Transformer specs and calculations not adding up?

Good Day All.

First post here, So, I took an existing toroid, It said 168va 30-0-30Vac I removed one of the windings completely and then rewound the other one, calculated I should be close to 12-0-12Vac, sure enough, I measured after the rewind and was sitting pretty at 12.2Vac per winding. This is where I am now left scratching my head.... If I take 12.2Vac x 1.41- 1.2V diode drop I get 16.02Vdc on the calculator, when I measure this with the meter at the caps I only get 15.48Vdc...... The rectifier is a repurposed RS1505m and I have 2 x 4700mfd @63V..... Does this sound right? I do not know so would rather ask.

Thanks in advance.
Regards
Charles
 
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Well the RS1505M has a maximum Vf of 1.1V per diode, so the diode drops could be upto 2.2V (two diodes conduct in a full bridge rectifier, so two times the drop). Also there are other losses, and the waveform isn't a perfect sinusoid, so the factor of √2 isn't necessarily right for your mains.

An unloaded 12Vac winding should show about 13.5 to 14.5Vac, the voltage rating of a secondary winding is by convention for full power resistive loading and a droop of 10%--15% is common between unloaded and full load for a small transformer, due to winding resistances. If you plan to load this transformer heavily you'll have an even lower output voltage.

Talking of loading, did you use thicker wire for the new secondary windings? If not it won't be a 168VA transformer any more... The 12V winding should use wire with about 2.5 times the cross-sectional area as the 30V winding had - ie the same total weight of copper, and thus the same thermal power losses at a given load point.
 
Hi Mark......

Oh, Um, so, what I did was take one loop of the existing enameled copper wire and wound it once around the bear core which still had the primary on it and read that with the a true RMS meter, then I calculated the number of turns needed to get to the 12vac I wanted..:unsure: I did not think the voltage needed to be any higher..... ? The cross section of the existing wire is 0.98mm.... So, you are saying that I actually under wound the transformer...... I only want 5A from the transformer so I should be ok with the wire gauge or no?

Thanks for the reply.
Regards
Charles
 
Another factor perhaps not yet mentioned is that power companies don't guarantee exact line voltages. Usually there is some spec such as, they try to keep the voltage within +-5%, at least 90% of the time.

It also depends where a user is on the power grid. If close to a power station which regulates line voltage then voltage may be more stable. If at the end of a line, particularly one with industrial customers, then IR losses between the power station and a customer at the end of the line may vary at different times of day depending on aggregate customer demand. Also, even if close to power station, there is no guarantee that a customer power feed won't be rerouted on the distribution grid if power company decides to do so for whatever reason they may have.

As a result, if a precise voltage is needed for a particular end-use load then its pretty much standard practice to put a local regulator there. IOW use a regulated power supply, rather than rely on a transformer to achieve an exact voltage.

If in some cases its better to use a transformer for regulation then a ferro-resonate transformer may be used, but they are not all that precise either.
 
Also, here is an equivalent lumped model of a transformer:
1694869457384.png


As can be see there is more to it than the turns ratio of the ideal transformer shown in the middle. Lm is the mutual inductance, which results from the magnetic flux which is linked between the primary and secondary windings. L1 and L2 represent leakage inductance which results from magnetic flux which is not linked between the primary and secondary. Rf, shown in parallel with the primary winding, includes the effect of core losses in the iron core. R1 and R2 represent primary and secondary winding resistances. In particular, depending on which port you look into (measure from), impedances on the other side of the transformer appear stepped up or down in value according to the square of the turns ratio of the ideal transformer in the middle.
 
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Today I have a different question, I have a built up a distribution PCB, schematic attached, this has a 15-0-15 transformer one part of the board has a dedicated sub voltage delivery and a main voltage delivery, when the first tap is selected everything works as intended, but, when the relay switches tap C7 goes bang. Only the ground on this schematic is "commend" this can only be because of back feeding voltage from the higher voltage tap, my question is: Why, how and can this be prevented with out having to use a separate transformer?

Would be nice if there is someone that could explain this please?
Thank you again.
Regards
Charles.
 

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As a rule, you should never tie the DC out lines of bridges if the AC sides aren't completely floating and independent: there is always an unwanted conduction path through the diodes.
For your case, the solution is simple and requires just a single rectifier bridge:
1695114170799.png