inrush transformer current..........?

inrush current transformer

inrush transformer current............?
i know this that inrush transformer current is transient current in transformer.... but i am looking for more details...........

how to calculate inrush transformer

You can consider a transformer as an inductance (the magnetizing inductance of teh transforemer, equal to the primary inductance), in parallel with a resistor (the reflected load).
Now you are applying an aC voltage to this circuit. You know how to calculate the current and you know it will be sinusoidal, which means it will have an average value of zero.

But what happens if you apply a DC voltage (actually, a step voltage) to this circuit? The current will ramp up exponentially to a value determined mostly by the resistance in series with the circuit. When you remove the DC, which is like applying a negative step, the current will ramp down exponentially to zero, with a time constant given by the resistor in parallel with the inductor.

So, you have a fast exponential when appling the voltage and a slow exponential when removing it.

Well, when you apply an AC voltage, in the beginning it looks much like applying a DC voltage, because only one part of the sinewave is "visible" to the circuit when you apply it. As time goes by, the actual voltage does look like an AC voltage, after at least one cycle, because now the average is truly zero.

So, when you connect an AC voltage you will see much the same phenomenon, which means the current will go up rapidly (although it maintains the sinusoidal shape) and then come down as a sinusoid riding on an exponential.
That makes the first peak larger than the steady-state current peaks.

To minimize the inrush current, you should switch the voltage on at 90°.

Take a look at the pictures.

VVV, can you explain: inrush current of the toroids (O-core) tranformer is much bigger than that of common (E-I core) transformer of the same power.

I have always to consider this matter when using a toroids transformer.

I have realized that but have no instruments to watch or recording the processes, and have not found out the book about this sympton either.

nguyennam

Frankly, I was not aware of that.

The only way I can imagine why that would happen is by considering the core characteristic. Since the toroid has a well closed core (little or no airgap), it is much easier to drive the core towards saturation, than it is to drive a laminate one (this inherently has some airgap).

My simulation did not take into account any core saturation effects.
I am not saying that the core actually saturates, but that at power-up it can be driven way higher on its hysteresis curve, which effectively lowers its permeability, resulting in higher inrush.
A core with a significant airgap will have its characteristic much elongated and would behave almost linearly, resulting in little difference from my simulation.
That kind of core would be approximated relatively well by my model.

For the time being, model the transformer as a inductive only. Apply a voltage v(t)=Vmax*sin(w*t) switched on at t=0. The current is i = 1/L *int(v(t)dt)
This gives i = =(-Vmax/L/w)*cos(w*t)+ C.

Evaluate C from the initial condition i(0)=0 gives c = Vmax/L/w

i(t) = Vmax/L/w * (1 - cos(w*t)).

The solution for this pure inductive circuit closed at this point of time if a sinusoid with a large dc component Vmax/L/w.

If you add in the resistance, then the dc component starts at the same worst-case initial value and decays according to exp(-t * L/R).

But look at the initial period before the dc decays away. Now you have an exciting current i(t) which is up to twice as high as normal. Actually this represents the flux up to the saturation point. When you cross the saturation point, the flux now only increases very slowly with current and you need much higher di/dt to match the applied volts. Or if we neglect this we can also consider that now the current increases the point that it is limited by the resistance. This creates very high current peaks. The peaks are assymetric ...ie. not the same for the positive peak as the negative peak. In this example only the positive peaks force the core into saturation which dramatically increases the current further. The positive peaks are very high and the negative peaks are normal. The waveform will be rich in harmonics. I have done a matlab simulation of this... will dig it out if I get a chance.

In summary I would say that neglective saturation would give much lower result for inrush current than actually occurs. The dc component forces the core into saturation which greately increases the inrush current.

Yes, what electricpete and VVV say is completely true. In fact, when the transformer reaches saturation it can act much like a rectifier of sorts. I would also add that the realitive permeability of a core in saturation is 1 (i.e. the same as an air core inductor).

Another thing to consider, if the core was switched off at the positive peak then the core will retain some magnetic field because the domains do not disorient themselves (i.e. you are starting with a residual flux in the core). If you then switch back on at the negative peak then the flux will increase and the core will saturate quicker than if you started with no flux. As a result, You will have a much higher inrush. It is easiest to think about this by drawing an H-B curve for a transformer with a steep slope up to the saturation point and a shallow slope beyond the saturation point and think about what happens as you move along the curve with different starting conditions.

Generally, inrush is dampened by the load on the transformer and the losses in the transformer and eventually you end up with steady state load. In rush on a loaded transformer will decay quicker than on an unloaded transformer. When choosing a fuse to protect the transformer, inrush must be considered so that the fuse won't blow during the inrush.

In summary, the magnitude and decay of the inrush current depends on the residual flux in the core (i.e. when it was switched off), the starting voltage (i.e. when it was switched on), the transformer resistive load and losses, the L/R ratio, and the core saturation characteristics.

Hope that helps clarify the excellent information already provided.

-jonathan

i thank each and every reply placed here.......
i have exam at my doorsteps and i wish this question appears in my question paper............

Thank you mannnish for openning this topic and thank you all for the answers. I have been facing the inrush currents of transformers and have experience about that, but I have never understood why and how they are.

nguyennam