Transformer Inductance

The Electrician · Nov 15, 2013

smijesh said:
Please find the estimation of your small transformer.
Primary impedance Z = 120V / 0.0353 = 3399Ω

Z = 2 x pi x 60 x Lp+DCR

This is not correct. The correct relationship is:

\[Z= \sqrt{(2 \pi 60 Lp)^2+(DCR)^2}\]

smijesh said:
Primary Inductance Lp = (3399 – DCR) / 2 x pi x 60

Approximate DCR = 1500 (For small transformer < 3VA)

Lp = 5.04H

As I said in post #27, this is a 5VA transformer. The measured primary DCR is 138 ohms. The measured secondary DCR is 2.01 ohms.

The correct expression for Lp is:

\[Lp=\frac{\sqrt{Z^2-DCR^2}}{2 \pi 60}\]

Even if DCR were 1500 ohms (which it isn't), the result of the correct calculation would be Lp=8.09H, not 5.04H. Using the measured value of DCR, we get Lp ≈ 9H

smijesh said:
Voltage ratio = 120/13.9 = 8.633 Square of voltage ratio 74.53

Secondary inductance Ls = Lp / 74.53 = 68mH

The correct calculation of the secondary inductance gives Ls = 9/74.53 = 121mH, which is not too far off my measured value of 135mH

All these calculations verify the small effect that the real part of the impedance has on the magnitude of the impedance. This was earlier determined by the ratio of the apparent power to the true power as discussed in post #27.

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gabi_pds said:
For the primary side (with the secondary opened), using a resistor of 9.7 ohm, I`ve got 241 mV (24.85mA) with a phase shift of -63 and the voltage in the terminal of the transformer was 228V.
And for the secondary side, I`ve got 0.96V around the resistor (98.97mA) with a phase shift of -62. The voltage in the terminal of the transformer was 18,5 V.

Something is wrong here. Your turns ratio, using the rated voltages, is 230/18 = 12.78

The ratio of the exciting currents, if the applied voltages are in the proper ratio (the turns ratio), should be the same as the turns ratio. But the ratio of your exciting currents is 98.97/24.85 = 3.98. This is not even close to 12.78.

You really need to measure the open circuit secondary voltage with 230 volts applied to the primary, and use that as the voltage applied to the secondary for the exciting current test.

smijesh · Nov 15, 2013

FvM said:
To get equal flux, you would apply voltages according to the windings ratio, with corresponds in a first order to the open circuit voltages rather than rated voltages. But the difference isn't so large that it explains the observed results. It would be easier to reproduce the calculation if we know the actual voltage and current values.

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In post #34

Code:

Z = 2 x pi x 60 x Lp+DCR

should be corrected to

Please refer to basic AC network theory.
The derived calculations are respectively wrong.

Yes FvM You are right,

Please find the estimation of your small transformer.
Primary impedance Z = 120V / 0.0353 = 3399Ω

Z = Sqr[(2 x pi x 60 x Lp)²+DCR²]

Primary Inductance Lp = Sqr[(3399² – DCR²)] / (2 x pi x 60)

Approximate DCR = 1500 (For small transformer < 3VA)

Lp = 8.09H

Voltage ratio = 120/13.9 = 8.633 Square of voltage ratio 74.53

Secondary inductance Ls = Lp / 74.53 = 109mH

The Electrician · Nov 15, 2013

smijesh said:
Yes FvM You are right,

Please find the estimation of your small transformer.
Primary impedance Z = 120V / 0.0353 = 3399Ω

Z = Sqr[(2 x pi x 60 x Lp)²+DCR²]

Primary Inductance Lp = Sqr[(3399² – DCR²)] / (2 x pi x 60)

Approximate DCR = 1500 (For small transformer < 3VA)

Lp = 8.09H

Voltage ratio = 120/13.9 = 8.633 Square of voltage ratio 74.53

Secondary inductance Ls = Lp / 74.53 = 109mH

You should also read post #41

It's not necessary to guess the DCR of the transformer; the DCR of the primary of the transformer from which these measurements were taken, is 138 ohms.

gabi_pds · Nov 18, 2013

The Electrician said:
You are in Germany, but the text in the image is Polish. Apparently this isn't an oscilloscope capture that you made yourself.

This image is doesn't look quite correct for the magnetizing current of a real transformer winding which is energized by a sine wave. Since the flux in the core is proportional to the integral of the applied voltage, the flux peak occurs at the end of each half cycle, not at the peak of the applied sine.

The exciting current consists of a sine wave of current due to the resistance of the wire, plus a peaky waveform with the peak occurring near the end of each half cycle. The image you show is too symmetrical to be representative of a small real transformer. It's what you might expect from some sort of an ideal winding with no resistance.

Very small transformers in particular tend to be designed to operate further into saturation than larger ones, because their regulation is poorer, and their surface area to volume ratio is more favorable to dissipate heat; this gives an exciting current waveform that is even more peaky than larger transformers.

I found a small (5 VA rating) transformer to make some measurements with.

Here's a scope capture of the applied voltage (blue), exciting current (orange) and the instantaneous product of the two (red):

The average of the red waveform is the true power dissipation (1.12 watts). The exciting current is much more peaky than shown in the image you posted. The third harmonic is 37% of the fundamental.

The apparent power is the product of the applied voltage (121 VAC) and the exciting current (35.3 mA), which is 4.27 VA, 3.8 times the true power. The apparent power is sufficiently larger than the true power to justify characterizing the exciting current as a mostly reactive current.

From this we get the primary inductance as 121/(2*Pi*60*.0353) = 9.09 henries. Compare this to the inductance measured on an LCR meter:

Code:

Applied Inductance Voltage .01 3.02H 0.1 3.29H 1.0 4.56H 10 8.30H

Do you know how to physically explain this peak in the exciting current? Is this because of when the core is saturated the exciting current increases more quickly?

FvM · Nov 18, 2013

In the peak flux region, the core isn't yet saturated, but differential permeability (H/B curve slope) already decreases, so yes, the current rises faster.

smijesh · Nov 19, 2013

The magnetizing current waveform is shifted because primary current lags the applied voltage by an angle 90°.(Transformer act as an inductance in AC circuit)

Transformer core made up of non linear magnetic material. Excitation current follows its non linear property.

Applied peak sinusoidal voltage cause the core to operate at its maximum flux density (Flat point of BH Curve). At this point permeability of the core (B/H slop of the curve) will be low. This low permeability of the core leads to low primary inductance followed by low primary impedance hence more current flow in to the winding.

But half of the sinusoidal voltage (Valley point), at this voltage the B will be half of the B max but H = NI/l will be very less. So permeability of the core (B/H slop of the curve) become more so less current flow in to the winding

Transformer Inductance

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