Transformer vs inductor

[mbrooke]

Why doesn't a transformer behave like an inductor? A running transformer has a PF near unity unless the load is otherwise, yet an inductor produces a current shift. Why not an unloaded transformer?

 

[dionysius]

Why doesn't a transformer behave like an inductor? A running transformer has a PF near unity unless the load is otherwise, yet an inductor produces a current shift. Why not an unloaded transformer?

A trafo is like a single coil magnetically coupled to a second coil. This is a process called electromagnetic induction. Direct comparison is invalid since two different devices. I am keeping the answer very simple for starters.

 

[mbrooke]

A trafo is like a single coil magnetically coupled to a second coil. This is a process called electromagnetic induction. Direct comparison is invalid since two different devices. I am keeping the answer very simple for starters.

I can agree with that. But both have a coil (think unloaded secondary), have laminated sheets and work on electricity/magnetism.

 

[Smart $]

Why doesn't a transformer behave like an inductor? A running transformer has a PF near unity unless the load is otherwise, yet an inductor produces a current shift. Why not an unloaded transformer?

Who's saying that the power factor of an unloaded transformer approaches 1 rather than 0?

 

[gar]

170119-0954 EST

mbrooke:

An unloaded transformer looks like an inductor. Look at current relative to voltage with a scope. Or measure power input and compare with the VA input.

Next short the secondary, and excite the primary with a low voltage that does not exceed the primary current rating. Still looks like an inductor, but this inductance is much lower than the inductance seen when looking at the unloaded transformer.

If you do not have a Variac, then use a resistor or incandescent bulb in series to limit current, but you have to do the voltage and current measurements at the input terminals of the transformer.

I suggest that you run your own experiments, and see the results.

.

 

[ron]

A delta wye transformer does have a shift from primary to secondary. The is no direct electrical conductivity from primary to secondary in a regular transformer.

An inductor is somewhat closer to a auto-transformer which is a single winding.

 

[mbrooke]

170119-0954 EST

mbrooke:

An unloaded transformer looks like an inductor. Look at current relative to voltage with a scope. Or measure power input and compare with the VA input.

Next short the secondary, and excite the primary with a low voltage that does not exceed the primary current rating. Still looks like an inductor, but this inductance is much lower than the inductance seen when looking at the unloaded transformer.

If you do not have a Variac, then use a resistor or incandescent bulb in series to limit current, but you have to do the voltage and current measurements at the input terminals of the transformer.

I suggest that you run your own experiments, and see the results.

.

No scope with me ATM, but any idea what to expect?

 

[mbrooke]

Who's saying that the power factor of an unloaded transformer approaches 1 rather than 0?

My lack of understanding :lol::thumbsup::ashamed1::dunce:

I saw transformer PF mentioned in another thread but did not want to hijack that one already in progress :angel:

 

[winnie]

Just a simple jump in:

An inductor stores energy in a magnetic field, and this energy gets returned to the rest of the circuit as the applied voltage cycles.

A transformer also stores energy in a magnetic field, and to the extent that it is storing energy it looks like an inductor. The secondary coil is acting to remove the energy from the magnetic field, making the device look less like an inductor. The greater the load on the secondary, the more the transformer looks like a scaled version of the load attached to the secondary.

-Jon

 

[Smart $]

Who's saying that the power factor of an unloaded transformer approaches 1 rather than 0?

My lack of understanding :lol::thumbsup::ashamed1::dunce:

In that case, the PF of an unloaded transformer approaches 0. There is some nonreactive current associated with the unloaded condition which will keep the PF above 0.

 

[GoldDigger]

... There is some nonreactive current associated with the unloaded condition which will keep the PF above 0.

:thumbsup:
Basically because the phase angle will not exactly 90 degrees because of the resistance component of the input impedance. Hysteresis losses in the core contribute to that resistive component along with the actual winding resistance.

 

[kwired]

Low power factor when the true power is low to begin with is not going to be a very high KVAR level compared to what the KVAR will be when you drive a low power factor motor from it.

 

[gar]

170119-2129 EST

mbrooke:

Here is some data from another thread for a transformer I am runnibg some tests on.


As an illustration my Signal Transformer A41-175 type, 175 VA, has the following no-load values as measured with a Kill-A-Watt EZ:

Volt ---- Amp -- Pwr -- VA ---- PF -- Calculated -- IR loss as % of Input power loss
---------------- Loss ---------------- Pri IR

120.1 -- 0.21 -- 4.2 -- 25.6 -- 0.16 -- 0.071 --- 1.7%
110.7 -- 0.14 -- 3.3 -- 16.5 -- 0.21 -- 0.031 --- 0.8%

Note: Kill-A-Watt's calculated VA is slightly in error. While volt, Amp, and Power are measured by Kill-A-Watt.

Note that at 120 V input the no-load loss is 2.4% of rated VA.



The above are for no-load on the secondary. Following is a scope measurement of the same transformer. Same 120 VAC input. The photo is input AC voltage and input current. The voltage calibration is correct, but current is unscaled because I used a fractional part of a 1 ohm Ohmite dividohm resistor for a currrent shunt. The current waveform is the important part and its phase relation to voltage.
.



The voltage is blue and shows the typical peak distortion of line power these days.

The current is red, and not a sine wave, it is non-linear, but you do see a lagging phase shift of the current.

You may not understand the following statement. This non-sinusoidal waveform results from the instantaneous variation of the inductance of the inductor as one progresses thru the cycle. This results from the magnetic core being driven into saturation.

Near zero current the slope of the current waveform is more like what would result from the inductance of the primay if measured at a low excitation level far away from saturation.

So I don't run out of time the next photo is in the next post.

.

 

[gar]

170119-2215 EST

To continue:

If we next short the secondary two elements are added to the equivalent circuit.

The previous experiment and waveform was the result of a series combination of a linear resistance, the primary DC resistance and some small external resistance, and a moderately large non-linear inductance.

In the next experiment with the secondary shorted we will ignore the inductance of the unloaded primary because we assume in this case that its inductive reactance is large relative to the impedance of the leakage inductance from primary to secondary. Again we have a series circuit of an inductance and resistance. In this case it is the sum of the primary and secondary resistances. When the transformer ratio is not 1 to 1, then when viewing the circuit from the primary side the secondary resistance has to be reflected to the primary by the square of the turns ratio.

Assume the transformer is 1 to 1 and both coils are identical. Then the series resistance is 2*R of one of the coils. The inductance in this series circuit is the leakage inductance of the primary to secondary. And its impedance is much lower than the primary inductance. For me to run this test I do it at a much reduced input voltage.

The voltage and current viewed from the input to the primary are shown below.

.


.
Very heavily inductive, almost 90 deg phase shift. Because this is very inductive you see a much cleaner sine wave current than the input voltage,

.

 

[GoldDigger]

170119-2215 EST

To continue:

If we next short the secondary two elements are added to the equivalent circuit.

The previous experiment and waveform was the result of a series combination of a linear resistance, the primary DC resistance and some small external resistance, and a moderately large non-linear inductance.

In the next experiment with the secondary shorted we will ignore the inductance of the unloaded primary because we assume in this case that its inductive reactance is large relative to the impedance of the leakage inductance from primary to secondary. Again we have a series circuit of an inductance and resistance. In this case it is the sum of the primary and secondary resistances. When the transformer ratio is not 1 to 1, then when viewing the circuit from the primary side the secondary resistance has to be reflected to the primary by the square of the turns ratio.

Assume the transformer is 1 to 1 and both coils are identical. Then the series resistance is 2*R of one of the coils. The inductance in this series circuit is the leakage inductance of the primary to secondary. And its impedance is much lower than the primary inductance. For me to run this test I do it at a much reduced input voltage.

The voltage and current viewed from the input to the primary are shown below.

.

View attachment 16534
.
Very heavily inductive, almost 90 deg phase shift. Because this is very inductive you see a much cleaner sine wave current than the input voltage,

.

At first the 90 degree phase shift puzzled me, since I expected the low secondary resistance to cause the secondary voltage to follow the primary voltage and then cause an in phase secondary current. Then I realized that your test has effectively converted the transformer from a potential transformer to a current transformer.
An inductive current in the primary would produce and in phase inductive current in the secondary.

Confusing. Thanks for doing the experiment.

 

[gar]

170119-2350 EST

GoldDigger:

When the secondary is shorted we essentially have a source voltage, in this case moderately low, where the scope voltage is measured, loaded with a series circuit consisting of a first series resistance (the primary resistance), a series inductance (the transformer leakage inductance), and last a second resistance (the secondary resistance scaled as viewed from the primary side). The current is the current thru this series path.

The assumption is made that the open circuit primary inductance which parallels the leakage inductance and secondary resistance has a large enough Xl so that it can be ignored.

Essentially the scope is looking at a source voltage in combination with the current in a series load circuit of one resistor in series with one inductance.

If we put a resistive load on the secondary to approximately fully load the transformer when the transformer has its normal full input voltage, then we may see some of the core excitation current in the primary current. Later I will try that experiment. Also I might expect to see some apparent change in the apparent leakage inductance impedance.

.

 

[mbrooke]

In that case, the PF of an unloaded transformer approaches 0. There is some nonreactive current associated with the unloaded condition which will keep the PF above 0.

Makes sense.

 

[mbrooke]

170119-2215 EST

To continue:

If we next short the secondary two elements are added to the equivalent circuit.

The previous experiment and waveform was the result of a series combination of a linear resistance, the primary DC resistance and some small external resistance, and a moderately large non-linear inductance.

In the next experiment with the secondary shorted we will ignore the inductance of the unloaded primary because we assume in this case that its inductive reactance is large relative to the impedance of the leakage inductance from primary to secondary. Again we have a series circuit of an inductance and resistance. In this case it is the sum of the primary and secondary resistances. When the transformer ratio is not 1 to 1, then when viewing the circuit from the primary side the secondary resistance has to be reflected to the primary by the square of the turns ratio.

Assume the transformer is 1 to 1 and both coils are identical. Then the series resistance is 2*R of one of the coils. The inductance in this series circuit is the leakage inductance of the primary to secondary. And its impedance is much lower than the primary inductance. For me to run this test I do it at a much reduced input voltage.

The voltage and current viewed from the input to the primary are shown below.

.

View attachment 16534
.
Very heavily inductive, almost 90 deg phase shift. Because this is very inductive you see a much cleaner sine wave current than the input voltage,

.

Gar, Thank you I am busy now, but will come back to discuss this in depth. My apologies for the delayed replies.

 

[gar]

170120-1013 EST

I created an instrumentation error in trying to reduce noise pickup on the current measurement channel. A new plot for 120 V excitation with no secondary load for the A41-175 is:
.


.
Now you see the peak of the current waveform is exactly at the voltage zero-crossing as it should be. Also the slope of the current waveform is more flat before saturation starts.



Still with the A41-175 transformer as setup last night with the secondary shorted.

The apparent un-loaded impedance at 120 V is 120.1/0.21 = 572 ohms. I use the word apparent because the load is non-linear. Can this be ignored below, I believe so.



At the 10.96 V input excitation level, the low level of excitation below for the shorted secondary test, the unloaded input current is 4 mA, or Z = 10.9/.004 = 2700 ohms, and a fairly good sine wave. Easily ignored.

With the secondary shorted the input to the primary is 10.96 V at 1.49 A. The calculated Z is 10.96/1.49 = 7.36 ohms. From previous measurements the primary DC resistance is 1.6 ohms. Double this and calcuate the phase angle. The Xl component should be about (seen looking into the primary) = sq-root of (7.36*7.36 - 3.2*3.2) = 6.63 ohms. The arctan of 3.2/6.63 = 26 deg. Sub from 90 and the calculated phase shift is 64 deg. This calculation is subject to a number of assumption errors.



Looking at the scope waveforms the phase shift looks to be about 369*2.5/16.67 = 54 deg. If we use 54 deg and the 7.36 impedance value, then as seen looking at the primary the components look like R = 4.32 ohms,. and Xl = 5.95 ohms. There is a disparity, and the two are in the ballpark of each other. But they both point to a large phase shift and highly inductive.

I don't plan to try to solve this difference at this time.

A corrected and time expanded scope plot of input voltage and current to the A41-175 with shorted secondary follows:
.

.
You can easily see the 2.5 mS phase shift here.

Before I get into troubles with pictures I will close this post and continue in the next post.

.

 

[gar]

170129-2440 EST

Next I shall go back to full voltage excitation and put a large resistive load on the secondary, but not to full VA rating.

Some side information --- with 119.4 V input the open circuit secondary V = 27.1 V. Loaded with 5 ohms the load voltage is 25.7 V. Iload calculated is 25.7/5 = 5.14 A. A very rough estimate of source impedance is (27.1 - 25.7) / 5.14 = 0.27 ohms (viewed from secondary side). Just reflecting DC resistance we might expect R to be around 0.16 ohms. Thus, source impedance apperars to be quite inductive.


For the full voltage and large secondary load ( 5 ohms at 25.7 V ) the input results are:

119.4 -- 1.16 -- 140 -- 142 -- 0.98 from Kill-A-Watt EZ

Note 5 ohms is not a precision resistor. Calculated load power = 25.7*25.7 / 5 = 132 W. Estimated IR loss 2*1.16*1.16*1.6 = 4.3 W. Core loss about 4.2 W. Total power loss 8.5 W and this is close to the difference between 140 and 132.

The waveforms with this resistive load of 80% of full load rating follows:
.


.
The apparent phase shift is 0.6 mS or 360*0.6/16.67 = 13 deg. There is substantial distortion of the current waveform. This results from the very large peaking of the primary magnetizing current. If one looks at the secondary current it is a quite good sine wave, and there is less phase shift of the current zero-crossing.

Using the assumption of sine waves and linear components is probably the major cause of differing results when using different aqpproaches to obtain amswers.

.