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I am trying to understand the datasheet of a small transformer (SMT transformer, ref B82801B from EPCOS, available for example here: https://www.elfadistrelec.no/Web/Downloads/_t/ds/B82801B_eng_tds.pdf?mime=application%2Fpdf ).

I am not an electrical engineer, and here is my understanding of how a transformer works (I write this here as it influences how I read the datasheet). There is a primary, resistance \$R_1\$ and inductance \$L_1\$, and a secondary, resistance \$R_2\$ and inductance \$L_2\$. Considering perfect coupling between those, the coupling coefficient is M = \$ \sqrt{L_1 L_2} \$. Therefore, I expect that the transformer follows equations (with \$V\$ and \$I\$ tension and intensity, and index indicates primary / secondary):

\$ V_1 = R_1 I_1 + L_1 \frac{d I_1}{dt} + M \frac{d I_2}{dt}\$,

and symmetric formula for \$V_2\$ (the sign before M may vary, depending on how the transformer is wired).

When I look in the datasheet page 3, I get to know which pins are to the primary and secondary coils. This is great, I think I get it right.

When I look at the datasheet page 5, I get:

  • The turn ratio. This is great, I understand (it will give me amplification if no load).

  • The typical max DC resistance \$R_i\$ for the primary and secondary. This is great, and I get the right value when I measure with the multimeter.

  • A \$L_{min}\$ value. What is that? Is it \$L_1\$, \$L_2\$, or \$M\$? Comparing the scaling of \$L_{min}\$ with the number of turns in the secondary, it looks like \$L_{min} \propto N_2^2\$, so I would guess actually \$L_{min}\ = L_2\$ as in general for a coil, \$L \propto N^2\$. Am I right?

  • What is Voltage-Time product? I guess it is maximum value of (Voltage) * (time when applied), am I right?

  • Why is there a recommended \$R_T\$?

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  • \$\begingroup\$ Are you aware that this is a current transformer and, given your question, I'm not convinced you know what you may want. Try and stick to asking a simpler more application-based question at first. \$\endgroup\$ – Andy aka Jun 2 '17 at 12:43
  • \$\begingroup\$ @Andy aka: Thank you for your answer, I think I may start to understand now. I was not aware of the difference current transformer / voltage transformer ;) \$\endgroup\$ – Zorglub29 Jun 2 '17 at 13:14
  • \$\begingroup\$ @Andy aka: I thought I had missed something by not seeing that this is a 'current transformer', but after a bit of searching online, transformers are always built on the same principle with 2 coils coupled together, and there is no theoretical difference between a current and a voltage transformer, or am I wrong? If I am right then my question should be valid (but it may be that I am not aware of some convention used when describing specifically 'current' transformers). \$\endgroup\$ – Zorglub29 Jun 2 '17 at 13:27
  • \$\begingroup\$ A current transformer needs a burden resistor if you are to operate it as a current measurement device. If you are just trying to make a few hundred volts it might be OK but also it might break down the insulation on the secondary and then it's lost its purpose. Don't use that specific device for that purpose is my recommendation. \$\endgroup\$ – Andy aka Jun 2 '17 at 15:27
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This is made for measuring the current. It can be used in switch mode power supplies or in other high frequency circuits 50 kHz...1 MHz. It has no use in audio or normal mains power frequencies. The primary coil is in series with the circuit whose current you are measuring. The secondary coil must have a load resistance. You connect the voltage measuring device in parallel with the load resistor.

The given equations are for preventing the saturation of the ferrite core and for calculating the load resistor for the wanted conversion ratio from Amperes in the primary to Volts in the output.

Leaving the load resistor off gives a high output voltage. It can be disastrous for the insulation and the connected voltage measuring circuits.

The voltage-time product is one simple way to characterize how long pulses with a given voltage can be applied before the unwanted transversal current (=an useless part of the current in transformers) grows too high. I believe that the voltage to be used here is the output voltage.

Addendum due the comments: This definitely is an ordinary transformer in terms of the transformer theory, but it has only 1 turn in the primary coil. Let's assume you have the 100 turns in secondary -model. Its secondary inductance is 2mH. The primary inductance is only 0,2uH. If you connect 5V to it, the unwanted transversal current grows at rate (5V/0,2uH) = 25A/us. If your Arduino could supply say 25mA, it takes 1 nanosecond before the used output is shorted. You have no way to generate say 100 picosecond pulses. And if you had it, it's still useless because the frequency range of the transformer ends to 1 megahetz and you needed say 10GHz.

If you need high voltage pulses, get a transformer that is designed for it.

You may arque, that I just wrote that the secondary voltage can grow harmfully high without a proper loading resistor. But this is in applications where the primary current is amperes and in any case there's other stuff in series to keep the primary current reasonable. The one turn primary alone is a short circuit and it should be, because otherwise it would remarkably disturb the measured current.

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  • \$\begingroup\$ So if I understand well, 'current transformer' means that it is a transformer which secondary should not be let opened without risking damage (I would guess due to the kind of insulation on the secondary?), but that apart from that it is a completely usual transformer. Or am I wrong? ;) What I want to do is put a LC (RLC in practise) circuit on the secondary, and drive the primary with the PWM of an arduino (at the resonance RLC frequency), to get a high voltage around the capacitor and generate sparks (hence the choice of this transformer, because of the ration 1:100). Sounds reasonable? \$\endgroup\$ – Zorglub29 Jun 2 '17 at 13:47
  • \$\begingroup\$ PS: this is why I want to know what \$L_2\$ is, and if it has something to do with the \$L_{min}\$ in the datasheet: it will be part of the RLC I want to make resonate, so I should know it to be able to tune the C value I take and the PWM frequency. \$\endgroup\$ – Zorglub29 Jun 2 '17 at 13:48
  • \$\begingroup\$ PPS: would it be ok to summarize it in this way. A current transformer has a very low impedance (resistance and inductance) on the primary, and a high \$N_2 / N_1\$ ratio. It means it will disturb very little the circuit on the primary side, and extract very little energy from it, while still giving a measurable tension on the secondary (but with very low intensity). Then the problem I will face is that probably my PWM will feel short-circuited if not I work at very high frequency. \$\endgroup\$ – Zorglub29 Jun 2 '17 at 14:19
  • \$\begingroup\$ Thank you, now the comments here + your addendum make everything clear :) I think there was a lot of confusion on many discussion feeds about current transformers I visited, I think your post makes it much clearer. And now I know that it is \$L_2\$ that was given in the datasheet :) \$\endgroup\$ – Zorglub29 Jun 2 '17 at 14:27

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