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filter inductor and DC

filter inductor and DC

filter inductor and DC

(OP)
Hi,

I have a 3phase 8.8mH inductor rated for 4A and switching appications (3 columns regular laminated iron core), used for output filter in a VFD.

In some motor configuration VFD fails starting in about 10s and trips by overload.

I took some measurements, and it happens that in the initial freq. ramp-up, it stucks at almost zero freq.,
phase current is then being about a triangular wave at switching freq. of 0,2A DC level, 5A peak to peak, until it trips.

I did a very simple simulation of an equivalent circuit, and got a similar current wave, but using 1mH inductor instead of 8mH.

The question is, ¿could that small DC level seen saturate the inductor rated for 4A AC?

Regards
 

RE: filter inductor and DC

It probably would not saturate an inductor, maybe a transformer. But the air gap in the inductor covers a lot of sins.  However, even if it had saturated the inductor, that shouldn't prevent the motor from running. When the core saturates, the inductance goes to the air core inductance of the inductor (typically well below 10% of rated inductance).

Neil

RE: filter inductor and DC

(OP)
Maybe is just coincidence, but 10% of 8.8mH is about 0.9mH that is the value that matches the behaviour in the simulation.

I passed the question to the inductor manufacturer, lets see what they say...

One thing that I do not fully understand is that if inductor is rated for 4A AC, then it will move each cyle in the magnetizing B-H curve up to a B1 field level, but certainly a non saturated level.

Why a small level of DC could cause so much effect if I<<4A?.
In this case, H field will be much less than with 4A (independently of the freq.), so my understanding (probably wrong) is that B value should be also in non saturation zone.

Please let me know what I am missing.

Regards
 

RE: filter inductor and DC

Well, like I said, it probably isn't saturating but keep in mind that the inductance changes (drops) as the core nears saturation. This is because the permeability of the steal drops with flux density. So effectively as the inductor reaches saturation, the inductance is falling and the effect is that it assymtotically approaches saturation. Volt*Seconds in the core is causing saturation and volts change more slowly as inducance falls. This holds true nicely for a time varying waveform because every half cycle the flux get's zeroed. But when using DC current, the flux never gets driven to 0. Remember Volt*Seconds. at 60hz, the flux is only rising for a very short period of time. How long is the current on with the DC amps? and when it turns off, is there a circuit or mechanism to eliminate the residual flux?  If not then the next time the current is turned on (and in the same direction, DC), the flux builds or integrates.

Neil

RE: filter inductor and DC

(OP)

So for DC, change in Volts*seconds is actually zero as dv/dt is zero, so you get the DC flux level corresponding to the DC current level (independently of the freq.)

it remains however the DC bias and cumulative residual flux phenomenon,
that remanent flux, is it really so significative in % of the saturation flux?
(sorry fo the open question, but just want to have an idea about how much could affect)

Regards
 

RE: filter inductor and DC

With regards to residual flux. This gets to my point about the likelihood of a transformer saturating much more quickly than an inductor.  Inductors typically have much lower permeabilit, made so by air gaps. The air gap allows the residual flux to seek a very low level.  Unlike transformers which usually have no intended air gap ( the only air gap is that which is needed to assemble the core) thus the transformer has much higher permeability and residual flux does not reset.

 

Neil

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