Shorting transformer leads

transformer

Sorry for a trivial question. I have a transformer with a tapped secondary coil: 12(blue wire)-0(black wire)-12V(2nd blue wire). I need to use it as a 12V supply source, so can use 0-12VAC (either side of a tap, one blue and the black-ground).I understand that in this configuration I only get a half of the rated current. However, what happens if I short both sides of the tap (both blue wires together). Can I get a full current? or because of the phase difference between the blue wires I get nothing (or maybe even both windings blown?).
Is going for a bigger transformer – the only option? (I am restricted in space inside the electronic enclosure).

Best Answer

You definitely do not want to short the +12 and -12 outputs together. I don't see any reason to believe that using the center tap with one of the others will reduce available current.

If you need DC output from the transformer, you can use all three outputs to get full-wave output with (marginally) less circuitry than a full-wave bridge requires.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

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Electronic – Transformer has continuity but no power on secondary coil

Your update tells me that your initial statement "...110V Ac is present to the terminals that supply power to the transformer..." is misleading. It implies that you are measuring at the transformer terminals - which you are not.
It is clear to me, that there is some safety circuit between the board terminals (were 110V comes in), and the transformer's input terminals.
Your safety circuit is detecting an unsafe condition and disabling power to the transformer primary terminals. There is nothing wrong with the transformer. Most likely, your power output transistors are shorted/blown, causing the safety circuit to remove power to the transformer.

Electrical – How to find step down transformer secondary output current rating

My standard way to initially estimate the total VA rating of a transformer is to weigh it, and then compare the weight with published figures from a transformer catalogue. From for instance, RS conventional EI transformers weighing 500g are rated at 20VA, whereas toroidal transformers of the same weight are rated 30VA.

This is the total VA, what you would expect the primary to handle. You have to divide this amongst the secondaries.

You have the measured voltage \$V\$ and the measured resistance \$R\$ of the windings.

There are two limitations to the max current that can be drawn from secondaries, regulation (voltage drop) and temperature rise.

Regulation is easy to handle, as it can easily be estimated from the current we want to draw \$I\$ as \$V_{drop} = IR\$. This doesn't damage the transformer, only affects our load, and whether the voltage is sufficient at the load.

Temperature rise is more difficult, and can damage the transformer. We can see how warm the transformer gets to the touch, but that doesn't tell us whether one particular winding is getting too hot or not.

If we assume that all windings are cooled to ambient in the same way (which is not too wrong, especially for a first stab, and especially for a toroidal) then the VA of a winding is proportional to the mass \$m\$ of copper used in it, regardless of the number of turns \$N\$, length \$L\$ or wire area \$A\$.

As \$V\propto N \propto L\$, and \$R\propto \frac{L}{A}\$, we can see that the mass varies as \$m \propto \frac{V^2}{R}\$

I'll leave it as an exercise for you to prove this, hint, dimensional analysis helps (don't forget the dimensions of the constants of proportionality). Note that the expression itself has units of power, which we would expect as it's supposed to help estimate VA.

So, for each secondary, calculate \$V^2/R\$, and apportion your total 20 or 30VA in proportion, this is your estimate of VA for each winding.

Having got initial estimates for the VA of individual windings, fire the transformer up from a fused supply. Load each winding with a resistor to draw half your calculated VA, and measure the voltage drop. Make sure this is reasonable (a few percent for big transformers, possibly 10% for small ones) for all windings before you continue the test.

Load all windings with half your calculated VA, and let it run for 30 mins to reach thermal equilibrium. Now disconnect everything, and quickly measure the resistance of each winding before they have had a chance to change temperature. You can estimate the temperature rise of each winding by knowing that copper has a tempco of 0.4% per degree C at room temperature. For example, if your 2.5ohm winding went to 2.75ohm (+10%), that indicates a 25C rise above ambient. You may need to do a four-terminal measurement to get differences accurate enough to be worth using at the ohm level.

If any winding is particularly hot or cool, you can vary the proportional of the total VA going to it, and try again. The maximum temperature a winding can reach is governed by the insulation used on the transformer. I don't like to go above 70C (remember if you box the transformer up, its ambient will increase) without knowing more about the specific wire used.

Before you finally test the transformer at full VA, remember the temperature rise of the transformer is proportional to \$I^2\$, so the half VA temperature rise you measured was one quarter of the final, not half!

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