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I want to prove that a voltage signal I apply in the primary of an ideal transformer(or 2 coupled coils) will produce the same voltage signal in the secondary, e.g. a square wave(+- V) in the primary yielding a square wave in the secondary by using the fact that V=L*(di/dt).

So let´s suppose I have a square wave generator. At the moment the square wave reaches +V then V=L1* (di1/dt) and the current I1 in the primary ramps up linearly, causing then V/(L1)*t = i(t), and that means (di1/dt) = V/L1.

This increasing current will then create a magnetic field B= k1 * i(t), where k1 is a constant and the induced voltage in the secondary is going to be V2= -(d(phi)/dt) = - k1 * (di1/dt) = -k1 * (V/L1).

The outcome shows the secondary is going to have the same wave shape signal as the primary, which in this case is a square wave. So assuming I have a V(t) applied to this ideal transformer, the secondary signal is going to be the proportional to this V(t).

Finally, my question is: Can I prove the secondary signal will be alike the the primary by the way I did above? Which is by using the inductor formula V=L*(di/dt)?

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  • \$\begingroup\$ The copy-voltage from input-to-output of a 1:1 ratio transformer depends upon the near-perfect cancellation in the core, of primary-winding flux being bucked (opposed) by the secondary-winding flux. Some tiny resistance in the voltage source powering the primary (or some tiny impedance at the frequency of interest) is needed; the tiny resistance/impedance implements a HUGE SERVO-LOOP REGULATION action. The core summation provides the negative-feedback behavior. Want a poor transformer? Add a large source impedance. \$\endgroup\$ – analogsystemsrf Aug 10 '18 at 8:00
  • \$\begingroup\$ @analogsystemsrf I think your comment is valid for a real transformer but the question asks about an ideal transformer. The common model of an ideal transformer cares only about the turns ratio and not resistance or second-order effects. \$\endgroup\$ – Elliot Alderson Aug 10 '18 at 14:02
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Ideal Transformer By Faraday's law of induction:

Vs = − Ns dΦ / dt Vp = − Np dΦ / dt

Where V is the instantaneous voltage, N is the number of turns in a winding, dΦ/dt is the derivative of the magnetic flux Φ through one turn of the winding over time (t), and subscripts p and s indicate primary and secondary.

Combining the ratio of equations 1 and 2: Turns ratio = Vp/Vs = −Np/−Ns

Now to your question: You can not simply determine i1 with the inductor equation V1=L*(di1/dt) on the primary side because i1 is also dependent on the magnetic flux through both windings and i2 on the secondary side. Also, L2 is missing in your equations.

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  • \$\begingroup\$ An ideal square wave alternates between two constant values. But a transformer only generates induced emf's in response to changing inputs, and the only changes are the instantaneous transitions between constant values. So a square wave output will not be achieved. To produce a square output, the input current needs to be triangular. \$\endgroup\$ – Chu Aug 10 '18 at 7:08
  • \$\begingroup\$ @Chu I think the critical issue is that the OP asked about an ideal transformer and the common definition of an ideal transformer says that the voltage ratio is equal to the turns ratio at any frequency, including dc. I agree, however, that this definition has very limited usability. \$\endgroup\$ – Elliot Alderson Aug 10 '18 at 13:59
  • \$\begingroup\$ @Elliot Alderson, I’d like to see the definition that says an ideal transformer will work at DC. \$\endgroup\$ – Chu Aug 10 '18 at 14:45
  • \$\begingroup\$ @Chu Try Wikipedia, as most students will. The "ideal transformer identity" does not include frequency as a factor; it states that the ratio of voltages is equal to the turns ratio. You can argue that this does not imply operation at dc, but it does imply operation at 0.000...01 Hz which is effectively dc in my book. Find the "Ideal Transformer" page at www.mathworks.com and you will learn that the model can also be used for an ideal DC-DC converter. I could cite many other sources but you only asked for one. I didn't say I was happy about it, but that's the way it is. \$\endgroup\$ – Elliot Alderson Aug 10 '18 at 15:13
  • \$\begingroup\$ @Elliot Alderson, the wikipedia article on the ideal transformer starts by saying '... A varying current in the transformer's primary winding creates a varying magnetic flux in the transformer core...'. Clearly, DC is not varying. \$\endgroup\$ – Chu Aug 10 '18 at 16:43

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