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This appears to have been asked in some way before (at Transformer: loaded vs open-circuited vs short-circuited) but both the question and the answers there are rather vague to me, so I'll try to be more explicit.

As anyone who was toyed a small transformer might have noticed, the output/secondary voltage in "open circuit" (actually seeing the megaohm-level resistance of the DMM/voltmeter as load) can be much higher than when the transformer is under significant load. For example, I get 8V on a "5V" transformer (it's built on an EI-30 core, so you get an idea of its approximate power rating; somewhere around 3VA). So, my question is: what's the load typically used to rate the voltage "faceplate" output of such transformers? (I assume this might be mentioned somewhere in IEC and/or US transformer standards, if you know more precisely please let us know).

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The nominal output voltage should be the rated voltage with nominal input voltage and the full load (resistive) as rated.

In other words, a 12V 300mA transformer should have 12V RMS output with nominal input and a 300mA resistive load. For loads less than the rated load, the voltage, of course, will be higher.

Edit:

The regulation of a transformer is typically defined as:

Regulation(%) = \$ \frac {V_{open} - V_{full.load}}{V_{full.load}}\$

Large high power transformers might have regulation of a few percent, cheap small transformers maybe 5 or 10x worse.

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  • \$\begingroup\$ I guess this is the right answer, but it might help to elaborate how the max/open-circuit voltage relates to this faceplate/full-load voltage. I guess that to deduce the open-circuit voltage from the faceplate voltage one would need to know/measure the winding resistance of the secondary at the very least. \$\endgroup\$ – Fizz Nov 25 '14 at 2:55
  • \$\begingroup\$ @user3588161 or simply measure it. See edit above. \$\endgroup\$ – Spehro Pefhany Nov 25 '14 at 3:58
  • \$\begingroup\$ It's actually rather hard to find concrete (real-life) data on transformer regulation vs. power. I found zen22142.zen.co.uk/Design/dcpsu.htm which says 12% at 12VA and 25% at 6VA. So I guess ~50% at 3VA is not so outlandish. \$\endgroup\$ – Fizz Nov 25 '14 at 21:45
  • \$\begingroup\$ As I found this topic in another thread, it reminded me to add here an example datasheet from one the [few, IMHO] manufacturers that specify the open-loop voltage (for a series of transformers of various powers, so it's actually a useful dataset): tme.eu/en/Document/505ab7a151e6fa111ec7ce1e42277c4f/55xxx.pdf \$\endgroup\$ – Fizz Sep 22 '15 at 23:52
  • \$\begingroup\$ @RespawnedFluff When you buy transformers in bulk from manufactures the regulation (and loss, magnetizing current etc.) are fully specified. The premade ones- less so, but for example this Tamura 2.4VA part shows regulation 24% "typical" from full load to no load. As I said in my answer, around 10x worse than the few percent a high power transformer might present. It's typically going to be worse on split-bobbin transformers than the toroidal type you linked. \$\endgroup\$ – Spehro Pefhany Sep 23 '15 at 3:10
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To add to what others have already said, I should stress again that the nameplate voltage is assuming full load with a resistive load. This is a very common pitfall for people designing power supplies for the first time, and I've even seen similar mistakes in commercial gear. When you put a rectifier and capacitor on the output of the transformer, the peak current is very high - so the voltage can actually drop below the nameplate voltage. If you look at the transformer voltage on a scope, you'll see the peaks of the sine wave have been lopped off by the heavy current pulses charging the filter cap.

I don't know what the best way to calculate the voltage droop is - so far, my preferred method is to use circuit simulation. I calculate the virtual impedance of the transformer by subtracting the off load voltage (RMS rating * transformer regulation) from the nameplate voltage. Then, I divide that by the maximum RMS current rating of the transformer. You can then simulate the circuit by using an AC source in series with a suitable resistor. The peak voltage of the AC source should be RMS rating * transformer regulation * 1.4. But you also need to account for mains voltage tolerance (+/- 10%), so multiply that by 0.9 to get a worst-case minimum voltage.

Then you have additional sources of voltage drop - the rectifier diodes, and the ripple. I usually assume 1V peak voltage drop for a silicon rectifier (not schottky), so a bridge could lose up to 2V. The ripple might be 1 or 2 volts peak to peak (depending on the size of the capacitor), so there's another 1-2 volts gone. Then consider the minimum dropout voltage of your regulator (3V for a 7805), and you'll have to use a much, much higher input voltage than you might first expect. A 9V AC RMS transformer seems to be suitable for a 5V regulated power supply taking this into account, using a bridge rectifier and a 7805. A 6V AC transformer certainly isn't sufficient (even without rectifier voltage drop, ripple and transformer droop, the maximum peak voltage with -10% mains is only 7.6V).

The worst-case maximum input voltage will be the RMS voltage rating * transformer regulation * 1.4 * mains tolerance (1.1), then minus the minimum voltage drop of the rectifier diodes (0.6V for a full wave rectifier, 1.2V for a bridge). Your capacitor and regulator needs to be able to handle it.

I usually calculate the regulator dissipation by calculating the voltage drop across the transformer. To do that, I multiply my RMS load current (less than the transformer's rated current) by the transformer's virtual impedance (calculated earlier). Note that the transformer's AC RMS load current will be about twice the DC output current with a bridge rectifier. Having calculated the RMS voltage drop, I then subtract it from the transformer's highest RMS output voltage (RMS voltage rating * transformer regulation * mains tolerance). Subtract the minimum voltage drop of the rectifier diodes - and half the ripple voltage - and you can then determine the typical average input voltage to the regulator under maximum load. Subtract the output voltage of the regulator and multiply by the DC load current to determine power dissipation.

Finally, I calculate the required heatsink by dividing the allowable temperature rise by the wattage dissipated in the regulator. Allowable temperature rise needs to stay below max junction temperature (typically 125C) at the maximum expected ambient temperature. So for a 35C ambient, you can tolerate a 90C rise. Then, you subtract the junction to case thermal resistance from the calculated C/W rating of the heatsink. For a 7805, the junction to case is 5C/W. Then you have to subtract the case to heatsink thermal resistance, which could be as much as 1.4C/W if you don't use thermal paste. You might be surprised by how large a heatsink you need!

Take these calculations with a grain of salt, because I could well be wrong. I'd appreciate the input of anyone more knowledgeable in this area if there are any errors!

If you don't have access to the datasheet of the transformer, then I'm afraid it's going to be more difficult. You'll have to measure the incoming AC mains voltage, output voltage of the transformer under no load, and calculate the regulation from that. I've seen some small transformers with regulation nearly as bad as 60% - so it's not an insignificant factor!

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