT)) it happened Scott > <smbaker(a)gmail.com> wrote in > <d721ef17-ccf7-44b4-9cf2-b6e83883da51(a)t5g2000prd.googlegroups.com>: > >> I have a toroidal power transformer that I'm considering using for a >> power supply project. It has two secondaries -- 18v-0-18v and 0-48v. >> Part number is 7-40-0011 (already tried googling it, no luck). I'm >> pretty sure I purchased it from one of the online surplus places, >> maybe All Electronics, probably about a decade ago. >> >> What's the best way to determine how many amps the secondary is rated >> for? Can I just throw a resistive load on it an measure the current? >> or does this risk damage? > > Wire diameter? Not a very good indicator..wire could be undersized for the MFG rating, or oversized, like i do: circular mils is current rating in milliamps.
From: Grant on 11 Aug 2010 02:02
On Tue, 10 Aug 2010 17:45:44 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >On Wed, 11 Aug 2010 10:25:37 +1000, Grant <omg(a)grrr.id.au> wrote: > >>On Tue, 10 Aug 2010 09:29:53 -0700, John Larkin <jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote: >> >>>On Tue, 10 Aug 2010 08:32:06 -0700 (PDT), Scott <smbaker(a)gmail.com> >>>wrote: >>> >>>>I have a toroidal power transformer that I'm considering using for a >>>>power supply project. It has two secondaries -- 18v-0-18v and 0-48v. >>>>Part number is 7-40-0011 (already tried googling it, no luck). I'm >>>>pretty sure I purchased it from one of the online surplus places, >>>>maybe All Electronics, probably about a decade ago. >>>> >>>>What's the best way to determine how many amps the secondary is rated >>>>for? Can I just throw a resistive load on it an measure the current? >>>>or does this risk damage? >>> >>>This will get you started. >>> >>>ftp://jjlarkin.lmi.net/XfmrScatter.JPG >> >>Thanks for that. Quite useful, though our 50Hz xfmrs would be a >>little heavier, I think. > >I think all these were 50/60 Hz transformers. All the ones we buy are. > >> >>I been wondering lately how to tell the power split between two >>windings on a nice looking transformer in the 'junk box'. > >You could measure the resistance and the open-circuit voltages. You >could assume that each winding could be loaded to, say, 90% or 95% of >its open-circuit voltage. > >> >>Xfmr looks around 120VA with 9V and 33V secondaries. >> >>I was thinking measure no load, then load to 5% or 10% voltage >>drop? Work from those results. > >Oops, yes. > >> >>Tim's resistance measure would shed some light too. >>> >>> >>>Or just load it progressively and see how hot it gets. >> >>Good double check of the other methods. Bit slow? >> >>This xfmr is a good quality E+I one with earth screen, got the >>outside shorted copper turn and iron turn to reduce external >>field, switch and fuse in a partial enclosure. Destined for >>a bench power supply, I think. Once I have an idea where to >>set the power limits. >> >> >>Using SLA batteries for testing stuff has produced some fun >>moments, watching the magic smoke escape ;) >> >>http://grrr.id.au/image/74HC74-let-go-magic-smoke-with-a-bang.jpg >> > >I don't suppose it still works. Nah, everything connected to 5V blew up, except a TL431 'cos it had a current drop resistor ;) > >>That was an odd situation where a line of five 'HCMOS chips >>started emitting smoke from their middle pins, until the one >>in the centre one popped its top. Didn't blow the 35A battery >>fuse. > >Wirebonds make nice detonators, but don't pass a lot of I^2*T. The >first atom bombs used exploding wires. Later ones used gold thinfilms. > >I spent the weekend blowing up DC/DC converters, not too bad at only >$4 each. > >I unpotted one, but lost the transformer in the process. > >ftp://jjlarkin.lmi.net/VASD1_top.JPG > > >Next project is to breadboard a current limiter circuit, and blow them >up together. Better you blowing them up than your customers :) Enjoy! Grant. |