An inductor is a passive electronic component which is capable of storing electrical energy in the form of magnetic...

# IMPEDANCE AND AUDIO TRANSFORMER DESIGN

One of the most “mysterious” transformers to design are impedance and/or audio transformers. From here out I’ll refer to both types just as impedance transformers.

If you go searching online, you’ll find very little useful information on how to design such a part.

Now often impedance transformers are quite low in power and have large PRI inductances, so they are less likely to saturate because typically the turns are large compared with the applied volt-sec’s.

Now a simple approach that often yields fairly good results is as follows.

You’ll often have a desired PRI impedance you want to see for some attached load impedance on the SEC. We call this the reflected impedance.

Let’s take the standard equation for the voltage across an inductor as a function of inductance and the time rate of change in current and solve for the impedance. We will assume the applied voltage is sinusoidal in nature…..we get the following:

I have found that making the inductance be about twice that needed at your lowest frequency of operation for a desired reflected impedance often yields good results….

You can solve for your PRI inductance as:

Ok so now it’s time to find the needed PRI turns…..

If you know what your PEAK excitation current should be, along with your core area and the PEAK flux density you’d like in the core, you can solve for the turns as follows:

Where the inductance, L, is what you found in equation (1)….

Now in most cases you do not know or have a PEAK excitation current in mind, but instead you will often have the applied rms PRI voltage…so let’s find a different equation we can use instead….we will get this equation from equation (1)….

And from equation (2a)….

Now at this point you should solve for your peak excitation current just so you know what it is….it would be needed if you plan to do an efficiency analysis after you are done…plus you want to confirm it is small…your impedance transformer will reflect better at the low frequency end with a small excitation current.

Setting equal equations (2b) and (3)….

Ok so now you need to see if your turns are enough to give you the inductance you need found in equation (1)…

Let’s use the equation for solenoid inductance….it will be accurate enough…

Now you need to know your core area, the path length and the permeability…this is where it becomes “tricky” because your core is likely a good grade of steel, perhaps even an 80% Nickel steel core…the permeability of such cores vary greatly with applied amp-turns….you will need the B-H curve for your core….

Find your H value (magnetic field intensity)…this will be amp-turns per meter…the current should be your magnetizing current but this will be very close to your found excitation current, so just use this…you know your turns and your path length so you have what you need to get your H value.

Find this on your B-H curve and then determine the flux density at this value.

Take the ratio of B to H and this is your operating permeability….

You can then put this, your turns squared, area and length into the solenoid inductor equation and see what you get….how does it compare with the required inductance found in equation (1)? If it is low then increase your turns until you meet the needed inductance. If it higher than what you need then you’re fine….it’s ok to have a higher PRI inductance than needed.

Ok so what if you don’t have a B-H curve for your core? Do you have the core? Can you wind it with the turns you found and see what you get for an inductance? Do you have the AL value for the core you want to use? If so, solve for inductance knowing the turns you found….

Lastly if you have another part (using the same core material) who’s inductance you know, even of a different size, wound with different turns, you can compensate for this to determine the new inductance….let’s call all the items of the part you have as “old” and the ones you are designing as “new”…

Let’s show that math….

Ok, so by now you should have the needed PRI turns and you should know if you are meeting the needed PRI inductance with those turns….

Next come finding the needed SEC turns….

To find the SEC turns we will derive the equation assuming the power in is the power out….this is not true but it is often a good approximation….nonetheless I will add a factor which accounts for efficiency….let’s assume the efficiency will be 95%…..

Now, if your impedance transformer is very low power there’s a good chance your efficiency is even better than 95%…perhaps 98%….if so feel free to replace the 95 in equation (5) with 98…it’s going to make only a small difference in turns…

Ok so now you have your PRI turns, your SEC turns and your PRI inductance. You can calculate your SEC inductance if you wish…this would just be as follows:

Next let’s find your currents…you already had the rms PRI voltage and you have your turns ratio found and used in equation (5)….so you can determine your rms SEC voltage. You have the power so you can set the input power equal to the product of PRI rms voltage and PRI rms current and your output power equal to the SEC rms voltage and SEC rms current…

Then solve for both the PRI and SEC rms currents….

So the last thing you need are wire sizes….

Given the DCR’s will probably be kind of high because the transformer will likely be small with many turns, I would suggest using a 2 Amps per squared milli-meter rule of thumb….just plug in the winding current, I and your wire diameter can be found as follows:

Now….check the skin depth at your highest desired frequency…what is it compared with the wire diameter? You really don’t want the skin depth to be less than half this diameter…if it is consider using multiple strands of smaller gauge wire who’s total area is the same as the area you need….or consider using Litz wire….

Last thing to consider is your high frequency end…what distorts your reflected impedance at the high frequency end? It is mostly leakage inductance….and with more turns you will have more leakage inductance, but if you decrease turns you will increase your excitation current and lower your PRI inductance; both of which hurt your low frequency end response…plus it may move you too far toward core saturation….

Instead consider winding techniques to lower leakage inductance…

Interleaving your windings can greatly help with the leakage inductance and so will greatly help with your high end frequency response…..

Feel free to reach out to CET for any of your audio/impedance transformer needs! We have the design expertise and manufacturing ability to get you exactly what you want!