Confused about transformer saturation

Ghar

Joined Mar 8, 2010
655
We're still sort of dodging each other.

I agree the DC current is limited by the DC circuit, that's what I've been saying when I mentioned series resistance - but is that current generally low enough not to saturate a transformer?
If as you say there is 1A of DC current, how much room is there left before saturate? Basically I'm wondering if either the DC current is very small due to very small offsets or whether most transformers are just so well below saturation. Or I have something fundamentally messed up in my understanding...

Let me put down some equations I'm familiar with, under some assumptions of course...

Assuming constant H along a path length l:
\(H = \frac{I_{enc}}{l}\)

Assuming constant cross section area A:
\(\phi = BA\)

Permeability is a function of H, hence the B-H hysteresis curve.
\(B = \mu(H) H\)

Combining and generalizing for 2 windings, where both currents should be positive:

\(\phi \frac{l}{\mu A} = n_1i_1 - n_2i_2\)

Defining reluctance:
\(\Re = \frac{l}{\mu A}\)

Rewriting:
\(\phi \Re = n_1i_1 - n_2i_2\)

Faraday's Law for the first winding:
\(V_1 = -n_1\frac{\partial \phi}{\partial t}\)

Plug in the equation for flux...
\(V_1 = -n\frac{\partial}{\partial t} \frac{n_1i_1 - n_2i_2}{\Re}\\
V_1 =\frac{n_1^2}{\Re}\frac{\partial}{\partial t}(i_1 - \frac{n_2}{n_1}i_2)\)

This equation is equivalent to an inductor (magnetizing inductance) in parallel with the primary winding, with:
\(L_M = \frac{n_1^2}{\Re}\\
i_M = i_1 - \frac{n_2}{n_1}i_2\\
i_M = \frac{1}{L_M}\int v_1dt\)

Going back to the beginning, we have:
\(I_{enc} = n_1i_1 - n_2i_2\\
I_{enc} = n_1i_M\\
H = \frac{n_1i_M}{l}\\
B = \mu \frac{n_1i_M}{l}\\
\)

Or, to rewrite it in a more familiar form...
\(B = \mu \frac{n_1}{l}\frac{1}{L_M}\int v_1dt\\
B = \mu \frac{n_1}{l}\frac{\Re}{n_1^2}\int v_1dt\\
B = \mu \frac{\Re}{n_1l}\int v_1dt\\
B = \frac{l}{\mu A}\frac{\mu}{n_1l}\int v_1dt\\
B = \frac{1}{n_1A}\int v_1dt\)

So, uhh, I think that's about it.

Magnetizing current is directly related to the flux, because it's the current that creates the H.
The core, through permeability, defines B. Permeability changes with H of course.

And, that magnetizing current is the current through an inductor parallel with the winding. So, that DC current should be set entirely by the DC circuit (DC sources and resistance).

Which goes back to my question...
In general we have one of:
  • Small DC offset voltage, so small DC current
  • Significant DC offset voltage, fairly high resistances
  • Transformers are generally well below saturation so even a moderate DC current doesn't lead to saturation
  • I'm missing something.

[That was some serious Latex abuse]
 
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Basically I'm wondering if either the DC current is very small due to very small offsets or whether most transformers are just so well below saturation. Or I have something fundamentally messed up in my understanding...


Magnetizing current is directly related to the flux, because it's the current that creates the H.
The core, through permeability, defines B. Permeability changes with H of course.

Which goes back to my question...
In general we have one of:
  • Small DC offset voltage, so small DC current
  • Significant DC offset voltage, fairly high resistances
  • Transformers are generally well below saturation so even a moderate DC current doesn't lead to saturation
  • I'm missing something.

[That was some serious Latex abuse]
In magnetics, as in electricity, one can ask whether it's the current through a resistor that causes a voltage drop across it, or if it's an applied voltage that causes the current.

One can similarly ask whether, in magnetics, B causes H, or does H cause B?

I'm not going to get into that here.

In normal commercial transformers, the iron is usually operated somewhat into saturation; ordinary transformers are not operated well below saturation.

Here are some scope captures. It's possible to display the B-H characteristic (hysteresis loop with AC excitation) of a transformer's silicon steel core. What you do is put the scope in X-Y mode and display the current into the primary winding on the X-axis, and connect an integrator consisting of a high value resistor (470k) in series with a few μF film capacitor. The R-C combination is connected across the primary so that it sees the applied AC voltage. The voltage across the capacitor is then proportional to the integral of the applied voltage, and is proportional to the total flux in the core.

All the images are taken with no load on the transformer secondary.

The first image shows the line voltage applied to the primary (green), the current in the primary (orange), and the voltage from the R-C integrator (blue, which is proportional to flux). We see that the magnetizing current (orange) is not proportional to the flux (blue), due to hysteresis and non-linearity of the core.

The second image shows the hysteresis loop traversed as the applied AC voltage makes it excursions; the applied voltage is 120 VAC. The core is somewhat into saturation.

The third image shows the loop for an applied voltage of 100 VAC. In this case, the core is just beginning to saturate.

In the next post, I'll show some more captures.
 

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Here are 3 more scope captures.

The first shows the hysteresis loop with 140 VAC applied.

The second shows the hysteresis loop with 300 mA of DC injected into the 24 volt secondary of the transformer. I didn't want to inject it into the primary for safety reasons, but the effect is the same. You can see that the core now traverses a hysteresis loop that is pushed off to one side. The core saturates in one direction but not in the other direction.

The third image shows the same as the second, but with 500 mA DC injected instead of 300 mA.

This evening I again looked at the DC offset present at an outlet in my kitchen; I saw values of a couple of millivolts max. There are 5 homes served by the distribution transformer that serves my home. I asked my neighbor to participate in an experiment. I asked her to operate her hair dryer at full power and at half power to see if I could detect any offset. We coordinated the experiment by phone, and when she said the dryer was running on half power, I couldn't see any effect. But, my own dryer caused a 100 mV offset at the same outlet where it was plugged in.

Maybe later, I'll see if I can measure the actual DC current in a transformer when it's plugged into the same outlet as the dryer on half power.

But, without an extreme source of DC offset such as a large half-wave load on the house wiring, you just won't see much effect.

The hysteresis loops I've shown were taken with the transformer energized from my house wiring, and you don't see the loop pushed off to the side by the 1 or 2 millivolt offset that I measure. The DC component of current in the primary might be a milliamp or so, and it isn't constant; it's constantly varying.
 

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t_n_k

Joined Mar 6, 2009
5,455
Wow - nice work on the equations Ghar!

I've attached a theoretical problem with a practical perspective - in which I aim to show that a primary winding voltage with DC bias on the AC component does not necessarily saturate the core. Given your interest in the magnetic theory equations you might like to check my solution using your own method.

Hopefully I have the numbers correct.
 

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Borrowing some of your latex:

Assuming constant cross section area A:
\(\phi = BA\)


\(B = \frac{1}{n_1A}\int v_1dt\)

Making one more small rearrangement:
\(\phi=BA= \frac{1}{n_1}\int v_1dt\)

This is essentially Faraday's law, and the permeability doesn't appear in it.

So, uhh, I think that's about it.

Magnetizing current is directly related to the flux, because it's the current that creates the H.
The core, through permeability, defines B. Permeability changes with H of course.
As I said in another post:

"One can similarly ask whether, in magnetics, B causes H, or does H cause B?"

I would say that it's the integral of voltage that creates the flux. Then the core, through permeability, defines H.

\(H = \frac{B}{\mu}\)

If when you say "Magnetizing current is directly related to the flux", you mean linearly proportional, or related by a multiplicative constant, it's not true when the core is made of iron.

It's H that is directly proportional to the magnetizing current. As you included earlier:

\(H = \frac{I_{enc}}{l}\)
So, the magnetizing current is not directly proportional to flux; μ is involved. That's why the magnetizing current in the first scope image shows the flux (blue trace) being sinusoidal, but the magnetizing current (orange) is not; it is non-linearly related to the flux.
 

Ghar

Joined Mar 8, 2010
655
When I said 'directly related' I was trying to avoid implying linear or proportional but trying to emphasize there's just one multiplication between them.

You have a good point though, I have long ago accepted voltage and current being interchangeable as a source but to me Maxwell's equations always seemed to imply a direction. It might be from my discomfort with inverting the vector calculus... they're written pretty concisely and it's hard to resist the feeling that they prefer to be that way.

I've actually done these B-H measurements in a lab once but without a doubt it was the most confusing lab in all of undergrad. That course is still notorious for destroying hopes and dreams :)

So now that we have concluded transformers are in fact near saturation and it is easy to shift the curve even further into saturation, why doesn't it ruin everything? Why aren't the grid voltages all distorted and why aren't there huge currents (AC and DC), since in saturation μ is reduced and the magnetizing inductance appears smaller which is of course a shunt.

----

I suspect you can't measure the DC offset from your neighbour because you have very little common impedance between your homes. If your hair dryer is taking that 1 A DC from the transformer you might have 100 mlliohm resistance, giving you that 100 mV offset.
The resistance of AWG 14 is about 8 milliohms per meter from the table I looked at and that's what my house has at least.
Your neighbour's house will have its own lines from the transformer, so you might share only 10 milliohms from the transformer itself and a short length of wire, giving you a lot less offset in your house.

----

t_n_k, your example does cut through to the issue. I have not convinced myself yet that the offsets are low enough and the resistances high enough in general to prevent high DC currents and saturation. This then makes me doubt everything else.
Is half an Ohm a reasonable value for a transformer DC winding resistance? If it is that high then it things make more sense.

It's funny, I've done those calculations too but you never get an idea of how reasonable the numbers are until you get more experience.
The procedure looks good to me, I didn't re-calculate to check though.
 
So now that we have concluded transformers are in fact near saturation and it is easy to shift the curve even further into saturation, why doesn't it ruin everything? Why aren't the grid voltages all distorted and why aren't there huge currents (AC and DC), since in saturation μ is reduced and the magnetizing inductance appears smaller which is of course a shunt.
Why would there be huge DC currents? As I've explained, only a few special loads such as hair dryers draw DC anyway, and it's isolated from the grid at large by the distribution transformer at homes. These half-wave loads are usually only active a small fraction of the time, and constitute only a small fraction of total loads.

Even though a typical transformer is in saturation at the peaks of the magnetizing current, that current is still small compared to the rated load of the transformer; saturation is gradual. Look at the first image in post #24. The peak magnetizing current is only 150 mA and the transformer is rated at 200 VA. Typical load current would be much larger than the magnetizing current, and load currents reduce the flux in the core somewhat so the transformer doesn't go so far into saturation under load.

I suspect you can't measure the DC offset from your neighbour because you have very little common impedance between your homes. If your hair dryer is taking that 1 A DC from the transformer you might have 100 mlliohm resistance, giving you that 100 mV offset.
The resistance of AWG 14 is about 8 milliohms per meter from the table I looked at and that's what my house has at least.
Your neighbour's house will have its own lines from the transformer, so you might share only 10 milliohms from the transformer itself and a short length of wire, giving you a lot less offset in your house.
I didn't really expect to be able to measure it, but I thought I'd give it a try anyway.

t_n_k, your example does cut through to the issue. I have not convinced myself yet that the offsets are low enough and the resistances high enough in general to prevent high DC currents and saturation. This then makes me doubt everything else.
Is half an Ohm a reasonable value for a transformer DC winding resistance? If it is that high then it things make more sense.

It's funny, I've done those calculations too but you never get an idea of how reasonable the numbers are until you get more experience.
The procedure looks good to me, I didn't re-calculate to check though.
t_n_k's example is not applicable to typical actual transformers. They are already somewhat in saturation when excited with rated voltage, so any DC will push then further into saturation in one direction.

The 200 VA transformer I made the scope captures with has about 1 ohm primary DC resistance. A 2 kW transformer on hand has .093 ohms primary DC resistance.

The golden ears audio guys who complain about DC offsets causing a problem are using toroidal power transformers. These toroidal cores have essentially no air gap in the core, unlike the standard E-I lamination cores. The hysteresis loop for the toroidal cores is more nearly a so-called "square loop", so the onset of saturation is more sudden. Even so, their only complaint is a buzzing noise rather than that the transformer is drawing excessively large primary currents.
 
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t_n_k

Joined Mar 6, 2009
5,455
t_n_k's example is not applicable to typical actual transformers. They are already somewhat in saturation when excited with rated voltage, so any DC will push then further into saturation in one direction.
Your point is well made Electrician. Indeed the expectation is that mains transformers always operate into the saturation region - for sound economic reasons. Imagine the cost of transformers which only operated in the quasi-linear region of their cores' B-H characteristic.

To take up Ghars' point on AC waveform distortion - it's worth emphasising the point that large mains transformers do operate into the saturation region and there is very little observable sinusoidal waveform distortion. My expectation is that the applied voltage and resulting core flux are mutually related and therefore sinusoidal for a sinusoidal voltage system. It's the excitation mmf (and hence primary exciting current) that shows increasing distortion with increasing saturation. However, I presume the series impedance drop on the primary side [with a situation of high primary harmonic current distortion] is what contributes to distortion of the transformed "sinusoidal" waveform. Perhaps this is what we observe in cheaper low power mains transformers which may be designed to operate even further into the saturation region than a well engineered high capacity mains transformer.
 
I've done some more measurements. The hair dryer I was using previously is a small travel dryer. I rummaged around in the bathroom cabinet and found a full size unit. Instead of relying on the built-in diode which enables half power, I wired an external diode so I could get a half-wave current draw even on full power.

I plugged the dryer+external diode in to an outlet and measured the offset at that outlet when the dryer was on high power setting. I was amazed to get a 1 volt DC offset! To make sure my Fluke 187 DMM wasn't freaking out, I got out my old Triplett analog VOM. It also indicated a 1 volt offset!

I arranged to plug in at the same outlet the primary of the 2kW transformer I mentioned in an earlier post; the one with .093 ohm DC primary resistance; the secondary was not loaded. I measured the DC current passing into the primary. It measured 2.5 amps!

I then left the transformer plugged in at that outlet, and moved the dryer to another outlet on the other side of the room; the DC current in the primary was 1.9 amps.

I went to another room on the other side of the house and plugged in the dryer, still on high power setting. Then the DC current in the transformer primary was 1.19 amps.

Using the smaller 200 VA transformer I mentioned earlier, under the same circumstances, when plugged into the same outlet as the dryer (running on high power), the DC current into the transformer was .25 amps. This is the transformer with about 1 ohm primary DC resistance.

I reconnected the 200 VA transformer to the scope, with the RC integrator as before.

Attached to this post are 3 captures showing the applied voltage (green), magnetizing current (orange), and integral of the applied voltage (blue). I offset the magnetizing current trace to the bottom of the screen, and reduced vertical sensitivity of that trace.

The first image shows things with the hair dryer turned off.

The second image is with the hair dryer on low power. You can see the magnetizing current offset due to the DC in the primary, which was 50 mA.

The third image is with the dryer on high power. Remember that ordinarily at this setting the dryer would not be operating in half-wave mode, but I have an external diode so in this case, it is drawing DC from the line. The DC current in the primary measured 200 mA. The dryer and transformer are not plugged in to the same outlet for these measurements, so the DC isn't as high as 250 mA as it is when they're plugged in to the same outlet. You can see that the transformer is quite saturated in one direction.

Unfortunately, the RC integrator method can't respond to the DC component of flux. What I've done in the past when I wanted to visualize the DC flux, is to drill a small hole into the core and insert a Hall effect device to measure the flux, but I'm not going to do that here.

I measured the (AC) magnetizing current when the transformer was excited at rated voltage, but without the hair dryer causing a DC offset; that current measured 120 mA.

With the dryer on high power, the magnetizing current had an AC component of 380 mA, and a DC component of 200 mA.

I measured the unloaded transformer loss without the DC offset, and it was 5.0 watts. With the dryer on high power and the DC current of 200 mA in the primary, the loss increased to 5.9 watts. I expected the increase in loss to have been greater. These measurements were made with a precision analog wattmeter, which responds to true power, AC and DC components included.

So, the current peaks due to saturation are fairly large, but the total current under load would swamp out the increased magnetizing current when DC is present. The percentage increase in transformer loss isn't very much compared to what the transformer losses at full load are.

I've created a situation substantially more extreme that what would typically be the case, and it had a negligible effect on the total power consumption of my home.

Note that if a transformer with a low primary DC resistance is connected to the same outlet where the dryer is plugged in, the DC current that in-house transformer absorbs is kept out of the pole pig. However, the pole pig is so large that a couple of amps of DC on its secondary won't have any noticeable effect on its primary current.
 

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Here are the scope captures showing the hysteresis loops for the three situations described in the previous post. I've offset the display so that we can see the full extent of saturation.

Image 1: hair dryer turned off. This is the normal magnetizing current with rated excitation voltage.

Image 2 shows the case with the dryer on half power.

Image 3 shows the case with the dryer on full power. Remember that normally the dryer on high power would not draw any DC, but I've wired in an external diode so that half-wave current pulses are drawn in this case.
 

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To take up Ghars' point on AC waveform distortion - it's worth emphasising the point that large mains transformers do operate into the saturation region and there is very little observable sinusoidal waveform distortion. My expectation is that the applied voltage and resulting core flux are mutually related and therefore sinusoidal for a sinusoidal voltage system. It's the excitation mmf (and hence primary exciting current) that shows increasing distortion with increasing saturation. However, I presume the series impedance drop on the primary side [with a situation of high primary harmonic current distortion] is what contributes to distortion of the transformed "sinusoidal" waveform. Perhaps this is what we observe in cheaper low power mains transformers which may be designed to operate even further into the saturation region than a well engineered high capacity mains transformer.
It's interesting to note that the peak in magnetizing current occurs as a particular half sine of applied voltage is coming to an end. Since the flux is proportional to the integral of the applied voltage, those volt-seconds max out as the applied voltage is about to change polarity. You would think that the increased IR drop occurring at that time might be visible on the waveform, but I can't see it.

Edit. On the image named Xcur_DryHigh, attached to post #33, you can see this effect at the time of the peak in the orange waveform; a little flattening in the blue waveform is visible. This is with a DC current into the transformer of magnitude that just won't happen with a pole pig unless you're being really abusive.

What I do see is that the peak of the voltage is flat-topped due to all the millions of capacitor input power supplies in all the millions of consumer electronic devices (computers, TVs, etc.).
 
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Ghar

Joined Mar 8, 2010
655
Heh, I haven't read any of your new posts, I was out all day.
But in response to post #30...

Why would there be huge DC currents? As I've explained, only a few special loads such as hair dryers draw DC anyway, and it's isolated from the grid at large by the distribution transformer at homes. These half-wave loads are usually only active a small fraction of the time, and constitute only a small fraction of total loads.
Sorry I meant AC, not sure why I said "AC and DC".

The magnetizing inductance drops in saturation, but you're right, it's a matter of numbers again.

Looking at this:

120V_HystLoop.png

I'm going to do it assuming two linear sections.
Permeability is the slope of the B-H curve, so near the origin you have a slope of roughly 2.
At the peak, your slope is down to about 1/5.
That means the magnetizing inductance dropped by a factor of 10, and the current needs to increase... now I see one of my bad assumptions.
I was thinking the current would go 10 times higher but it's an inductor so it can't increase like that. The current rate becomes 10 times faster and only briefly.

\(\frac{di}{dt} = \frac{V}{L}\)

Let's see if that works:

TranMeas.png

Let me try to make a simple equivalent circuit, assuming a step change in inductance.
Using your value of Rp = 1

To estimate L I digitized the plot.
For the first segment of increasing current the integral of voltage is 712 V-ms
Current increases from 14 to 92 mA.
\(L = \frac{1}{\Delta i}\int Vdt\\
L_1 = \frac{712}{92 - 14} = 9.1 H\)

The second portion:
We get 82.4 V-ms and current from 97 to 147 mA.
\(L_2 = \frac{82.4}{147 - 97} = 1.6 H\)

So inductance changed by a factor of about 5.

I digitized the B-H curve now too... to get a better slope number.
So initially the slope is (using # divisions as co-ordinates):
\(\frac{1.44 - (-1.36)}{1.68 - 0.4} = 1.55\)
It then becomes:
\(\frac{2.04 - 1.72}{2.96 - 1.84} = 0.29\)
That gives a ratio of 5.3.

What do you know, it actually is consistent!!

Let me try to SPICE this. I'm going to just have an RL circuit and switch in shunt inductors to lower the inductance.
So inductor 1 is 9.1 H, then we need a parallel combination to become 1.6, so inductor 2 is 1.9 H.
I'm using your measurement of Rp as 1 ohm.
I want to start when current is zero, so the voltage source gets a phase of about 2.5/16.666 ms so 54 degrees.
The inductance drops after another 4 ms or so.
Flux keeps increasing though, so it looks like that falling current edge is actually faster for an even smaller inductance.
\(L_3 = \frac{90}{147 - 0} = 0.6 H\)

That gives the last parallel inductor to add as 0.97 H and it switches in at about 5.5 ms.

Let's try it...
Here's the schematic:

saturation_schem.png

And the plot:

saturation_curves.png

Now let's superimpose your measurements with this...

saturation_curves_imposed.png

Wow, it's really close. The only noticable difference is the current isn't 0 when I assumed it was.

So that 3rd inductance value is the returning path on the B-H curve... let's get that slope too.

\(\frac{2.12 - 1.72}{2.84 - 0} = 0.14\)

This is 0.14/1.55 = 0.09 of the original value, meaning the inductance should be 0.09 of the original.
I had:
\(L_1 = 9.1 H\\
L_3 = 0.6 H\\
\frac{L_3}{L_1} = 0.07 \)
That's pretty close for rough slopes on a graph.

So anyways...

The conclusion is that the AC currents don't get huge because even if the inductance drops by over 90% like we have here, it's only the current rate that increases and it only happens briefly.
Additionally, that magnetizing inductance should still be a much larger impedance than the load, like you say this is peaking at 150 mA while the load current can be 200 VA / 120 V = 1.6 A, or at least 10 times larger.

Edit:
I added the B-H curve with my linear approximations just to give you an idea. I didn't save the digitized lines so this isn't exactly it.
120V_HystLoop_lines.png
 
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Let's try it...
Here's the schematic
I mentioned in post #33 that the transformer loss (almost all core loss; eddy current and hysteresis loss) was 5 watts.

You need to include a 2880 ohm shunt resistor to account for those losses. Adding this resistor would widen the hysteresis loop to more nearly simulate the real core.
 

Ghar

Joined Mar 8, 2010
655
As it is right now there is no hysteresis loop... I manually put in the times to switch in the inductors.

Adding that resistor just adds a noticeable sinusoidal current which I don't see in your measurement.
It's probably non-linear as well.

I can try to actually model the loop but I won't do that right now.

saturation_Rcore.png
 
Adding that resistor just adds a noticeable sinusoidal current which I don't see in your measurement.
I searched my transformers and found one with a very high quality, low loss, high permeability core.

Here are 3 images showing the performance of the core with 120 VAC applied to the primary.

Green is the applied line voltage, orange is the primary current, blue is the integral of the primary voltage (total flux), and I've added a 4th trace in red which is the instantaneous product of primary current and voltage.

The first image shows the transformer alone with no load on the secondary. The red trace is the instantaneous power in the primary. You can see positive lobes where the primary inductance is absorbing power, and negative lobes where it's delivering power back to the line. Since the area under the positive lobes is slightly greater than than area under the negative lobes, we see that the core losses are net positive. You can tell that the core loss is small because the horizontal portions of the orange trace are almost coincident with the zero volt X-axis. Alternate horizontal portions are slightly above and slightly below the X-axis; this means there is a small loss.

The second image shows the case with a small resistive load applied to the secondary. The red trace now has substantially larger area under the positive lobes. Alternate horizontal portions of the orange trace are now further removed from the X-axis, indicating more loss. Compare to the file "TranMeas.bmp" in post #24, which shows the capture for a transformer with a much lossier core.

The third image shows the case with a larger load on the secondary. The primary current is beginning to show a noticeable sinusoidal current as your simulation showed. This is too much loss to be a good simulation of the lossy core in the 200VA transformer.

In the next post I show the hysteresis loops for these 3 cases.
 

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The three images attached to this post show the hysteresis loops corresponding to the three images in the previous post.

The first image shows the unloaded transformer case. The loop is very narrow, indicating low loss, and the vertical portions are quite vertical, indicating high permeability.

The next two are showing what happens when a load is applied to the secondary. The load widens the hysteresis loop, showing how a resistive load on the transformer can be used to simulate a wide loop.
 

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Ghar

Joined Mar 8, 2010
655
Cool stuff, and you're definitely convincing me to shell out for a 4-channel scope when I finally get one...

I'll have to try the resistor thing when I get around to modelling the loop.
Haven't done many functions in SPICE but I should learn anyway.
 
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