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If I build (a very poor) transformer (for pedagogic use) using a few windings of wire for the primary and secondary coils, and use a signal generator on the primary side, I can measure the voltage on the secondary side with an oscilloscope probe, but only if I don't ground the other side of the secondary. If I ground the other side of the secondary to the oscilloscope, I only see noise on the scope. If I use a commercial transformer whose secondary coil has a high inductance, then I can measure the voltage across the secondary in the usual way, by connecting one oscilloscope probe to one side of the secondary, and the other (ground) to the other side. This only seems to be a problem with small, make-shift transformers with low inductance. But I can't seem to understand this behavior. Thanks!

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    \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$ – Nick Alexeev Oct 2 '17 at 21:39
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The transformer must not load the impedance of the source more than 10% of rated load. We call this the excitation current.

Otherwise if primary impedance is low then the secondary impedance will also be low but determined by a turns ratio \$n^2\$ . Then if mutual coupling is low, you don't get much signal out.

Then capacitive coupling from the primary to secondary windings permits you to see a potential signal coupled to the secondary. But as soon as it is grounded or connected to a load, the signal would drop because of poor mutual coupling. then ensure Self Resonant Frequency is much greater than application from interwinding capacitance.

schematic

simulate this circuit – Schematic created using CircuitLab

See how grounding the other side in this case, shorts out the capacitance. Try reversing the secondary windings and see if it changes slightly.

Estimate the impedance or each component, R, L C and load to understand why you need to ensure the primary impedance is 10x higher than lowest load impedance (referred to input) for 10% load regulation. ( typical)

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  • \$\begingroup\$ I definitely think what I'm seeing is due to the capacitance! Thanks to you and others who pointed this out! But I don't completely follow what you are saying about the excitation current. I definitely considered that I needed the impedance of the primary inductor to be large relative to the resistance on the primary side, and I can use the signal generator to produce a high enough frequency that that is true. But now I realize that this also reduces the capacitive impedance, which explains the behavior I'm seeing. \$\endgroup\$ – user1247 Oct 2 '17 at 21:38
  • \$\begingroup\$ measure SRF then short secondary to measure mutual coupling , Lc.. Use an RLC nomograph chart to see Z(f) for each part and your impedance divider. \$\endgroup\$ – Sunnyskyguy EE75 Oct 2 '17 at 21:41
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5 turns, diameter = 2cm in a comment. I see that the coils are proper for tens of megahertzes. You have far too low frequency.

In the comments and other answers the signal is guessed to have a capacitive route to your scope input when the GND side of the secondary is disconnected.

But this is only one wire route. I must add that the second wire is the connection from generator's GND to oscilloscope's GND via the protective earth or capacitance from the power supplies to mains supply wires.

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  • \$\begingroup\$ It's definitely not external noise -- I'm picking up exactly the right frequency and waveform shape coming from the signal generator that is the source for the primary. And I only measure it on the secondary when it is right on top of the primary coil, and with about the expected voltage that depends on the winding ratio! \$\endgroup\$ – user1247 Oct 2 '17 at 21:06
  • \$\begingroup\$ @user1247 I see I misreaded the question. I'll fix the answer. \$\endgroup\$ – user287001 Oct 2 '17 at 21:09
  • \$\begingroup\$ (regarding your edit) I understand that these constitute a very poor transformer, but I'm still trying to understand the behavior. I think others hit the nail on the head regarding the capacitance between the coils. \$\endgroup\$ – user1247 Oct 2 '17 at 21:40
  • \$\begingroup\$ @user1247 one wire is not a circuit. I added something about it to the answer. \$\endgroup\$ – user287001 Oct 2 '17 at 22:24

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