Isolation transformer draws excessive current under no load condition

I don't see how a radar detector is going to help him.

On Fri, 10 Feb 2012 19:22:33 -0000, "Gareth Magennis" put finger to keyboard and composed:

My local Aldi has these for AU$13:

The meter also reports the power factor.

Same model number (GT-PM-04), probably made by Globaltronics:

- Franc Zabkar

--
Please remove one 'i' from my address when replying by email.

Yes. I've measured it with the same result using both a Watts Up meter and a Fluke 87.

Making a humming noise, yes. It's not getting hot, the external temperature rise of the case is only about 10c.

Nail. Head.

My test setup and how this all started:

120V in is being fed to a variac, then to the isolation transformer, then the load. On the secondary of the transformer I added a toggle switch that would connect the secondary in series or parallel depending whether I wanted 120V or 240V out. When I first wired up everything as I stated in my OP, being on the cautious side I placed a 1.5A fuse into the variac so if I mis-wired something there'd be no damage. Well, on first power up I slowly raised the voltage on the primary and at a little over 110V the fuse blew.

The actual wattage when the primary is at 120V is only 42W as measured by the Watt meter. Here's why - the power factor is really crappy at just about .15. This would explain everything. I wonder if this is typical of a

1KVA transformer? I never thought to check the wattage being drawn...

Noted, and thanks.

Also, thanks to everyone else for their input - some very good thoughts. It would appear that there's no problem after-all.

It's potted inside the box, so that's not possible. In any case it would seem that there's no problem (see my other post), thanks.

Out of curiosity, I took out my Tenma 72-545 1.5A isolation transformer.

On my Kill A Watt, it draws 60 mA unloaded, 3W / 7VA, with a power factor of

0.48.

I assume the current without a load varies with the primary inductance, not the rated capacity. So the 2.45A drawn by the OP's transformer seems to be unnaturally high.

=A0-

Adding caps will indeed have NO effect on the tranny, but WILL improve the PF as seen by your AC mains, which means the amount of power you drop in your wiring [and pay for] will be less.

Adding caps is a STANDARD way to adjust power factor to 1. Used by the industrial power consumers that are heavily penalized for lagging power factor. It is cheaper for them to add a building full of caps, just to shift their PF and pay less for power.

If you do the analysis of power consumption throughout a standard AC mains power distribution, you will find that a lagging power factor [caused by motors, etc] INCREASES the power required to simply get billable wattage to you. And, it's surprisingly large.

"Robert Macy"

** Is there any way to shut fools like you up ??

Adding caps will indeed have NO effect on the tranny, but WILL improve the PF as seen by your AC mains,

** Absolute BULLSHIT !!!!!!!!!

which means the amount of power you drop in your wiring [and pay for] will be less.

** Absolute IDIOCY !!!!!!!

Adding caps is a STANDARD way to adjust power factor to 1.

** But never used with an off load transformer - because that idea is 100% STUPID.

YOU are an obsessed IDIOT with a one track mind.

Piss off.

... Phil

"JW" "Phil Allison"

** Good.

** Sounds about normal for a large e-core tranny ** That is pretty high actually - a typical 1KVA tranny has about 6 to 8 % power loss at full load. ** Not everything.

You have to know that the current waveform is distorted - THIS fact is causing the poor PF when unloaded.

... Phil

100%

I stand by what I said as technically correct.

Many, many technical journals, text books, and supporting calculations based upon terms of definition confirm what I said.

Quit spewing vitriole, making erroneous statements, *and* misleading.

CONTRIBUTE, EDUCATE, and LEARN!

s 100%

What are your PFC capacitors doing when the transformer supplies a load? And if you were to switch them out when there was a load on the secondary, wouldn't it make more sense just to open the primary when there was no load? Then you wouldn't get any idle current draw at all.

e

is 100%

The cap in parallel will continue to 'remove' the inductive current - as far as the AC mains sees it. The inductive current remains whether the transformer is loaded or not.

Completely disconnecting everything from the AC mains will always save power.

As far as I know, no industrial power consumer adds caps just for transformer core currents, only for large motors that are extremely inductive loads.

"Robert Macy = Troll "

** Is there any way to this MORON up ??

( Load of totally absurd crap deleted )

... Phil

But, except for the extra power dissipated in the household wiring, is not measured by the meter, and not included in the billed energy units.

Sylvia.

=A0-

True!

Today, the only way that utilities companies can bill for this lost energy is by assigning a 'penalty' for bad PF.

I'll bet with today's smarter metering, they'll figure a way to measure it real-time and bill for it.

God knows, we've tried... but you keep posting.

I use google access so the thread line is NOT preserved.

William,

Is this email for me?

Is there an error(s) in what I posted?

And if we correct our power factor, our bills will be lower than they are now.

Oh, wait a minute, what was I thinking?

Sylvia.

to 8 %

Agree with that range.

Assume for this transformer the power loss is 8%, 80W. Further assume this is a 'well designed' transformer where the designer allowed half the dissipation in the windings and half in the core. That means 40W in the core and 40W in the windings, at full load. Four windings with each carrying half the full load current, 4.167A, implies 0.576 ohms/winding two in series would be 1.15 ohm and since most pri/sec are 45%/55% that would mean each pri winding 0.52 ohm each sec winding 0.63 ohm

Now unloaded the two pri windings are passing 1.27A each, dissipating almost 2 W and the core is dissipating around 40W, so JW would have measured power for the unloaded transformer at around 42W, which he said he measured.

JW, Did you ever measure the winding impedance? It would be interesting to compare prediction to actual values. [of course, preaching to the choir, for better accuracy put the two pri in series and the two sec in series]

On Feb 13, 5:45=A0am, JW wrote:

=A0-

A cap does improve PF...

To answer the question whether an external capacitor placed in parallel with the input of a transformer improves power factor or not, for both loaded and unloaded conditions; I started the following analyses and share with others who may be interested.

First, use a Linear Model to represent the isolation transformer. Simulate the iso transformer and circuitry by using LINEAR SIMULATION, that is, only using linear components to represent observed data.

Second, simulate using nonlinear components, specifically the chan model, which simulates BOTH hysteresis (coercivity) and saturation. TO BE DONE LATER.

(Model of the 1kVA ISO-Transformer wound for 120/120 operation)

LINEAR SIMULATION The following transformer model is accepted for low frequency operation. The transformer is approximated by representing the transformer as primary and secondary winding resistances connected to a core which has a parallel resistance to represent the core losses. The transformer's coupling coefficient was arbitrarily set to one, more likely to be as low as 0.98. But modifying coupling coefficient does not appreciably change results.

Values for core inductance and parallel resistor were determined from data provided by JW and winding resistance values were estimated based upon experience:

Core Inductance: Lcore =3D 0.128 H, based upon JW's measurement of current with NO LOAD Core Losses: Rcore =3D 360 ohms, based upon JW's measurement using Watts Up meter with NO LOAD Winding Resistances: Rpri =3D 0.26 ohms, Rsec =3D 0.315 ohms estimated from experience

The measurements that one would obtain with this linear model are Current NO LOAD =3D 2.54 Arms Power NO LOAD =3D 41.7 W PF =3D 0.137

All fairly close.

As a sanity check, what happens at FULL load with 14.4 ohms? Vout =3D 115V, Power =3D 922W, PF =3D 0.960, which is reasonable.

Load regulation is not that good in this model. The output voltage with NO LOAD is 120V, but when loaded with 14.4 ohms to get the expected 1kW output, the output voltage drops to 115V. This manufacturer probably did what everyone else does and not wind the transformer EXACTLY 1:1, but add a few extra turns on the secondary to compensate for this expected drop, more like 100:105 or such. JW could verify the winding ratio by running the transformer 'backwards'. He would then see he'd probably only get 110 out the pri with 120 into the sec. Anybody who has used two doorbell transformers, one down and one back up just to make an isolation transformer has discovered this 'feature'.

For simplicity, I leave the winding ratio as 1:1, which will not appreciably change results.

In a previous post I said add a parallel capacitor, 55uF, to adjust the PF closer to 1.

When 55uF cap is added in parallel to the input of the transformer, the PF does indeed change. NO LOAD: PF =3D 0.997 FULL LOAD: PF =3D 1.00 And, as I claimed adding a cap improved for both NO Load and FULL Load.

I've included a copy of the simulation model below, watch out for word wrap, but feel free to try it in FREE LTspice and see for yourself.

CONCLUSION: Adding a parallel capacitor to the input of a transformer does help power factor both loaded and unloaded.

Now, this was LINEAR simulation. I will now more accurately model the non-linear characteristics of a real transformer and compare to see if my assertion is still true.

LINEAR SIMULATION:

Version 4 SHEET 1 3060 1012 WIRE 1216 752 1184 752 WIRE 1344 752 1296 752 WIRE 1456 752 1344 752 WIRE 1584 752 1456 752 WIRE 1728 752 1664 752 WIRE 1808 752 1728 752 WIRE 1904 752 1808 752 WIRE 2080 752 1984 752 WIRE 2384 752 2160 752 WIRE 2432 752 2384 752 WIRE 1728 768 1728 752 WIRE 1808 768 1808 752 WIRE 1904 768 1904 752 WIRE 1984 768 1984 752 WIRE 2432 768 2432 752 WIRE 1456 784 1456 752 WIRE 1184 800 1184 752 WIRE 1728 880 1728 848 WIRE 1808 880 1808 848 WIRE 1904 880 1904 848 WIRE 1984 880 1984 848 WIRE 2432 880 2432 848 WIRE 1184 896 1184 880 WIRE 1456 896 1456 848 FLAG 1184 896 0 FLAG 1808 880 0 FLAG 1904 880 0 FLAG 1984 880 0 FLAG 2432 880 0 FLAG 2384 752 OUT FLAG 1344 752 IN FLAG 1728 880 0 FLAG 1456 896 0 SYMBOL voltage 1184 784 R0 WINDOW 0 48 52 Left 0 WINDOW 3 47 113 Left 0 WINDOW 123 49 82 Left 0 WINDOW 39 0 0 Left 0 SYMATTR InstName Vac SYMATTR Value SINE(0 170 60) SYMATTR Value2 AC 120 SYMBOL res 1200 768 R270 WINDOW 0 32 56 VTop 0 WINDOW 3 0 56 VBottom 0 SYMATTR InstName Racmains SYMATTR Value 0.01 SYMBOL ind 1792 752 R0 SYMATTR InstName Lcore SYMATTR Value 0.128 SYMBOL ind2 1888 752 R0 SYMATTR InstName Lpri SYMATTR Value 10 SYMATTR Type ind SYMBOL ind2 1968 752 R0 SYMATTR InstName Lsec SYMATTR Value 10 SYMATTR Type ind SYMBOL res 1568 768 R270 WINDOW 0 32 56 VTop 0 WINDOW 3 0 56 VBottom 0 SYMATTR InstName Rpri SYMATTR Value 0.26 SYMBOL res 2064 768 R270 WINDOW 0 32 56 VTop 0 WINDOW 3 0 56 VBottom 0 SYMATTR InstName Rsec SYMATTR Value 0.315 SYMBOL res 2416 752 R0 SYMATTR InstName Rload SYMATTR Value 10MEG SYMBOL res 1712 752 R0 SYMATTR InstName Rcore SYMATTR Value 360 SYMBOL cap 1440 784 R0 SYMATTR InstName Cc SYMATTR Value 55=B5F TEXT 1808 936 Left 0 !K1 Lpri Lsec 1 TEXT 1152 624 Left 0 !.ac LIN 201 50 70 TEXT 1152 552 Left 0 ;.ac LIN 201 50 70 TEXT 2672 800 Left 0 ;NO Load Rload =3D 10MEG\nFULL Load Rload =3D

14.4 ohms TEXT 1648 624 Left 0 ;PF =3D RE(V(in)*I(Racmains))/(V(in)*I(Racmains)) TEXT 1400 936 Left 0 ;NO PF Correction Cc =3D 55pF\nWITH PF Correction Cc =3D 55uF