We don't see iron losses equal to copper losses in a transformer.
Iron losses are due to the input voltage, so are the same (roughly) at no load and full load. In fact, they're slightly lower at full load as the primary voltage drop requires less flux swing.
Copper losses are due to the current flowing, so are small at no load and large at full load.
Checking SMPS transformers is always a bit hard:
You probably have no information to information that would tell you what the intended winding ratios are; and without, shorts between traces on either side are hard to detect.
Even worse, if it's e.g. a flyback configuration, it'd be non-trivial to reliably test isolation non-destructively, since these ...
Many old literature can be found. This is a magnetic amplifier, those days, the power amplifiers were made similarly as that. Book
If the magnetic cores are isolated and there is no common magnetic path, then the result is much simpler:
Both fluxes (...
No, do not get vinegar on this part or you could cause it to fail. Vinegar and things like Naval Jelly (a thixotropic concoction of dilute phosphoric and sulphuric acids) are highly conductive electrically.
The rust causes no harm, just ignore it.
and the secondary current so produced creates a flux equal and
opposite to that produced by the primary winding
I think there may be some misconception here. The secondary current that flows is because of the load on the secondary. This produces a flux that is entirely cancelled by the increase in primary created flux caused by that load.
In other words ...
If you are controlling the pulse width but not using a DC restore circuit, it will mean that at a high value of duty cycle, the peak voltage from the transformer will not be enough to properly turn on the MOSFET.
Fact: the output of a transformer must have an average value of 0 volts and if the original pulse shape is 10 volts for 90% of the time, it will ...
Are Lp and Ls only due to leakage flux or they generally represent
self inductances of both the coils? If yes, what about the rest of the
flux, won't that induce emf?
Lp and Ls are leakage inductances due to imperfect coupling between the primary and secondary. Lm is the prime mover when it comes to flux because it is the magnetization inductance: -
The following is largely complementary to Dave Tweed's answer
The welds are generally along the surface and can be removed with an angle grinder "fairly easily". The laminations can be rebuilt using a clamp to hold the core together. Epoxy helps prevent lamination chatter but DO NOT introduce any additional air gap (or "epoxy gap") between laminations.
Why is it driven with rectified mains which equals to 375 VDC?
It isn't, the rectified mains is alternately switched onto the primary of the flyback transformer (in order to store energy) then it is disconnected thus allowing most of that stored energy to be transferred to the secondary winding. U1 does that switching: -
the duty cycle does in fact, like ...
Is AC in general by-default expressed in RMS values?
Yes. An unqualified AC voltage is usually understood to be RMS. Otherwise, it would be specified as peak-to-zero (Vp) or peak-to-peak (Vpp).
Is the "120VAC" here in the US actually closer to 170V peak-to-peak?
120VAC mains is about 170V peak-to-ground. It's actually double that, close to 340V peak-to-...
Is AC in general by-default expressed in RMS values?
Yes. AC voltages are usually in RMS. So when we say the wall outlet is 120VAC, it means 120Vrms, which is 170Vpk or 340Vpk-pk.
The reason is because it's more straightforward to do power calculations with RMS. If I tell you the wall voltage is 170Vpk, you have to jump through some extra hoops in power ...
If you're going to use a MOT as a 'normal' transformer, then be aware it's been optimised to the max for low cost and high power output, when blown with a fan. This results in the flux and consequently the magnetising current being very high, comparable with the load current. It's good to put another 25 primary turns on (50 outside the US) to reduce the ...
For this I can increase the number of turn on the primary side N1. Nevertheless I will increase the magnetic excitation H which is equal with some appoximation to H = N1*I1/l. It will then saturated even "higher" my core.
That would be entirely true if the magnetization current remained the same but, it doesn’t...
Increasing the number of primary turns ...
Yes technically this is possible ... although I would not recommend bringing this to practice without a firm understanding of the basics!
You say low current, but depending on who you ask this might be different by orders of magnitude. (ECG specialist vs welding engineer. You get the idea.)
Can you elaborate on the current requirement or application?
It’s a common-mode choke (noise filter, not a transformer per se.)
That said, it might make more sense to replace the entire power supply than to attempt to repair it yourself. That unit was subjected to a severe stress - overvoltage surge probably - and likely has other damage that you’ve yet to find.
Power supplies are cheap; your personal safety is not.
What that formula is computing is the approximate effective airgap area, the area that the magnetic flux passes through to complete its circuit round the core.
In the limit of large area pole pieces and small gap, the airgap area is just the area of the poles, W*d. As the lg increases however, this becomes a less good approximation.
The reason we need to ...
Well AC voltages are usually rated in RMS, because if you put 10VAC or 10VDC into a resistor, it dissipates equal amount of power and thus generates equal amount of heat.
But it is approximately right that 25 VAC can approximately be converted to 25 VDC. First of all, there is voltage loss at the rectifier diodes, if roughly 1 V per diode, the drop on full ...
So if the current through Lp is not null, as the current to the
secondary is proportionnal to the current through Lp, it is not null
The current in the secondary is NOT proportional to the current in the primary; the voltage on the secondary is proportional to the voltage on the primary. Of course any secondary current modifies this but, until the ...
The inductance of the whole primary winding (pins 1 to 3) is given as >475uH. The inductance of a half winding is therefore (going as N^2) >119uH.
As the turns ratio is 1.3:1, and again inductance going as N^2, the inductance of half of a secondary winding is >200uH.
This is the inductance of the winding, not the inductance that the transformer presents to ...
The reason I'd like to know the inductance at the secondary is to
calculate some RLC filter with it (to match the output filtering to my
You are misunderstanding how this forward converter works; it appears you might be thinking it operates like a flyback converter, which it doesn’t.
The inductance that can be used in the secondary in terms of ...
I would like to know why leakage inductance is function of the air gap
transformer. How does it influence the leakage flux ? What is the
relation between them ?
With a transformer core of high effective magnetic permeability, you could say that the flux produced by one winding 99.9% couples to the other winding. The closer the coupling is to 100%, the ...
It's probably a switching supply built to drive a neon tube (they require a high output impedance since the tube itself has a negative resistance characteristic). The device would rectify the 230VAC input to create about 300VDC and then chop that to get 10kV using a high-frequency ferrite-core transformer. I would expect if it worked you might get about 1/20 ...
Sounds similar to a telephone ring tone generator, for example like this one, based on the LT1684:
The LT1684 is a more complex chip that actually generates a PWM sinusoidal wave.
Generating a square wave is easier, you can just use four MOSFETs or IGBTs and bootstrapped MOSFET gate drivers (eg. from IR). A full H-bridge with ~ +95VDC supply can be used. ...
Take transformer design one step at a time.
Do you need an air gap or not?
It depends what type of transformer you want. Do you want a normal, forward, power transformer, with output voltage always a turns ratio of the input voltage? Or do you want a flyback transformer, which stores energy, only to release it into the load at a voltage defined by the load ...
If you bought the cheapest transformer you could find, the full-load losses might be as much as 240 watts and the no-load losses as much as 60 watts. That estimate is made with a small amount of internet research.
As you correctly pointed out, a quasi-square-wave resonant converter colloquially called a quasi-resonant converter (QR) is a self-relaxing converter: there is no clock in the controller except for clamping purposes. The switching frequency varies with operating conditions. It is low at low line and high power, it is high at high line and low power.
You need to pay attention to the wiring, the way you have drawn it in LTspice is wrong. Then again, the source picture is terrible, too. It's always a good thought to double check where you get your schematics from.
You should also know this schematic is a very, very crude one that should not be used in practise.