Transformer

T

tonyelectric

Heres another one for you..

Have got a Cutler Hammer LoVo light transformer with 415V primary, 12 & 24V secondaries, so if I run this on 230V I should get somewhere in the region of 6V & 12V (slightly more actually), so with a modest resistance should be able to use a 12V lamp on the 24V tapping.

Brain tired, comments please (polite ones) :LOL:
 
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About 13.3V. Why do you want to use it?

Yes, adding some resistance to it would work, but do you really want to go to all that trouble, not forgetting the voltage to the bulbs will increase if one of them pops?
 
T

tonyelectric

its in a bandsaw i've aquired, I have changed the motor to a single phase but the work light is quite nice and I have about 40 12V 60W bulbs kicking around from the days of centralised emergency lighting :)
 
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i've done the same before and it will work, but the impedance of the primary coil will be high so when you run it on 230v you'll only be putting roughly half the power in hence the bulb may be rather dim, give it a go can't hurt.
 
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i've done the same before and it will work, but the impedance of the primary coil will be high so when you run it on 230v you'll only be putting roughly half the power in hence the bulb may be rather dim, give it a go can't hurt.

I dont quite understand... I always understood it was the ratio of the turns that decided the output voltage? The power being drawn by the primary being the same as drawn by the secondary ( Plus iron and copper losses of course ),
 
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I dont quite understand... I always understood it was the ratio of the turns that decided the output voltage? The power being drawn by the primary being the same as drawn by the secondary ( Plus iron and copper losses of course ),[/quote]


Yes the secondary voltage is determined by the ratio of turns on primary to secondary coils, and as as the draw on the secondary increaces so dose that on the primary to a point, the impedance of the primary limits the amount of power the primary can take, so any further draw by the secondary will not result in any more power being drawn.
 
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the impedance of the primary limits the amount of power the primary can take, so any further draw by the secondary will not result in any more power being drawn.
Indeed, what I forgot to mention and what Mr Tinker rightly points out is that the transformer will not output the rated kVA value at a reduced voltage, as there will be an increased current on the primary. Think about P=I²R.
 
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the impedance of the primary limits the amount of power the primary can take, so any further draw by the secondary will not result in any more power being drawn.

It is not quite that simple. The impedance of the primary winding is a complex, part resistive and part inductive. With no load on the secondary the inductance of the primary is ( in theory ) infinite so no current flows in the primary.

Current flowing in the secondary has the effect of reducing the inductive part of the primaries impedance to allow current to flow in the primary. The reduction in impedance is such that the amount of energy in the primary is equal (in theory ) to the energy taken from the secondary.

The transformer action starts to fail when the resistive component becomes significant compared to the inductive component. The transformer itself fails when the inductive component is close to zero and the current in the primary is limited by the resistance of the winding invariably leading to a burnt out winding if there is no protective fuse.
 
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the impedance of the primary limits the amount of power the primary can take, so any further draw by the secondary will not result in any more power being drawn.
Indeed, what I forgot to mention and what Mr Tinker rightly points out is that the transformer will not output the rated kVA value at a reduced voltage, as there will be an increased current on the primary. Think about P=I²R.

P=I²R . Isnt that for resistive circuits ? I did my theory a LONG time ago LOL.

And for some bedtime reading :-

http://www.maintenanceresources.com/bookstore/electrical/jppdf.pdf

That should put anyone to sleep....
 
T

tonyelectric

Sadly the transformers are no longer made, there was a 230V version, the transformer has an extended top coil insulation piece that supports a bakelite block containing the switch mechanism and input/output fuses and fixings for the top cover of the aluminium housing which itself contains the rocker switch contact springs and rocker, it would be a very major task to fit any transformer other than the one made for the unit. Eaton discontinued the design only a few years back.
 
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the work light is quite nice and I have about 40 12V 60W bulbs kicking around from the days of centralised emergency lighting :)
And the reason you can't get any old 12V transformer of suitable rating, mount it somewhere appropriate (band saws being fairly static bits of kit, you hope) and run a cable to the light is..... :confused:
 
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I can remember in Uni we had to calculate the number of turns required and voltage did come into this calculation as did the type of core used. Not got books to hand as can't say exactly what changed but it did change so there could be excess losses and of course that means transformer gets hotter than it should. It's not just a ratio.

I know it will work but you would need to de-rate it so it will not overheat.
 
T

tonyelectric

Yes and no is the answer, yes it delivers 13.9V on the 24V output but the efficiency is reduced as suggested so it only lights the 60W bulb to half power and gets hot quickly, so I went for a plan B. I could not remove the transformer as the switch is integral to its insulation and the box cover fixes to the transformer which in turn supports the lamp, however I found a suitably small 60W rated SELV transformer that would fit above the original in the box, re-wired the switch to the SELV and lamp to the output, job done.

Thanks for all the posts guys :D
 

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