Running 240V transformer at 120V

[dnbdmr]

Hi all,

I got my hands on some Signal Transformers DP-241-8-48, which are dual 115/230V primaries with 48VCT @ 2A secondaries. My plan is to use them for a power supply. I'm thinking of switching between full wave rectified and full wave bridge to get two voltage ranges, like in the attached schematic. But, I don't need 48V and I am worried about dissipating 60W+ in my main regulators.

My question is: What will happen if I wire the primary as 240V with the primaries in series? I know I will have half the voltage on the secondary. Will I need to derate the current also? My guide is http://www.hammondmfg.com/pdf/5c007.pdf. So I want to have ~12/24v on the secondary with ~2A/1A. Will this work?

[coromonadalix]

You thread title is misleading ?  youre talking of switching the secondary outputs not the primary side ?

If you respect the serial primaries to get a 240v and put them in parallel  for 120vac   your transformer VA's will stay the same


all your calculations are on the pdf sheet  ??

2 diodes
V D.C. = 0.45 X Sec. V A.C.
I D.C. = 1.27 X Sec. I A.C.


rectifier bridge
V (Peak) D.C. = 1.41 X Sec. V A.C.
V (Avg) D.C. = 0.90 X Sec. V A.C.
I D.C. = 0.62 X Sec. I A.C.

[james_s]

I think it's obvious what he's asking. He wants to configure the transformer for 240V by wiring the two primaries in series, then power that from 120V which will result in the secondary voltage being half that of the rating. I see no issue with doing this, the secondary current rating will stay the same, so half the voltage means you will get half the rated VA out of the transformer.

[Siwastaja]

The transformer may even run a bit cooler (for the same output current) because sometimes the transformers are designed to operate very close to saturation increasing magnetization current near the peak flux and the related losses. By running at lower voltage, especially no-load power consumption decreases. So you definitely do not need to derate the current. You need to derate output power, though; if the transformer was capable of outputting 48V*4A = 192W, you definitely cannot output 24V*8A = 192W, but need to limit yourself to 24V*4A = 96W or maybe a tiny bit more.

[Kleinstein]

The no load loss will be lower even if not running close to saturation at normal voltage. Usually the core loss goes down with the magnetization (and thus the voltage) to a power of 2 to 2.6. So expect the no load loss to be lower by about a factor of 5-6.
The voltage drop due to loading is a little lager, as it would be relative to a lower voltage. So the voltage at the nominal current may be a little lower than 24 V, but this should not be much.

[Zero999]

I Googled the part number and found it on DigiKey.
https://www.digikey.com/en/products/detail/signal-transformer/DP-241-8-48/1118046

Yes, it will work. The transformer is rated for 2A, so will give you 24V or 12V at 2A on the secondary and should run a little cooler. The main disadvantage is you're using a transformer which is nearly twice as bulky as necessary.

[dnbdmr]

Thanks for the answers, everyone. I'm not super familiar with transformers and wasn't seeing VAin = VAout. This is all helpful.

[David Hess]

It will work fine; the current rating depends on the wire sizes so remains the same.  The magnetizing current will be 25% of normal reducing core losses, however the primary side resistance will be doubled doubling restive losses.  The only practical result other than higher efficiency will be that the doubled primary side resistance will cause the secondary side voltage to sag a little more under load.

[james_s]

Thanks for the answers, everyone. I'm not super familiar with transformers and wasn't seeing VAin = VAout. This is all helpful.

That is one of the fundamental principals of transformers, VA out is always equal to VA in - losses. The voltage on any winding is directly related to the voltage on any other winding according to the turns ratio. At some point if you keep increasing the voltage you'll cause the core to saturate but you can always reduce the voltage. Since the current capacity is determined primarily by the size of the wire used to wind it, the power will vary according to the voltage.