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2

The text defines Ai as inductance in mH per 1000 turns and that it's only for cores with gapped centre leg. The sentence confusing you is: "Shimming core halves will not yield reproducible Ai values over time, temperature and production spread." The manufacturer wants to mass-produce cores, making large numbers of them. They want all the cores ...


2

Shimming is a mechanical engineering term. It means to stick a thin material between two things to separate them. What is the meaning of "reproducible Al values over time"? It means it doesn't changes over the life of the product. What is "production spread"? Production spread is how much a specification varies between individual ...


2

If you connect them together correctly you'll have the equivalent of a center-tapped winding. You can think of the "dots" on the windings as the equivalent of (say) + polarity on a DC source such as a battery- you want to connect + to - to get them to add. Note that it's equivalent to a center-tapped winding of \$2\cdot V_{Sec}\$, not \$V_{Sec}\$.


1

Yes, the two secondaries can be connected together (e.g. black and yellow) to have a center-tapped output.


0

What is the difference between open center and closed center toroidal transformer? There is almost no difference. Just one is ready for mounting. Large toroidal transformers are usually bolted this way simply because it is convenient. Smaller ones may be glued into a base, either horizontally or vertically, with nothing in the center. Consider that the ...


0

What you call it closed is a mounting accessory. You drill the plate where you want to fix the transformer and then you bolt the transformer, the metal cap is a retainer.


0

i) ON THE LOW TENSION SIDE: The base for low tension side is 75MVA, 5.42KV Actual Value = 0.6 ohm Base Value = [(5.42)(5.42)]/(75) = 0.3916 ohm Per unit value of RL = (0.6 / 0.3916) = 1.52 ii) ON THE HIGH TENSION SIDE: The base for high tension side is 75MVA, 94KV Actual Value = 0.6[(9494)/(5.425.42)] = 180 ohm Base Value = (94*94)/75 = 117.81 ohm Per unit ...


8

Any given core material, cross section and frequency has a maximum volts per turn. If you want to use the winding at a certain voltage, then you need enough turns to support that voltage. For instance, low frequency mains transformer steel will only run up to a peak field of 1.7 T or so before it saturates. If you had a core that was 10 mm x 20 mm, and ...


26

Number of turns on a transformer matters greatly. The turns ratio is one of many considerations in designing a transformer. The following are high-level considerations when designing a transformer. As Tobalt stated, magnetizing inductance is important. Too few turns will consume excess current, even without a load. You need a minimum of turns to prevent ...


2

Transformer design has to optimize a lot of different variables. And I am not an expert. But the transformer primary needs to have enough turns so that it acts like a big inductor and prevents excessive current from flowing. If you take a particular transformer core, you can either wind the primary with many turns of fine wire (high voltage primary) or wind ...


5

Obviously, B has some advantages indeed. Otherwise all transformers would use a tiny number of turns. More turns generate more magnetizing inductance \$L_M \propto N^2\$. Imagine a transformer with the secondary open, which is essentially a large inductor. If you connect an AC voltage with frequency \$f\$ to the primary, it will see an impedance of \$2 \pi f ...


0

All datasheets are to be read like devil reading the bible. In your particular case, you can derive the inductance from the impedance, postulating it is the impedance at 1kHz. As you are trying to build a resonant whatever, you may as well think about +/- 10% variance of the impedance, as well as the pretty much unknown non-linearities.


0

There's not that much to get really. The info is not that complete, they don't actually say at what frequency the impedance is measured, but given that the device seems designed for telephone BW (300-3.4kHz) and they talk about impedance variation at 1kHz, it's probably at 1kHz.


6

The output current from two isolated power sources in series will be limited to smaller of the two constituent supplies. A simple '4 diodes one capacitor' supply has a lot of ripple. The output ripple will be the sum of the two constituent supplies. Depending on your application, the current limitation may be temperature rise in the transformers or diodes, ...


0

C3 and C4 overheats C3 and C4 overheat because their impedance is too high at the operating frequency, so they dissipate some of the energy passing through them as heat. You need to use capacitors with lower impedance. Those, incidentally, tend to be physically bigger, and also have higher capacity. There was a reason why the original design used large 10mF ...


1

An alternate solution would be to have only one DC power supply and an electromagnetic DPDT relay or contactor to change it over from mains power to DG.


1

The proper way to drive a flyback transformer is to drive it into saturation from the low voltage side —as much flux as possible—, then cut the driving current and let the high voltage side try to sustain the current from the flux. You don't want that in your speaker application as it creates horrible distortions.


2

I think it is not the case. There are 3 inductors coupled (one primary , and two secondary), thus the values of all inductors must be 1 H. Regarding the 2 secondary as one ... you must have 1H (ex : 1000 turns) for primary, and 4H for secondaries (which are coupled). Because of the turns number (2000 turns), and coupling is perfect. So LS1 and LS2 must be ...


0

There is a flat in the Schema. You have to connect the Battery to the middle contact of the Coil. And the Drain has to be there where the Battery was.


5

Note, the question has been substantially edited making this portion of my answer seem irrelevant. No. Never split loads across two sockets. If a plug comes out the pins may be live and present an electric shock hazard. You may already have 220 V available but you have no location code in your user profile or in your question so further help is not possible. ...


4

The word "transformer" is derived from "trans" meaning "across". "Across" can be simplified further to "cross". We now a have a literal cross, using the shorthand X. XTAL is another abbreviation by same logic, used for crystals.


0

It seems that The circuit works at frequencies of around 60000 Hz and it would be near impossible to give someone a shock without the use of a diode and rectify the high voltage output, there would be a very large skin effect and most of the current will pass through the skin. We feel an electric shock when current passes through our nerves.


3

To solve the problem we used a UPS with a pure sine wave output You don't mention any problem beside the one using a UPS. Possible reasons of the issue: The transformer heats up due to iron core losses. The output of the UPS I guess is a high frequency PWM modulated sine wave. You would perhaps need a filter or at least two chokes. The output if the UPS ...


3

The only unanswered question is why 32V. This design may be using a simple unregulated voltage from which to draw a regulated current. Thus taking the peak sine rectified voltage with no voltage drop on the diode bridge at 32, the take 0.7 of this for average with barely adequate storage caps to get 22.4V for minimum peak minus 2x0.7V bridge diode drop to ...


9

This is a constant-current output off-line switching power supply. In normal operation the output voltage is limited by the LEDs to between 10 and 21VDC. You should not connect it to an LED array with a range of forward voltage that is outside that range, in other words, so a single LED or two in series, in most cases is not acceptable. If you disconnect the ...


2

Here is the picture you are referring to. You are correct: 5 to 8 (or 6 to 7) gives you a center tap. If you connect 5 to 7 and 6 to 8 the windings would be in parallel for higher current.


0

I'd like to expend on Luiz answer. The transformer is made of turns of wire around a magnetically conductive core. You have the primary and secondary winding. As Luiz mentioned, the insulation of the wire may degrade over time, especially with the heat of an overloaded transformer. This causes some winding to be short-circuited, this will reduce the output ...


0

Based on my experience, I would say: transformer resistance does not increase significantly with time. Most of the resistance comes from the resistivity of the conductors (aluminum or copper) and this does not change over time, however there are some internal contacts that can become loose over time; These bad internal contacts could justify this ...


1

The load resistance referred to the high-voltage side is just, $$R_P=R_L*N^2=0.6\Omega*10^2=60\Omega$$ In the per-unit system we have the luxury of picking one, and just one MVA base for our entire system. The question tells you that it has been selected for you and is 75 MVA three-phase. Once you pick the voltage base of 1 bus, all of the rest of the buses ...


0

Yes that is correct. Due to conservation of energy, \$P_{in}\$ must always be greater than \$P_{out}\$. And, on an ideal converter, they will be the same. Of course, this might seem strange at first, specially on a DC-DC converter that lowers the voltage (buck converter, i.e.), because the current might be greater in the output. Nonetheless, electrical ...


3

For steady state conditions, the output power cannot exceed the input power so your calculation is correct. However the principle of conservation of energy (not power) allows for the output power to exceed the input power for a finite time. Thus you can store energy in a capacitor and then release it quickly so the output power can be high but only for a ...


3

The energy out won't ever be more than the energy in. You can have a high instantaneous output power if the DC-DC converter has energy storage, as long as the long term continuous output power is lower than the input power. Consider a flash-gun. It charges the storage capacitor over a few seconds, perhaps pulling a watt from the small batteries. For the ...


0

Depends on the possibilities of your motor driver. Each PWM motor driver itself acts like a buck converter, reducing the input voltage with regard setpoint speed. But this still doesn't provide a mechanism to limit the current and power. You have to have a motor driver with a current limit that can be adjusted. Introducing yet another step down converter won'...


0

NOTE: Line-operated appliances have lethal voltages inside. No warranties expressed or implied ... The original transformer almost certainly is a custom part designed for that appliance. Almost anything that meets or exceeds its specs will work - electrically. The 117 Vac primary can be anything from 110 - 125 Vac. The secondary current can be almost ...


0

I'm curious about the design of RF transformers with purposely designed non-unity coupling coefficient. If you are looking for some degree of precision in the value of k then use near-unity coupling transformers and use an external shielded inductor to create the uncoupled inductance you require. For example, if I made two air core inductors and placed ...


1

For example, if I made two air core inductors and placed them axially aligned in close proximity, what kind of ranges of coupling coefficient could I hope to achieve? And importantly, how accurate could I get that coupling coefficient to be? Way back in my early engineering days, we needed to implement a 50 MHz and a 500 MHz bandpass filter, with very good ...


1

In the transformer we have copper and iron losses. The copper loss is due to the electrical current I^2*R, while the iron loss is due to eddy currents and hysteresis of Weiss domains. They both produce heat, but the copper loss is relative to the electric current, while the iron loss is relative to the frequency. But what about the eddy currents don't they ...


2

... is Eddy current loss a heat loss? Yes, it is. Since the eddy currents flow "through" the core itself, the eddy current loss is still a kind of I2R loss (where I is the eddy current and R is the resistance of the eddy currents' return path). Thus, the eddy current loss produces heat and increases the temperature of the transformer. For the iron ...


5

Well, I will try to translate this Russian text at least to justify my Slavic name Cyril:) By the way, Russian is very close to my Bulgarian. Тисо-Продакшин is Tiso-Production - maybe the producer EМ-87 - I don't know what it is... maybe 87 is the year Rобм is Rcoil ("обмотка" is "coil"), i.e. this is the coil resistance = 23 Ohm W = ...


0

Turns out the tip31c was burned, and the primary was shorting to the secondary.


3

The first problem I see is there is no ground at all, spice needs a ground and there needs to be grounds on both sides of the transformer (or it makes the system of equations very hard to solve).


-2

Ferromagnetic materials have an extra layer of electrons which hold the momentum induced by the fluctuating flux and is transferred by induction itself again through electromagnetism


4

The changing flux in the core creates a 'volts per turn' around the core. This is the voltage that perfectly opposes the input voltage on the primary, and generates the secondary output voltage. When a secondary current flows in phase with this voltage, power is abstracted. The same power is delivered by the primary current flowing in phase (or to keep sign ...


6

The net (total) flux remains constant. When you allow a current to flow in the secondary, it creates its own flux in the core that partially cancels the flux originally produced by the primary. More current flows in the primary to create flux that "makes up" the difference.


0

A transformer can have many times the normal current on startup as it is unlikely that it get turned on at zero crossing. Depending on the actual rating of the transformer, this might not be an issue. So the general answer to your question is, yes, it does make a difference what side of the transformer you put the Thermistor


3

That type of flash-cap charging transformer will invariably be a flyback. You can reduce the power throughput, and hence the average current from the power supply, by lowering the pulse repetition rate. With a low primary inductance, the pulses will still need to be short. However, you only need to repeat them as often as is needed to get your power ...


1

You could try a parallel or series LC circuit running at or close to resonance by putting a capacitor in parallel or series with the primary of the transformer. However, you will need to simulate the circuit.


4

My question is, why there was this big current absorption if there was no load connected Because you connected a low voltage winding to 230V. It's not the resistance of a winding on a core that stops excessive current but its inductance. Overvoltage saturates the core before a half-period is over and before the core gets demagnetized again by the negative ...


1

Why is the magnitude of mutual flux in a transformer independent of the current in primary? In a transformer with no leakage inductance, and no coil resistance, the emf induced by the changing flux in the core must equal the voltage applied to the primary., at all times. If, in such a transformer, the voltage in each cycle is fixed, then the flux must also ...


0

I understand this fact mathematically. But why is this fact true, intuitively or physically? (I'm not really sure what I mean by intuitively, but if there's a different perspective on this, please do share) With the transformer unloaded on the secondary, all you have is one winding, that being the primary. That primary winding is just an inductor and your ...


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