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IamSmooth

Sun Jan 03 2010, 05:51PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

I am hoping a few of you here can clear up a few questions. I have, for the purpose of the questions, a 10:1 step down transformer, and the primary inductance is 0.5H and it runs off of 120vac/60hz.

Does this mean that the most current it can consume, no matter how great the load, is 120vac/(2pi*60*0.5)?
If the core saturates before this point, and I increase the number of primary (and secondary turns to maintain 10:1), will this drop the theoretical maximum current the transformer can deliver by decreasing its maximum current draw?

Steve Conner

Sun Jan 03 2010, 05:57PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

No.

If you measure the inductance of the primary with the secondary open-circuited, the result is called the magnetizing inductance referred to the primary. This determines the current drawn by the transformer unloaded, which is called magnetizing current.

What limits the output current is the leakage inductance. To see the leakage inductance referred to the primary, you would measure the primary inductance with the secondary shorted.

To a first order, saturation only affects the magnetizing inductance. Therefore, the output current of a transformer is not limited by saturation.

Sulaiman

Sun Jan 03 2010, 06:15PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141

An 'ideal' transformer has infinite inductance for primary and secondary,
real transformers have an equivalent parallel inductance (in this case 0.5H)
which as Steve said represents the 'magnetizing' current.
This current is 90 degrees out-of-phase with the input voltage
so will not move your electricity meter/cost money.
There is also some real power lost in the core due to magnetic hysteresis.

I agree that for hf transformers the output current limit is determined by the 'leakage inductance' but for practical 50/60 Hz transformers
(low frequency laminated steel core transformers)
the long-term output current is limited by heating of the wire/windings (P=I^2 x R)
and the short-term current is limited by winding resistances.
_____________________________________ __________________________
Commercial transformers usually operate close to saturation for economy,
with an oscilloscope look at the transformer output with no load
if it looks close to a sinewave the core isn't saturating.

If you have a Variac you can plot Vout vs. Vin ... worth doing at least once a lifetime.

If the core is saturating then you need to reduce the volts-per-turn,
to get more turns in the same space you need thinner wire,
with higher resistance (thinner and longer)
hence it will heat up at lower currents.

IamSmooth

Mon Jan 04 2010, 06:42PM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

So if a transformer was ideal with infinite inductance, magnetization current would be zero.

Now it is this leakage inductance that has me a little puzzled. If my transformer has an inductance of 0.5H, then the magnetization current is

120/(2pi*60*0.5) = 0.63A

right?

Now if I put a large load on the secondary, how is more current working its way through the primary past this inductor? The primary is an inductor with a set reactive impedance. Inductor primaries are even used as ballasts to limit current drawn on the line. So how come the coil's inductance is unable to do this when a secondary is linked to it?

Steve Conner

Mon Jan 04 2010, 06:56PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Magnetizing inductance acts as if it's in parallel with the load, and this is why it doesn't limit current. (It's Xm in the diagram linked.)

As you can see, the transformer model includes several inductances, so you can't say "the inductance of a transformer" without specifying which one you mean. The simplest model has two inductances, magnetizing and leakage, as we discussed above.

The answer to your specific question is that when a load is placed on the secondary, the secondary current creates a MMF that opposes the primary flux, so it can't travel around the core so easily. This causes more primary current to flow.

In the limit, with the secondary shorted, the flux can't pass through it at all, and hence can't complete a circuit of the core. The resulting primary inductance is now the leakage inductance, and since the core isn't doing anything, it's usually about the same as the inductance of the primary in air with the core removed.

Things that are in series magnetically appear in parallel electrically and vice versa. The "shunts" on a MOT or NST form a parallel path that the primary flux can take without passing through the secondary. So, it can still flow relatively easily, even if the secondary is shorted. This doesn't affect the magnetizing inductance, but it increases the leakage inductance.

Note, in the above discussion I assumed the windings to be perfect conductors with zero resistance, but for a decent-sized transformer, the resistance doesn't make much difference to the argument. In small transformers the resistance can make them behave quite differently.

Sulaiman

Mon Jan 04 2010, 07:04PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141

"real transformers have an equivalent parallel inductance"

For further info....... Google

Steve Conner

Mon Jan 04 2010, 07:10PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

I've just learnt something new!

I didn't realise that leakage inductance came in two varieties, academic and industrial.

radiotech

Tue Jan 05 2010, 04:12AM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
Location:
Posts: 1546

The primary current will never exceed the voltage applied/primary resistance. The %Impedance is the percent of rated voltage needed to circulate rated current with seconday shorted.

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