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Jump a head to the question if you aren't interested in the background of my project

Background of my project

I want to create objects of metal. So I need to be able to cast a metal. Aluminium seems great as its strong, light weight, easy to obtain, but most importantly: its not toxic. So I want to be able to melt aluminium (933 Kelvin is the melting point).

I was thinking about induction. Its effective and easier than using liters of gas.

I tried to make my own induction heating circuit, but I keep blowing MOSFETs (royer oscillator design) when I ramp up the voltage to increase the power (my 12Vac versions work, but prepared for 48Vac they break down, let alone 230Vac. Now I spend enough money on fancy MOSFETs, so I bought myself a 2 Kilowatts induction cook-top that should get me melting some aluminium.

The cook-top has a 0.33uF (MKPH marked) capacitor in parallel with the original flat spiral coil, running at a varying frequency of 20-24Khz, depending on the distance of the steel pot on top of the cook-top. the specs of the original flat spiral coil: 24 windings of Lytz wire, 58mm inner diameter, 153mm outer diameter.

Now I replaced the flat spiral coil that I found inside the cook-top with a self wound solenoid of isolated wire (1.75mm diameter) with the following specifications: 143 mm height, 50mm radius, 32 windings => yielding an approximate induction of 53.75uH according to my calculations.

This works! I heated up a piece of heavy iron with aluminium inside until the aluminium melted. Though it was fun, my setup is far from practical. I used a water bath to cool the coil and glass jar inside the coil to keep the water away from the work piece and put the work-piece inside the glass jar. Finally the glass cracked, and the water bath in combination with a coil @ 230Vac isn't typically safe. The self wounded solenoid must be cooled because if the isolation would burn or melt, the chance of a short in the solenoid is high, probably breaking down the circuit of the cook-top.

So I need a solenoid that needs no isolation as the windings are distant from each other. Now I have a nice wounded solenoid of copper pipe with the following specs: 110mm height, 28mm radius, 11.5 windings. It should have an inductance of 3.03uH according to the same calculations I applied for the self wounded solenoid.

53.75uH / 3.03uH = Factor 17.73 less inductance. So can I apply this solenoid with less inductance if I replace the capacitance with 17.73 greater value? Which would be 0.33uF * 17.73 = 5.85uF.

The question:

If I replace a solenoid in a LC parallel circuit with a solenoid with X times less inductance, and increase the capacitance with the same factor X, keeping the resonance frequency the same, does the current stay the same as well?

EDIT: I tried calculating impedance of the overall circuit. But with 12Vac royer oscilator test circuits I build, the measured current differs from my calculations. I used the "Example No1" calculation from this website: http://www.electronics-tutorials.ws/accircuits/parallel-circuit.html

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  • \$\begingroup\$ Now when I understand the Q factor, looking back at my question, I feel a bit stupid. Less inductance invokes less heat in the work-piece, meaning less energy is lost while resonating, meaning less current is drawn. Of course, its so logical! \$\endgroup\$ – Mike de Klerk Aug 28 '13 at 4:51
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The Q of a tuned parallel L-C circuit stands for quality factor and is a predictor to how much current flows in the tuned circuit at resonance. Here's the formula: -

Q = \$ \frac{1}{R_S}\sqrt{\frac{L}{C}}\$

\$R_S\$ is in series with the inductor, L

This basically means that if \$R_S\$ is zero then Q is infinite and the circulating currents in L and C are infinite for a non-zero voltage applied. Of course this doesn't happen because resistance is never 0 other than in a superconductor but that's another story!

Significantly for the question, if R stays the same and L is decreased by ten and C increased by ten (to retain the same resonant frequency), Q reduces by 10. In other words the high peaky tuned circuit you once had has become rather shallow and not so peaky in comparison.

Reducing inductance by ten does not usually mean \$R_S\$ reduces by ten. For a typical core wound with say 1000 turns, inductance might be (say) 1H. To reduce this to 0.1H means the turns reduce to 316 and the resistance therefore only reduces by about one-third.

Remember inductance is proportional to turns squared.

So after reducing inductance by 10 (resistance decreases by about 3) and increasing capacitance by 10, the net effect is a tuned circuit that has a Q that is 3x smaller than before and not as peaky or resonant.

Regarding the link - this is for an RLC parallel circuit where the resistance is represented as a parallel component and this is less-often used because the dominant loss in an RLC circuit is in the inductor (as series resistance). A parallel resistance could depict dielectric loss in the capacitor but this will be a tiny fraction (in most cases) of the inductor losses.

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  • \$\begingroup\$ Thank you so much for clarifying the Q factor. A factor I clearly overlooked. I am reading about it on wikipedia: en.wikipedia.org/wiki/Q_factor and it states Q = \$ \frac{1}{R_S}\sqrt{\frac{L}{C}}\$ for in series and Q = \$ \frac{1}{R_S}\sqrt{\frac{C}{L}}\$ for parallel circuits. Minor difference, but in my case it increases Q, meaning less energy per cycle is lost in the workpiece, meaning less current, which means it won't kill the cook-top :) But slower heating the workpiece as well of course. \$\endgroup\$ – Mike de Klerk Aug 28 '13 at 4:44
  • \$\begingroup\$ @MikedeKlerk A parallel tuned circuit with R in series with L has all three components in series and so the formula I've used is correct for predicting Q. Because induction comes from the inductor (clue in the name!) the power transferred represents a loss in the inductor arm of the circuit and this is modeled by a series resistor. \$\endgroup\$ – Andy aka Aug 28 '13 at 7:30

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