# Transformer question

#### hensle

Joined Dec 31, 2011
46
According to basic transformer theory, the voltage in the secondary is given by MdI/dt. That is the mutual inductance times the change in current of the primary. This would mean the higher the frequency, the higher the voltage. However, that is not observed in transformers. Their voltage is independent of frequency. What am I missing?

#### ZCochran98

Joined Jul 24, 2018
273
Inductance is a (mostly) constant thing with frequency. It's inductive REACTANCE that varies. So, as frequency increases, the inductance in both coils stays the same, but the reactive element of the impedance increases. That leads to an effective drop in the current, given a voltage, so the ratio stays the same.

To understand this more clearly, let's look at the simplest form of the math (ignoring any resistances or load mismatches or whatnot):
$v_1 = L_1\frac{di_1}{dt}$
$v_2 = M \frac{di_1}{dt}$
Converting to frequency-domain via a Laplace transform, we get:
$V_1 = j2\pi f L_1 I_1$
$V_2 = j2\pi f M I_1$
The ratio of the two gives:
$\frac{V_2}{V_1} = \frac{j2\pi f M I_1}{j2\pi f L_1 I_1} = \frac{M}{L_1}$
We can even do this in time domain!
$\frac{v_2}{v_1} = \frac{M\frac{di_1}{dt}}{L_1 \frac{di_1}{dt}} = \frac{M}{L_1}$

So, the ratio between the two voltages is constant, assuming zero resistance and a few other factors.

---

As a side note, we can actually get the turns ratio version of the equation from this! Assuming perfect coupling between the two sides, we have the mutual coupling, by definition, being:
$M = \sqrt{L_1L_2}$
So:
$\frac{M}{L_1} = \sqrt{\frac{L_2}{L_1}}$
The equation for an inductor coil is:
$L = \frac{\mu_0\mu_r N^2 A}{l}$
So if the two coils have the same core, length, and cross-sectional area, but have numbers of turns $$N_1$$ and $$N_2$$, the ratio above simplifies to:
$\frac{M}{L_1} = \sqrt{\frac{L_2}{L_1}} = \sqrt{\frac{N_2^2}{N_1^2}} = \frac{N_2}{N_1}$
Which is exactly as expected for an ideal transformer.

#### hensle

Joined Dec 31, 2011
46
Inductance is a (mostly) constant thing with frequency. It's inductive REACTANCE that varies. So, as frequency increases, the inductance in both coils stays the same, but the reactive element of the impedance increases. That leads to an effective drop in the current, given a voltage, so the ratio stays the same.

To understand this more clearly, let's look at the simplest form of the math (ignoring any resistances or load mismatches or whatnot):
$v_1 = L_1\frac{di_1}{dt}$
$v_2 = M \frac{di_1}{dt}$
Converting to frequency-domain via a Laplace transform, we get:
$V_1 = j2\pi f L_1 I_1$
$V_2 = j2\pi f M I_1$
The ratio of the two gives:
$\frac{V_2}{V_1} = \frac{j2\pi f M I_1}{j2\pi f L_1 I_1} = \frac{M}{L_1}$
We can even do this in time domain!
$\frac{v_2}{v_1} = \frac{M\frac{di_1}{dt}}{L_1 \frac{di_1}{dt}} = \frac{M}{L_1}$

So, the ratio between the two voltages is constant, assuming zero resistance and a few other factors.

---

As a side note, we can actually get the turns ratio version of the equation from this! Assuming perfect coupling between the two sides, we have the mutual coupling, by definition, being:
$M = \sqrt{L_1L_2}$
So:
$\frac{M}{L_1} = \sqrt{\frac{L_2}{L_1}}$
The equation for an inductor coil is:
$L = \frac{\mu_0\mu_r N^2 A}{l}$
So if the two coils have the same core, length, and cross-sectional area, but have numbers of turns $$N_1$$ and $$N_2$$, the ratio above simplifies to:
$\frac{M}{L_1} = \sqrt{\frac{L_2}{L_1}} = \sqrt{\frac{N_2^2}{N_1^2}} = \frac{N_2}{N_1}$
Which is exactly as expected for an ideal transformer.
Thanks much for your very thorough reply. I think I understand. The back emf from L sets the primary voltage which also depends on frequency, so in comparing the primary voltage to the secondary voltage, the frequency cancels out.

This brings me to another question which you might find interesting. What if I put a capacitor in the primary circuit and drive it at resonance? In such a case the impedance from the inductance is cancelled out and there is only impedance from the base resistance. In this case, would the output voltage then be dependent on frequency?

I am actually doing tests of this sort and I am still not seeing a frequency dependence (other than what seems to be related to the ability of the ferrites to be magnetized).

#### ZCochran98

Joined Jul 24, 2018
273
Thanks much for your very thorough reply. I think I understand. The back emf from L sets the primary voltage which also depends on frequency, so in comparing the primary voltage to the secondary voltage, the frequency cancels out.

This brings me to another question which you might find interesting. What if I put a capacitor in the primary circuit and drive it at resonance? In such a case the impedance from the inductance is cancelled out and there is only impedance from the base resistance. In this case, would the output voltage then be dependent on frequency?

I am actually doing tests of this sort and I am still not seeing a frequency dependence (other than what seems to be related to the ability of the ferrites to be magnetized).
Good question. It could be dependent on frequency, especially depending on your load on the secondary. This is the principle on which Tesla Coils work, after all! To get them to work "just right," the frequency, capacitor, and primary all need to be tuned "just right." However, the frequency may be a bit challenging to reach where it becomes interesting, depending on the size of the inductor.

If, on the primary, we have a capacitor $$C$$ in series with the primary, then we can write the primary voltage/current relation as (for the voltage across both the capacitor AND the primary - not just across the primary):
$V_1 = j\left(2\pi f L_1 - \frac{1}{2\pi f C}\right)I_1$
And the secondary as:
$V_2 = j2\pi f M I_1$
The ratio of the two becomes (using $$\omega = 2\pi f$$ so I don't lose my sanity typing it out over and over):
$\frac{V_2}{V_1} = \frac{\omega M}{\omega L_1 - \frac{1}{\omega C}}$
$= \frac{\omega^2 MC}{\omega^2 L_1C-1}$
If we let $$\omega_1^2 = \frac{1}{L_1C}$$ and $$\omega_M^2 = \frac{1}{MC}$$, then we can write this as:
$\frac{V_2}{V_1} = \left(\frac{\omega}{\omega_M}\right)^2\frac{1}{\left(\frac{\omega}{\omega_1}\right)^2-1}$
As an example of what this might look like, if we take $$\omega_M=\omega_1=1$$ as an example, then the response would look like (in dB scale): So, you will get some kind of frequency response. At $$\omega = \omega_1$$, you'll get a peak, for $$\omega > \omega_1$$ you'll get a little rolloff but mostly flat response, and for $$\omega < \omega_1$$ you'll get a rolloff of about 40 dB/decade. As $$\omega\rightarrow\infty$$, the response approaches $$\left(\frac{\omega_1}{\omega_M}\right)^2$$.

Using the same stuff as before, we can actually approximate that limit base purely on the turns:
$\frac{\omega_1^2}{\omega_M^2} = \frac{MC}{L_1C} = \frac{M}{L_1}$
Which, earlier, we saw this to be:
$\frac{M}{L_1} = \frac{N_2}{N_1}$
The math checks out to be what it was before.
So, to summarize, all the "fun" stuff happens when $$\omega = \omega_1= \frac{1}{\sqrt{L_1C}}$$, and reduced gain occurs for when $$\omega < \omega_1$$.

Edit: and, yes: you'll see a hysteretic effect from the ferrites being magnetized, as the magnetic domains can only magnetize/demagnetize so much so quickly. If you use air core coils, you won't see that limitation anymore, which is why Tesla coils almost exclusively are air-core towers.

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#### ZCochran98

Joined Jul 24, 2018
273
As a note: the "real" response at $$\omega = \omega_1$$ wouldn't be an infinite spike, as would be predicted by the above equations, as resistance is ignored. Real resistance would cause that spike to "round off" and lead to some finite peak in voltage. Plus, loading effects get "reflected" back onto the primary, which would cause other effects (such as less-than-perfect efficiency). Plus, you'd see the peak of the "spike" wouldn't be at exactly $$\omega_1$$, but off to the side (translated by some amount dependent on resistance).

#### hensle

Joined Dec 31, 2011
46
Thanks for your input. It was very informative.

One note for others who might read this. You mentioned the Laplace and many may not be aware of the various transforms. Another way to look at it is to use the maximum value in alternating current ($$I_m$$). For example: $I=I_m \sin \omega t$
Then ${{dI}\over{dt}} =I_m \omega\cos\omega t$
So the maximum change in current would be $$I_m\omega$$ which results in your transformed equations.
In my experiments I approach it this way. Typically I only go to RMS with average power and one can simply divide the result by 2.

#### ZCochran98

Joined Jul 24, 2018
273
Thanks for your input. It was very informative.

One note for others who might read this. You mentioned the Laplace and many may not be aware of the various transforms. Another way to look at it is to use the maximum value in alternating current ($$I_m$$). For example: $I=I_m \sin \omega t$

Then ${{dI}\over{dt}} =I_m \omega\cos\omega t$

So the maximum change in current would be $$I_m\omega$$ which results in your transformed equations.

In my experiments I approach it this way. Typically I only go to RMS with average power and one can simply divide the result by 2.
Precisely, and, as you pointed out, ultimately, it'll get you the same results, with more-or-less the same amount of work (as it all works out to be some variation of algebra).

Laplace transforms are a nice tool to have in the "engineering toolbox," along side Fourier transforms, that allow you to easily go from time to frequency domain and back, simplifying some of the math (after first complicating it if you have to do a transform by hand). However, until you get to more complicated systems, they're not always necessary if you're comfortable with derivatives and ODEs.

The transforms also don't strictly work with RMS; you CAN use RMS values as your $$I$$ and $$V$$ terms in the post-Laplace equations, but you can also use the peak value (just remembering to throw an 1/2 in front of the power equation for AC: $$|P_{AC}| = \frac{1}{2}|I||V|$$, or $$|P_{AC}| = |I_{RMS}||V_{RMS}|$$).