There's some energy lost in the conversion, but not a lot. If you use 12V halogen bulbs you lose a large amount of energy as heat from the bulbs themselves, use LEDs and it's very little. High voltage fluorescent lighting is much more efficient than filament bulbs (high or low V). It's more what you are using to produce light than your power supply. L
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Not really, if the transformer was built properly. A step-down (AC to AC) transformer trades voltage for current: Vin x Iin = Vout x Iout. That equation is basically just energy conservation: power (energy per unit time) = V x I, which must be the same (neglecting small resistive heating) on both sides of the transformer. To keep the math simple, if you're going from 120V at the wall to 12V output, you'll use 0.1 amps of mains current to get 1 amp of output current (120x0.1 = 12x1). Now, once you've got your low-voltage AC, it gets put through a nice bridge rectifier (four diodes arranged in a diamond pattern) to turn it into DC. A straight diode would do the job, but half of each cycle would be lost (it would heat the diode, basically). The bridge rectifier gets you forward-going DC from both sides of the AC cycle.
How would a half wave bridge lose energy in the lost cycle ? The single diode isn't conducting, so it has no resistive losses.....
A half wave rectifier (only one diode is needed, so it's not a bridge) wouldn't cause any resistive losses. And there wouldn't be a load on the transformer for half the cycle, so the power consumption would be somewhat less. But would cause other problems. Transformers really like it when the magnetic flux is equally balanced throughout the full waveform. The collapse of the flux in one direction should be reinforced by building the field in the opposite direction. Removing the current draw for half the cycle messes with the operation... This will send the transformer core into saturation at lower current draw; creates a standing DC current on the trannie, and generally causes havoc on the rest of the AC system, due to the imbalance. And any lower current-consumption would be somewhat offset by the inefficient operation of the transformer in this mode.
Yes, sorry, its not a bridge. However, my point is that its not actually lossier than a full wave because of losses in the diode Saturation losses in the transformer are exactly why I always attack the idea of trying to do on-line AC/DC conversion, especially at high currents. Steve
Yes, I was trying to indicate you are correct about the diodes. Over-saturation losses, right? ;-) Slightly off topic--this is why single-ended output transformers are larger and "gapped," (compared to push-pull OTs) because the flux isn't symmetrically balanced...
The quality of transformer designs seems to have gone downhill over the last few years doesn't it ?
I'd agree. Transformers were once relatively cheap components in elec design, and often (not always) had a larger margin of safety (and spec.) But if they could eliminate a transformer from a design, they would; then or now. And now they are one of the most expensive (material cost, etc.) You can get really good transformers today, of course (but man, are they pricey...)
The regulation on small ones <20VA these days is AWFUL - we had some that were something like 50%.
Gad! How could anything be that far out of spec?
....and we had 100 of 'em, from Newark.
Steve and GMoon below are both correct. That's what I get for writing too late at night. Obviously the single diode isn't conducting in reverse bias (that's the point). There's no load through the transform during those half-cycles, so no power losses.
Thanks everyone, I take it if I had a solar panel on my roof, and used an inverter to go dc to ac to patch it into the grid, and then inside the house I used transformers to step down the voltage and convert to dc, then the energy loss would be minimal and the best way to do things?
. http://en.wikipedia.org/wiki/Transformer#Energy_losses . http://ecmweb.com/mag/electric_basics_transformer_voltage/