2
\$\begingroup\$

I've got an RLC meter which I can set so it's a fixed AC voltage amplitude source. I'm using a planar coil of 25 turns and in theory, when a secondary resistance increases, so should the primary. But this is not what I observe.

When the coil primary coil is isolated, the resistance is just the resistance of the wire, the reactance is high, so the current flowing is comparatively small.

But when I add a second planar coil on top of it, it decreases reactance, because the current in the secondary is flowing in such a way to create a magnetic field that is in the opposite direction from the primary, which reduces total magnetic field, reducing the reactance felt by the primary, which is why the current increases.

This is all fine, but I don't understand what is happening with resistance. Because it increases. And if I look at equations, it says that increase in secondary resistance should create an increase of primary resistance. Now if I add a resistor in series with the secondary coil, the reactance increases (less current is flowing which makes sense), but the problem is that resistance felt by the primary actually decreases. So adding resistance to the secondary actually DECREASES primary resistance felt, not increase it.

But in induction heating if we increase the temperature of the load (effectively increasing resistance) the resistance felt by the primary also increases. So why do 2 different experiments give 2 different results?

EDIT: I've done some more testing and found out that the resistance of the primary actually increases if I add resistors to the secondary to about the same resistance that the primary had on it's own. Any more resistance, and the primary resistance starts to fall. It has something to do with the power transfer but I don't know exactly how. And I'd like to find out.

\$\endgroup\$
  • \$\begingroup\$ You say that you are using the RLC meter as a fixed voltage source, but I can't see how it can generate an induced field in the transformer. Perhaps you could add a schematic and a drawing? \$\endgroup\$ – clabacchio Nov 16 '17 at 9:47
  • \$\begingroup\$ I don't see what you mean. Anything that generates current Will generate an induced field. RLC meter is keeping constant voltage with it's internal regulation. It is connected to a single planar coil. Then I bring the secondary coil to the vicinity of the primary one. That's the whole circuit. \$\endgroup\$ – MaDrung Nov 16 '17 at 11:00
  • 1
    \$\begingroup\$ This doesn't make sense to me either, aren't you forgetting Faraday's Law? The field that you're describing is unchanging. \$\endgroup\$ – replete Nov 16 '17 at 11:11
  • \$\begingroup\$ In order to have an induced EMF, you need a varying field, so a varying excitation current. Perhaps you didn't notice that the voltage generated by the LCR meter is not DC. \$\endgroup\$ – clabacchio Nov 16 '17 at 11:13
  • \$\begingroup\$ You need to know the excitation frequency of your RLC meter. Even measuring a single inductor, the value will increase, and then go capacitive, as you increase the frequency. With two coupled coils, it's more complicated. If it's autoranging, then the model it uses to report resistance will change between L and C behaviour. An auto RLC meter is probably too complicated to do meaningful experiments on this sort of transformer. You need full manual control, and then need to measure at many frequencies. \$\endgroup\$ – Neil_UK Nov 16 '17 at 11:33
1
\$\begingroup\$

This is all fine, but I don't understand what is happening with resistance. Because it increases.

I'll try to explain. RLC meter measures imedance Z of the coil and converts it to one of possible representations. What are talking about (I guess) is representation by series R-L circuit:
Z = R + jX,
where R is resistance and X is reactance.

schematic

simulate this circuit – Schematic created using CircuitLab

And if I look at equations, it says that increase in secondary resistance should create an increase of primary resistance.

Here you talk about parallel R-L circuit, where inductance of primary coil (with it's own small series resistance) connected in parallel with transformed impedance (resistance) of secondary. But this parallel R is not that resistance, that your RLC meter shows in series representation mode!

Try to switch RLC meter to parallel R-L representation (if it's possible). I also recomend reading Impedance Measurement Handbook, especially ch. 1.17-1.18.

EDIT:
At a given frequency the same imedance value Z may be achieved (or modeled) by various equivalent circuits, for example, parallel Rp and Lp, or series Rs and Ls. It's just an algebra. Which one to use it's your's discretion. In a frequency range it's likely that Z will be a function of frequency and equivalent circuit values may vary too (or not). So, the equivalent circuit that uses fewer componens, which values are frequency indepenent (or less dependent) will be more appropriate model of real object. Let's get back to the transformer. When it is idle, impedance of the primary is almost reactive (inductive). This can be modeled by ideal inductor in series with small Rs:
Z = Rs+jXl, or by ideal inductor in parallel with large Rp:
Z = 1/(1/Rp+1/jXl).
(If Rs much less than Xl, then Rs/Xl=Xl/Rp.) For idle transformer the series cirquit is more adequate, because Rs corresponds to the primary coil wire resistance. When active load appears it increases active (in-phase) part of primary current. This may be modeled by increasing of Rs or by decreasing of Rp. In case of heavy load (when wire resistance becomes negligible contributor to primary impedance) parallel equivalent circuit becomes more convenient, because Rp corresponds to (transformed from secondary-to-primary) load value.

\$\endgroup\$
  • \$\begingroup\$ I've read the whole guide, but it's not that well explained for my knowledge level. My LCR meter does have Rs and Rp mode and I now see the correct behaviour when I use the Rp mode. But I'm confused about what Rp and Rs represent. When I have a simple air transformer and meassure Rs, what do I see about the load? I see some load resistance added to the coil resistance when I put a conductor near the coil. And when I watch Rp I see some other value. I'm confused about what they tell me about the load. \$\endgroup\$ – MaDrung Nov 17 '17 at 11:48
  • \$\begingroup\$ @MaDrung, I slightly expanded my answer. \$\endgroup\$ – Eugene K Nov 17 '17 at 23:31
  • \$\begingroup\$ So if I understand this, no matter the system, if we use parallel or series meassurement setting we get the same impedance and the same power losses can be calculated using it. It's just a matter of how we want to Express the data. So using series calculation for air transformer with load would be more practical because I could calculate the losses of system by just P=I^2 * R, while in parallel mode, I'd first have to calculate how much current is flowing trough the Rp, and then use the same equation to in the end get the same power loss? \$\endgroup\$ – MaDrung Nov 20 '17 at 10:50
  • 1
    \$\begingroup\$ You are right! I just whant to add a bit to your reasoning. Losses, or an 'acive power': $$ P_{a}=Re\left ( U\cdot I^{\ast } \right )=Re\left ( U\cdot \left ( \frac{U}{Z} \right )^{\ast } \right )=\left \| U^{2} \right \|Re\left ( \frac{1}{Z} \right )=\left \| U^{2} \right \|\frac{1}{R_{p}} $$ Also, $$ P_{a}=Re\left ( U\cdot I^{\ast } \right )=Re\left ( I\cdot Z\cdot I^{\ast } \right )=\left \| I^{2} \right \|Re\left ( Z \right )=\left \| I^{2} \right \|R_{s} $$ \$\endgroup\$ – Eugene K Nov 21 '17 at 18:51
  • \$\begingroup\$ I'm having troubles with my induction heating meassurements though. I'm meassuring Rp or Rs of primary with a metal piece as secondary. The problem is that the Rs and Rp both increase when I heat up the metal (increasing resistance). But doesn't that contradict itself because if Rp is increasing the resistance of secondary should also be increasing, but when I meassure Rs I see the same increase in resistance, when it in fact should be falling? I think that if Rp is increasing, Rs should be decreasing. Isn't that right? \$\endgroup\$ – MaDrung Nov 23 '17 at 7:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.