Part 2-FARFISA ART-482/1 door entry intercom repair. About Transformer calculations
Well, after so much of “background noise” about the transformer of the previous article of mine here https://www.jestineyong.com/farfisa-art-4821-door-entry-intercom-repair/ and in order for me to help the readers of it to obtain a better understanding of this specific task of identifying the transformers overall parameters, I considered inevitable the fact of giving some more practical information on what I did.
The burned transformer is shown again below:
First of all, when I opened the (EMI suppressing) side covers of the transformer’s core I noticed that it was wound in two separated parts of the bobbin. The lower one contained the two (separate) halves of the primary winding, while the upper one had all the secondaries wound there. This separation made my job of winding it off much easier than what it would be if all the windings were wound all over the core area, in layers, one after the other. The first secondary winding, that one nearest to the core, was the bi-phase winding for producing the various d.c. voltages and above it was wound the lock mechanism winding.
So, after removing the insulating papers from both windings, I immediately realized what had happened there (apart from the terrible smell of it). The external winding for the lock supply was in terrible condition with its turns stuck together. It was really very difficult to wind it off. I had to apply noticeable strength in pulling the wire. Needless to say that it was totally discolored, almost black, because it suffered an enormous overheat. Anyway I removed it and counted both, its turns and its wire diameter. It counted 50 turns and its wire diameter was 0,9mm.
At this stage I ran a “lamp test” on it. Unfortunately it was negative. The lamp was lighting in full brightness.
Then I continued with the bi-phase winding which also was a bifilar one. It counted 2×47 = 94 turns and each wire (of the two wound in parallel) had the same diameter as the previous winding (0,9mm). So this winding consisted of two strands of 0,9mm diameter enamel wire.
After the removal of the last secondary ran a second “lamp test” just to see the condition of the primary which was obviously discolored as well. Although unexpected, this time the result was positive. No light in the lamp. A sign that the primary winding had no problem.
I could stop here and simply rewind the turns of the two shorted windings without performing any calculations at all. I could buy some new enamel wire of 0,9mm diameter and that would be all, instead of dismantling completely the windings, namely the primaries that were left on the core. But this would be a mistake of mine for sure. The reason is that this overheating had already caused discoloring of the primary winding as well (and this is quite natural. To be frank, I didn’t expect to find the primary windings intact when I first saw the transformer without its insulating papers, which were really burned out, especially at their inner side, and caused such a terrible smell). So this solution would be a “short-termed” one. Then I would very soon be confronted again with the same problem.
I was convinced that this idea was correct during winding off the primary windings. They were also stuck together and many times their wire, being of thinner diameter (0,6mm), was many times cut with relatively little pulling force applied to it in order to remove it. This effect confirmed to me the correctness of the need for an overall rewinding. The two primary halves counted 412 turns each, which makes a total end-to-end primary winding of 824 turns.
Now I had all the info available and I could either rewind the same core or order an exact copy of it by giving a local constructor the parameters I found by calculations (which I will give you below).
When I contacted a famous transformer constructor here and told him what I needed, he avoided to accept the rewinding solution claiming that in my try to dismantle the core (which I failed to do because of its stuck laminations and no time to use a bath of chemicals for that purpose) I had distorted the shape of some of them and they had to be “aligned”, which would a) need time and b) be costly. When I asked him about his pricing options, he told me: “more than 30€ for “alignment” and rewinding, 45€ for a brand new unit”.
In such a case one doesn’t need to think much about making a decision. I ordered a new unit, giving him the parameters I had found. Later on I received almost a copy of the original. The core of it was of different shape only (but of same wattage) and all I had to do was to drill two new holes in the metal case which holds it.
Now about the calculations…After the second and positive lamp test, seeing that the primary was still in good operating condition, I wound 4 turns by hand in the empty space of the secondaries, using a piece of normal cable. Then I fed the transformer with line voltage, using my little variac and supplying exactly 220V a.c. Next, I measured 1,068V a.c. at the cable ends. Just for the purpose of verification and greater accuracy, I left only one turn in the core and I took a new voltage measurement. It gave me 0,267V a.c this time. So, I had already the 0,276V/turn factor in hand.
The (taken as of unknown voltage magnitude) secondary with the 50 turns in it used for the lock mechanism would now produce:
0,267V x 50 turns = 13,35Va.c. output.
The second (bi-phase winding, which mean a winding with middle point tap) would produce:
0,267V x 47 turns = 12,549Va.c. per each half, namely 25,098Va.c.
Next, as regards verification for the working voltage of the primary winding, this should work at a voltage level of:
0,267V x 824 turns = 220,008Va.c.
This was indeed the exact input voltage I was supplying to the primary winding through the variac.
So the overall verification as regards the working voltages of all windings was already established and the picture was now crystal clear.
One might say: “O.K, but if you couldn’t take any measurements at all, just because the primary was shorted, what would you do then”?
This is a very reasonable question which had occurred to me as well during the preliminary tests I ran for detecting shorted windings.
The answer is that I would go on with the classic method. Let me explain it below.
Having myself counted the turns of the windings, I already had their ratio. This is given by the fraction:
N1/N2 = 824 turns/50 turns = 16,48
Therefore the first secondary with its 50 turns would produce a voltage of:
220V/16,48V = 13,349Va.c. or 13,35 Va.c. as calculated above.
This holds true because the turns’ ratio and the voltages’ ratio are always equal and they form a so called “analogy”. This means in turn that:
N1/N2 = U1/U2
We simply solve this equation for U2. This will result therefore in:
U2 = U1 * N2/N1 = 220V x 50 turns / 824 turns = 11.000V * turns/824 turns = 13,349 Va.c.
Or more simply, using the turns’ ratio as a decimal number instead of a fraction and solving again for U2:
U2 = U1/ 16,48 = 220/16,48 = 13,349Va.c. again, as above.
What is missing now is a correction that should be done, due to the fact that I was testing with a voltage of 220V exactly, while the primary winding was designed for a maximum input of 254 Volt.
Solving the same equation formula as above, we have:
U2 = U1 * N2/N1 = 254V x 50 turns / 824 turns = 12.700V * turns / 824 turns = 15,41Va.c. for the winding feeding the lock mechanism.
U2 = U1 * N2/N1 = 254V * 47 turns / 824 turns = 11.938V * turns / 824 turns = 14,48Va.c. for each half of the second secondary winding, or 28,96Va.c. end-to-end voltage.
When I ordered the transformer, the nominal voltages I gave (for 254Va.c. input) were 15,5Va.c. for the first and 14,5V+14,5Va.c. for the second winding.
I also gave the man a drawing of its internal circuit (drawn by hand) as shown below.
Now, as regards the currents of the windings:
Both the secondaries had the same wire diameter of 0,9mm, with the bi-phase one being also bifilar.
At this point it would be helpful I think to inform you about the industry standards related to the wire diameter in relation to the working mode of the transformer in its real life.
A transformer designed to work round the clock, as it is expected, has different design than another one of exactly the same type which is intended to work only for a few hours per day.
The provision usually applied on the purpose to increase the longevity of a hard working transformer is to give it a plus percentage in its turns’ number as compared to their exact calculated number. Both the primary and the secondary windings should be given this equal percentage.
Along with this and similarly doing, a plus percentage is given to the wires’ diameter as well. The total result will always be a much more comfortable round the clock operation of the transformer, in terms of the magnitude of the temperature it will develop under nominal loading. I think that the reasoning is easy. Increasing the wire thickness, its resistance decreases. This in turn decreases further the copper losses of the winding as the voltage drop in the Ohmic component of the winding is also reduced accordingly.
All of these above mean that in the second case, the current density standard we take for assessing the cross sectional area of the wire (and therefore its diameter) is 2,55A/mm2.
In the first case, where the transformer has to work round the clock, we take lesser current density, say 2A/mm2 or even lesser, depending on the case. About that you can find relevant data in various tables included in books or in the web.
Next, what we need first is to calculate the cross sectional area of the 0,9mm diameter wire used in this case, in order to assess afterwards the maximum permissible current flow through it.
So if we take the usual 2,55A/mm2 standard of current density factor in account, the cross sectional area of this wire is given by the formula:
S= π * R2 = 3,14 * (0,9mm/2) 2 = 0,636mm2.
And the permissible current flow through the wire is given by the formula: S= Io/δ, whereas δ = 2,55A/mm2 = current density factor.
Thus: S= Io/δ and then solving for Io:
Io = S * δ = 0,636mm2 * 2,55A/mm2 = 1,62A
This is the maximum permissible current of the door lock winding. Do you remember that the fuse rating of this winding was exactly 1,6A?
For the bi-phase and also bifilar winding, since two strands of the same wire diameter were used, we will have as maximum permissible current flow through them the double value found previously, namely 3,24A.
Now it is time to calculate the total secondary power:
1st winding: 15,5V * 1,62A = 25,11V*A
2nd winding: 29V * 3,24A = 93,96V*A
And the total Psec is: Psec = 25,11 V*A + 93,96 V*A = 119,07V*A
Assuming now an efficiency of 85%: 119,07V*A/ 0,85 = 140 V*A (worst case)
Note that the cross sectional area of the transformer core as I measured it was:
3,7cm * 3,9cm = 14,43 mm2.
The assessed power above calls for the use of a transformer having a cross sectional area of:
S = 1,2 * √Ptot, whereas Ptot = 140V*A
Thus substituting Ptot and solving for S:
S = 1,2 * √140V*A = 14,2 mm2
That is the cross sectional area needed for the transformer to be used.
Since the resultant value is smaller than the one established by the physical dimensions of the cross sectional area of the used transformer, this means in turn that its core could handle this power absolutely safely.
Now as regards the current flow through the primary winding, this will be of the order of: 140V*A/ 230V = 0.609A, or 609mA.
The wire diameter chosen by the factory was 0,6mm. Is this diameter suitable to handle the primary current shown above? Verification:
This diameter gives a cross sectional area of: S= π * R2 = 3,14 * (0,6mm/2) 2 = 0,283mm2.
Thus: S= Io/δ and then solving again for Io: Io = S * δ = 0,283mm2 * 2,55A/mm2 = 0,722A, or 722mA, which means that the wire size chosen for the primary is adequate in order to handle the reflected secondary load on it plus the total losses. In fact it will work with the 84,46% of its maximum permissible value.
Hopefully this analysis will answer all the possible questions you might have after reading my previous relevant article.
This article was prepared for you by Paris Azis from Athens-Greece. He is 59 years old and has more than 30 years’ experience in electronics repairs, both in consumer and industrial electronics. He started as a hobbyist at the age of 12 years and ended his professional carrier as a senior electronics technician. He has been a specialist in the entire range of consumer electronics repairs (: valve radio and BW TV receivers, transistorized color CRT TV, audio amps, reel and cassette tape recorders, telephone answering and telefax devices, electric irons, MW cooking devices e.t.c) working in his early stages at the official service departments of National-Panasonic first and JVC afterwards, at their premises in Athens.
Then he joined the telecoms industry, working for 20 years as field supporting technician in the sector of DMRs (: Digital Microwave Radio transmission stations), ending his carrier with this subject. Now he is a hobbyist again!
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