3VA PT/VT Measurement and Safety/Accuracy

7V primary, 70V secondary is an odd ratio. Does transfo plate give a definite VT/PT standard?

7V + 20% on "primary" side would usually cause saturation. The "knee"of saturation curve is usually taken as point at which 10% increase of voltage gives 50% increase in magnetising current on no-load test. 37 V gives severe saturation with high current [limited only by winding resistance] which could damage VT if sustained. There will also be extreme voltage spikes on the secondary "70V" side, say 700V, as the core goes out of saturation twice per cycle, which could damage your load.

You would be far better off to reverse the CT - 37V is no problem for 70V winding - & maybe put 7V windings in series, unless something you did not write, like Monitored side being at 1000V. One cannot assume primary & secondary have same insulation thickness to core which is earthed?? (VT secondaries are usually earthed on one side for safety)

The VT will give its rated accuracy at the rated burden, but the difference is usually small. You could find out by varying the resistive load e.g. 1 watt, 2W, 3W & reading with voltmeter.

If you short circuit a VT load winding, then because VTs have very low regulation (low impedance) you will get e.g. 100x rated current and cause damage (supposing source resistance of test supply is negligible).

I suggest you energise 7V winding through a 12V 3 watt bulb or 22 ohm 3W resistor with shorted 70V winding. Voltage/current on 7V winding wiil give short-circuit impedance Z from "7V viewpoint". Then 7V/Z gives current on short circuit @7V, 1/10 of that for 70V winding if turns ratio is 10:1. Z will also give a value for maximum regulation for a given current [Z is both resistance R & inductance L, so you have to measure phase angle to get R & L values to compute regulation with resistive load].

Rating plate reads 7V and 70V, and is custom made (to increase measuring resolution). Though i have been promised it will keep its class 0,5 at voltage up to 14V. He said he could have given me a input voltage up to 4x7V, but at that voltage he couldnt keep the accuracy level at 0,5%.

It seems my only option is to measure the transformer, with open circuit and short circuit to get my R' and X' as well as R'' and X''. Or is it even possible to approximate or calculate the short circuit currents ?

To be clear, there is no intention to measure at any voltage above 7V.

And im curious of the load compared to the burden on the secondary side. My burden should be 3x0,1VA plus cables and terminals. And im assuming it, for now to be far less then 3VA.

I am "in the dark" because I do not know what meters or instruments you have, or even your mains volts & frequency. It is better to make the measurement with a lamp or resistor as I suggested than to guess. I do know that you have a "high resolution measurement" at about 70V AC full scale but do not know how low it goes or if full scale.

You can as well short 7V side and apply a current to 70V side of near "3VA" current, which is about 0.42 amp at 7V or 0.042 amp at 70V.

Round here I would get that by putting Two 15 watt 230V pygmy filament lamps in series from 230VAC 50Hz house supply live to one end of 70V winding, other end of 70V winding to neutral. Better still put a 4700 ohm +/-5%, 10 watt resistor in place of one lamp - then you can measure resistor's actual value with ohmmeter and measure voltage across resistor with AC voltmeter (even cheapest VOM have 200V AC range), then calculate current V/R. The voltage on the 70V winding will be about 0.7VAC. A lamp will burn-out on overcurrent long before a copper wire of similar rated current so do not worry abou transfo.

An approximation to saturated current of 7V winding could be got by measuring its resistance [about 0.1 ohm] - any reactance would reduce current. Most VOM cannot measure 0.1 ohm, using 3.33 ohm resistor made of 3 x 10 ohm 1/2 watt metal film resistors (usually 1% tolerance) in parallel as dropper from D size 1.5V manganese battery will give about 50 mV across 0.1 ohm. Measure voltage across 3.3 ohm then "0.1" ohm and use ratio to get value e.g. 3.33 ohm x 0.100V/1.400V = 0.24 ohm.

You could make a bridge with 100 ohm 1% versus O.1 ohm & 4700 ohm 1 watt potentiometer versus 3.33 ohm - measuring pot with ohmmeter after balancing.

Of course, 0.1 ohm @ 7V is 70 amp - but what is possible fusing on "7V supply" to 7V transfo?

You asked two questions, what happens if you over-volt your VT & what is the error on a very small load 0.1 VA compared to 3VA rated.

A VT standard usually allows that the stated error is met over a range of load VA & power factor within its ratings. in practice most VTs are lightly loaded. If the error due to resistance & load current were very much it could vary a lot with winding temperature(load) so VTs usually have low magnetising & copper losses.

The main error is just the turns ratio & the error open circuit due to primary leakage inductance (flux not linking with secondary, small) & magnetising current flowing through it (small). Then there is error due to loading & its power factor.

The regulation of a transfo, in volts, for small regulations is...

E_r.cos∅ + E_x.sin∅ - where cos∅ is the load power factor.

Er and E_x are the volt drops across resistance in transfo windings and leakage inductance [both primary & secondary]. Of course, for resistive loads (cos∅ =1, sin∅ =0), it all reduces to Er, which is going to be very small at 0.1VA compared to rated 3VA. If your transfo maker measured the no-load ratio & recorded it, he should be able to tell you the value - this should be as accurate as his test equipment & much less than 0.5%.

I have done some measurement regarding short circuit test at HV side.

[1] Vin = 691mV @ 46,3mA

And at another measurement with voltage dividing the primary Vin by a quarter... but i think this would distort the measurement. Because i could not get a lower value then 691mV from the Variac. It showed 46,3mA at first power on.

[2] Vin = 400mV @ 42,0mA (voltage divided by serial resistors).

[1] Calculated: (691mv x 42mA) / 46,3mA ~ 0,63V

0,63V / 70V ~ 0,9%

[2] Calculated: (400mv x 42mA) / 42mA ~ 0,40V

0,40V / 70V ~ 0,6%

Open circuit test at LV side gave:

Unp = 7,02V

I0=0,788A

Uns = 72,2V (didnt see any phase angle compated to Unp)

This seems unreasonable high, given that In=0,42 on LV(7V) side. My wiring were quite long, and i also went through a 1A fuse... and the connections might not have been optimal. But considering 3VA rating @ 7V would give me 16,3Ω, what is seen here this with 0,788A is mostly iron losses then ? Something around 9Ω ? Correct me if im wrong here.

Given the short circuit test, should i expect 100%/0,9% = 111 times In (111x0,042~4,8A on the secondary side, and 111x0,42~48A on the primary side). Or even consider the 0,6%.

I am going to some more tests. Its interesting and a good lesson.

Hi Roger,

Whether it is 0.9% or 0.6%, the regulation is going to be negligible at your 0.1 VA loading - 0.6%/30 = 0.02%!!. It is only the ratio that matters.

It would be helpful to know the DC resistance of "70V" winding & 7V winding since the split of impedance between 70V & 7V windings is unknown, as is inductive reactance.

Likely, most of the drop is resistance.

The magnetising current looks high, but a small "plugtop transfo I measured, 220/24V 0.2 Amp - thus 4.8 VA out, draws 61 mA rms @ 244V (true rms meter) which is 15VA un-loaded! Primary resistance is 430 ohm, secondary 6R9. That is 0.56 amp if fully saturated. This was the inrush - 0.6A peak scaled.

As you can see, the settled current is very peaky, a sign of saturation. Admiittedly, it is 244V on 220V nominal.

I just checked a Freidland bell transformer. 30 years in porch, barely warm to touch.

10 mA primary current at 244V. 8V 1 amp out. primary 410 ohm, secondary 1R2, DC

67model

I checked the resistance, first by direct measurement.

R₍₇₎ ~ 0,1Ω

R₍₇₀₎ ~ 3,2Ω

Then in serial with 3Ω (4//12Ω) with a battery(1,5V) i got around:

3Ω = 745mV and R₍₇₎ = 25mV (also another test 740mV and 11mV) which would suggests R₍₇₎ = 0,1Ω (0,05Ω)

3Ω = 420mV and R₍₇₀₎ = 490mV (also another test 420mV and 440mV) which suggests R₍₇₀₎ = 3,5Ω (3,14Ω)

I would say the measurement was about +/-10% in accuracy, voltage was not stable, or i have measuring equipment.

Also did open circuit test. I0 was same, around 786mA.

And also another short circuit test. Which was a bit tougher, cause of the minimum of voltage output at around 600mV with variac. Though i saw it was a clear linearity with voltage applied towards mA drawn.

I got 41,4mA @389mV as well as 71,2mA @662mV. Which tells me at 1mV i get around 0,1mA on 70V side and 1mA on secondary side until saturation/overload occurs... where it should get distorted. This would also give me 0,5% and short circuit current at 200xIn at nominal voltage. At primary side around 420mAx200=84A and secondary side 42mAx200= 8,4A.

So the ratio of 1mV:0,1mA might be better of with 1mV:0,12mA. Ill check back, with some better conclusion, Just let me now if im far away with my recent thougths.

389mV/41.4mA = 9.4 ohms: 662 mV/71.2 mA = 9.3 ohms.

Dc resistance of 70V winding = 3.3 ohm. 9.3 - 3.3 = 6ohms is resistance transfered by 10:1 transfo from 7v winding, 7V winding resistance ≈ 6/(10²) = 0.06 ohm - not far off from one of your test results.

N.B. if the inductive reactance were half that of the resistance, the power factor would be 0.87 - but since the total impedance would be 1 compared to 0.87, the effect of the reactance on the impedance you measure is not much, hardly more than 10%. As might be expected with a small inefficient transfo, the resistance dominates.

The problem you seem to have is that the 7V winding, your primary, has very little resistance - but appears to be saturating and drawing a high current - which makes it difficult to protect the transfo and could affect your 7V source, whatever that may be.

Taking the example of the Friedland transfo I measured [size about 70 x 50 x 50 mm], the magnetising current, 10 mA @ 240V would be 10*240/8 = 300mA at 8V.

In comparison, your magnetising current is more than double, even though its VA must be smaller.

It would be useful to have a current versus voltage test [saturation curve], from 10V up] for 70V winding - easier to do than 7V ,if you have variac [unless you have a 230/12V or similar 20VA transfo to hand]. The standard definition of the saturation point is where a 10% increase of voltage causes 50% increase in current.

You may not have a wattmeter, but you could measure the "phase angle of short-circuit test current to voltage" by using a resistive divider/potentiometer from the voltage to common - to get a voltage equal to that across transfo winding, then measuring the voltage between the two to find the third side of a triangle - drawing a scale diagram will find the angle, Drawing on a right angle triangle will give a reactance to resistance ratio N.B. Since you are feeding current through a large resistance for the short-circuit test, the transfo current is practically in phase with source voltage. This works if the voltage & current are sinusoidal, which they should be, because voltage on transfo is too small to cause any distortion.

Given your picture i approximate/get:

R1 = 3,5Ω (from DC measurement)

R2 = 100*0,1 = 10Ω (from DC measurement on 70V side)

X1 + X2*100 = 9,3Ω (from short circuit test considering most i copper losses) But as i have not wattmeter, and most tests suggests having wattmeter. I dont see how i could find X1 or X2.

Considering i get 786mA@7V in open circuit test. I would get Z=7/0,786~ 8,9Ω which could be approximated to R1+R2(*100)... as 0,1 was not really an accurate measurement. It seemed lower. But it also includes Im (magnetization current)...

Ill copy paste what i have so far:

Is it possible to measure as an wattmeter without having one ? Using two ampmeters ?

You may find this of interest.

http://www.danyk.cz/wmetr_en.html

Op-amps could be added, to reduce the shunt volt-drop required & adapt to different input voltages.

Many thanks. Sounds interesting setting up "your own" wattmeter. I might dig deeper on that one rainy day.

For now, i looked around, and there are quite big different price tags on them. As im only going to use it on testing for Cos phi and the accuracy for now isnt needed to be high, i will settle with this:

https://www.banggood.com/PZEM-001-AC-80-260V-10A-2200W-Power-Meter-LCD-Digital-Voltmeter-Current-Meter-Monitor-Display-Module-p-1249964.html?gmcCountry=SE&currency=SEK&cur_warehouse=CN

And very nice price tag on that one. Accuracy 1.0.

Also i would like to thank you for your support. Its been an experience and a learning process, still to be commissioned and tested. But at least i got to know the step up transformer a bit more. Normally, one just buy one at standard rate and go with that. Step up transformer... in the end i might realize that buying a standard 12,5 VA rated 1:10 would have gone just fine, but just use it as a step up. That is another test to be done, for sure, if i find someone at a good price.

As you suggested, one can work out reactance X, without wattmeter. I made some more measurements on the bell transfo: magnetisation curve - as far as I could - and S/C test; and then drew a scale vector diagram, as below using 245V, 217.5V, 64.8V and compasses to draw arcs for 64.8 & 245V from 217.5V line....

The 217.5V, being across a resistor, is in phase with the current though the transfo primary. I then worked out transfo resistance in primary circuit by using measured DC values & transferring resistance from secondary using turns ratio² . I then drew the volt drop (40V) of that effective resistance @ 26.9 mA in line with 217.5V, to get the triangle for transfo S/C impedance. As you see, the third side, 65V, is not at "right angle" to 40V resistance vector.

At this point I realised, I had just used ratio 240/8, as marked, = 30 to transfer secondary resistance. So I went back to hardware and measured open circuit voltage ratio [16.89] & re-calculated volt drop 20.2.

As you see, this gives third side 61V, corresponding to inductance drop, convincingly at right angle to 20.2V resistance drop and lagging. That works out as X = 2270 ohms, or L = 2270/(2x3.142x 50) = 7.2 henries.

Note that the impedance for regulation is 64.8V/26.9mA = 2409 ohms in primary terms.

Since rated secondary current is 1 amp, this is primary current of 1/16.9 = 0.059 A, so worst volt drop would be 2409 x 0.059 = 142 volts. That leaves 103 volts for the secondary or about 6.1 volts, applying transfo ratio. In the usual percentage terms given on transfo rating plates that is 100% x 142/245 = 58% - so the resistance is 58% x 20.2V/64.8V = 18.1% and the inductive reactance 58% x 61V/64.8V = 54.6%.

That would only occur with a very low load power factor, equal to power factor of transfo impedance - which 20.2V/64.8V = 0.31 from the diagram.

In usual transfo parameter terms one has E_r = 18.1%, E_x = 54.6%, X/R = 54.6/18.1 = 3.0 and regulation (N.B. sin∅ = √(1 - 0.31²)...

E_rcos∅ + E_xsin∅ = 18% x 0.31 + 54.6% x 0.95 = 5.6% + 51.9% = 57.5%

N.B. for large impedance like this, there is a more complicated, accurate, regulation formula, but 0.5% is not much!

At 0.8 p.f. lagging one would get 47% regulation or 14.5 volt x 53% = 7.7 volts (8V nominal winding), while unity power factor would give 82% x 14.5V = 11.9 volts.

To complete the picture, I tried shorting the 8V winding with 245 volts on primary. One might expect 26.9 mA x 245/65 = 101 mA in proportion - surprisingly, I got 104 mA.

After 1 hour 20 min, primary winding temperature had risen 73'C from 30'C initial ambient [410 to 523 ohm]. One would expect insulation max temperature to be 120'C class B - it could well be class H, 180'C.

So it appears this double insulated safety transfo can also tolerate a short circuit output for a long time