transformer characteristic impedance

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No, the image impedance is NOT independent of frequency.

The reason a transformer's performance rolls off at low and high frequencies is precisely because the image impedances vary with frequency. In this case, the image impedance of my transformer is close to 50 ohms at 100 kHz, so the performance is good there. At higher and lower frequencies, the image impedance deviates substantially from 50 ohms, and the transformer performance begins to degrade, leading to a rolloff in frequency response.

If you read the referenced thread:

https://www.electro-tech-online.com...chat/124261-audio-transformers-two-ports.html

you'll find this in post #4:

"An interesting propertry of logarithmic plots such as the one in the post #1 is that at a particular frequency, 1 kHz perhaps, if you read off the value from the top curve and from the bottom curve, multiply those values and take the square root of their product, that value, if then plotted on the same graph, will be a spot exactly halfway between the top and bottom curves. If we were to perform this operation for each frequency and connect all those result points, the locus of those points would be a new curve, halfway between the existing top and bottom curves. Since the existing curves are the open circuit and short circuit impedances of the primary of the transformer, this new curve halfway between the open and short circuit curves is the image impedance of the primary winding of the transformer.

Notice that it will not be a perfectly horizontal line; a curve of constant impedance. The input image impedance varies with frequency.

The input image impedance is the impedance which will match the input winding of the transformer when the output winding is also matched.

The same two curves can be measured and plotted for the secondary winding of the transformer, and a new curve halfway between those two would be the output image impedance of the transformer.

When a transformer is driven by a source with an impedance equal to the input impedance of the primary, and loaded with the output image impedance, the losses are the minimum possible. (This assumes the image impedances don't have a non-negligible reactive part; they're purely resistive, in other words. If they do, things become more complicated.)"

This applies to the first image I posted in this thread.

The last part mentions the possibility of a non-negligible reactive part. Your transformer definitely has that, but it's still true that using an impedance level near the image impedance is close to optimum.



zmkm1302022 said:
i found the upper cut-off frequency is at about 250KHz, if I get the transformer's characteristic impedance at a proper level comparing to the source and load impedance, i can also get a wider pass band than the one shown in picture?
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Maybe. It depends on the properties of your core and the details of how you wound the transformer.

My transformer has a lower 3 dB frequency of 800 Hz and an upper 3 dB frequency of 12 MHz. Notice that the geometric mean of those two is 97.979 kHz, very close to 100 kHz. As I said, I didn't actually design for that; I just grabbed a core and threw on some wire. It was just luck that it turned out that well.
 
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Thanks so much, The Electrician. this is what i am looking for. now, i am only doing 1:1 turn ratio. my measurements on last Fri were tested with 20 turns on primary and 20 turns on secondary. today, i reduced my winding to 7 on primary side and 7 on secondary side. the image impedance at both sides are reduced from 334 ohm to 68 ohm at 100KHz. tested with the same load, and I found that the upper cut-off frequency increases from 250KHz to 800KHz.

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can you tell me how do you decide the lower and upper 3dB frequencies? I plot the frequency response as 20*log(Vsec/Vpri), i measured all points from 50Hz to 1MHz, there is no -3dB at low frequencies, only find the upper cut-off frequency at 800KHz. but also, i plot a curve as 20*log(Vout*2/Vsource), my teacher told me this is called attenuation plot. yes, i can find the lower cut-off frequency of 2.51KHz and a upper cut-off frequency of about 1.5MHz.
 

zmkm1302022 said:
can you tell me how do you decide the lower and upper 3dB frequencies?
Click to expand...

I drive the primary with a generator having a 50 ohm impedance, and measure the output across a 50 ohm load connected to the secondary. I adjust the frequency until the output drops 3 dB at the low end and the high end.

 

 

The Electrician said:
What is the core material of your transformer? I'm surprised that you're getting such a low frequency cutoff with so few turns.
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the model number of my toroid is Neosid 28-097-C36 (material:F9C). this is my latest measurement, ratio is 7 turns over 4 turns. load is a 16.2ohm resistor. low cut-off frequency is at 2.9KHz, upper cut-off freq. is at 1.8MHz. the geometrical central point is at about 72KHz, and I measured the image impedance at 100KHz, it's about 49ohm.

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do you have any material about the saturation of this toroid (F9C). i have to measure the saturation point, but i dont know how large the current is needed to get to saturation point for this material.
 

It looks like you're getting close to your desired result. Are you in Australia? I found Neosid information here:

**broken link removed**

**broken link removed**
 

When it is a power ferrite, it will be around 400...500mT. At high flux density, permeability drops.

You may know, B = ur*H, and I*n/Lmagn = H, I*n is the ampere turns product.

Let us say ur= 1000, then you need about H =400A/m magnetizing field to saturate the core.

"ampere*turns" = H*lmagn (lmagn = magnetic path length)

Regarding "saturation point". There is no single definition for saturation flux density. Mostly a strong magnetizing field (H) is applied (for example 250A/m) so that most of the material is polarized. The flux density under that contitions is mentioned as "saturation flux density" in the datasheets.

It would be nice to measure the BH curve (plot on oscilloscope). You may know that to measure the flux in the core, you need to get the V*s product (so you need to integrate, for example with a simple RC network or active [opamp] circuit).
 

thanks WimRFP and The Electrician. i am in Australia
 

Whoops

B = Ur*Uo*H = U*H (U = permeability). In my previous posting if forgot U0...
 

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can u have a look at the F10 material, i have the magnetic length is 0.159 m, i try to calculate the NI to get to the saturation state, and my result is 8 Amper-Turns, if I wind the coil with 16 turns, does it mean a 0.5A current can saturate the core??? obviously, my calculation is wrong, because i tested this coil with a Variac, I increase the primary voltage to about 24V, and get a secondary current linearly up to about 4A, it does not saturate.
 
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I feel so ignorant reading these post's.
What kind of thing are we building that needs to be so concerned with such small values?
 

When the toroid has ur =6000 (no air gap) and assuming linear behavior up to the saturation flux density of 380mT, you would require H = 50A/m When applying H= 500A/m, sautation should be clearly visible.

As you have ,le(magnetic) = 0.159m, you need 80 AT to get 500 A/m. 16T at 0.5A generates 8 AT. 4A should generate 64 AT and this is sufficient to see the saturation. By the way, how do you measure B versus H? Are you sure you aren't measuring voltages because of DC resistance instead of induction?
 

i want to find the current (50Hz main current on power line) which may saturate the toroid core, so I use Variac to connect to the primary side, and the secondary side is connected to a 6ohm load. winding is a bifilar winding with a ratio of 1:1, then I increase the voltage from Variac, and try to find corresponding secondary current, the core saturates when the secondary current does not increase any more, even if the primary side voltage is increased. but I tested with the secondary current pumping up to 7A, and i dont think it's saturated. also this material does not response well to 50Hz signal, i found the primary and secondary voltage ratio is far away from 1:1, but the ratio goes close to 1:1 when the primary voltage increases to a certain level (like >20V).
according to the F10 material parameters, H=50A/m can saturate the core, does it not visible at 50A/m, that is why you suggest a H=500A/m to make the saturation clearly visible?
so I think if I wind the core with 20 turns, and increase the primary voltage to get a secondary current of at least 4A, then suppose I can saturate the core???
please give me your suggestion on my plan of testing the saturation of this toroid.
 

If your driving voltage is very stiff (so has low output impedance), and primary wire resistance is low, you will see reasonable output (secondary) voltage even with core saturation.

If you want to see the effect of saturation in your setup, you may add a series resistance in the primary circuit. The total primary circuit resistance should be >> reactance of toroidal inductor using ur=6000. Measure the open circuit voltage of the primary winding (so do not put a load on the secundary side, just your oscillsocope probe).

If you want to do some precise measurement, best is to plot primary current versus flux density in XY mode, this will show you the BH hysteresis curve.

Primary current you can determine by measuring the voltage across a series resistor. based on this you can calculate the H-field. To find the flux density, you need to integrate the secondary output voltage, as output voltage is proportional to d(phi)/dt. To undo the d.../dt you need d...*dt. The result shown on the oscilloscope has to be corrected based on number of secondary turns and core cross section.

Try a search on measuring BH loop, you may find a complete calculation example.
 

I can't read much sense into your results. By definition, the induced voltage would be measured as voltage across an open winding rather than current in a resistor loaded winding, but I don't see how this would change much.

A more visual representation of core characteristic can be achieved as integral of induced voltage versus current. Some DSOs have integration function already built-in, otherwise it can be estimated by a RC circuit, as previously suggested.
 

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