Optimal coil specification connected to a 160W audio amplifier to produce an "intense" varying magnetic field [duplicate]

Question

Disclaimer: I’m computer scientist

I would like to generate an "intense" varying magnetic field by connecting a coil to a max 160W audio amplifier. The theoretical output frequency range is 20Hz to 20kHz. I plan to only output sinusoidal signals in the range 500Hz to 5000Hz.

The audio amplifier I have to do that is a 4Ω SMSL SA 98E 2x160W specifications in french, less detailed specifications in English from the manufacturer.

I would like to know what should be the optimal inductance of the coil I should use and the wire thickness.

I have so far tried a 0.5mH coil with 0.5mm wire, but the coil heats which indicates that the wire is too thin. 0.5mH is ~4.4Ω at 1kHz.

What is the most unclear to me is the 4Ω impedance required by the amplifier. At what frequency should it be 4Ω ? I could not find a clarification on that with google. But I did find plenty of warnings that it can harm the amplifier if the impedance of the HP is too low.

Since I don’t want harm my amplifier, I ask the experts here if the 4Ω impedance should be at 1kHz or 20kHz. In the later case the inductance of my coil should indeed be much lower.

Edit 1: as asked by StackExchange I must clarify why this question is different from this similar question. The titles are very close, but I initially badly formulated my question which ended on a different topic (measuring inductance). Tony Stewart Sunnyskyguy EE75 did provide a precise answer to the question in the title but I was unsure if it was safe for my amplifier. This question is now more on topic and ask about the risk for my amplifier to use a low inductance. I apologise for the inconvenience.

Answer 1

Your amp has a max output current of 10A (datasheet) and a peak output voltage of 36V.

36V peak is 25V RMS, so if you want 10A into the load that's a maximum impedance of 2.5 ohms, which corresponds to an inductance of 80µH.

Adding a bit of safety margin, let's say 100µH.

Of course, at 500Hz, the inductive part of the impedance will be 0.25 ohms, so you won't be able to use full power or the current would simply be huge. But that's a class D amp, which means it has no Safe Operating Area restrictions for the output transistors, which means you don't have to add a resistor in series or do anything special for this extremely low load as long as the output voltage stays low enough to not exceed the maximum current.

So what you could do is just wire enough turns for 100µH, then set the voltage on the signal generator to have the AC current you want in your coil. This means you must set a voltage that is proportional to frequency.

Pros: you always get the same current in the coil no matter the frequency, and you don't have to add a resistor in series to waste power.

Cons: it's a bit more complicated since you have to adjust signal amplitude.

Answer 2

I would like to know what should be the optimal inductance of the coil I should use and the wire thickness.

Generally speaking a 4 Ω speaker load will have a DC resistance that is about 3 Ω. This is from tests I conducted many years ago on 8 Ω speakers; they tended to average out at 6 Ω DC resistance. OK that's my limited scope on what sort of impedance might be presented to an amplifier by the load of the speaker.

Based on my limited experience, I suggest that you have an external 3 Ω resistor (that will generate the heat) and wind the coil with thicker wires (in order to reduce the winding resistance) and aim for 0.5 mH as per your original design aim. However, if you think you might need more inductance then you might be on a slippery slope...

Inductance is proportional to turns squared (if all the turns couple to each other) whereas inductance is proportional to just turns (to the power 1) if there is no coupling. How to figure out the amount of coupling is difficult but, maybe aim for something a bit closer to 2 than 1 as the power.

I mention a slippery slope because as you add turns to get more inductance, for a given drive voltage and frequency, the current falls because the impedance has risen. However, inductance isn't the primary factor for magnetism; it's ampere-turns that create a stronger magnetic field, so in some respects, having a few turns of thick wire connected via a step down transformer to your power amplifier may be better.

For instance, a 4:1 step-down transformer connected to a load of 0.25 Ω (reactive) will look like 4 Ω to the amplifier terminals. But, it's difficult to guess what might be best for you; I'm just suggesting some options to try and boost the magnetic field.

Having said all of that, I would still have an external fixed value 3 Ω resistor feeding the inductor directly (or the transformer).

Answer 3

I understand you want to make an "intense" field, so, absent any other specifications, I interpret that to mean 'stronger than the one you have at the moment'.

How do you make your existing field stronger?

• less airgap

Does your existing coil have any air in the path that could be ferrite. If it's a solenoid, then can the return path be made to be ferrite? Halving the length of the air path round the winding will double the H field and so B field across the remaining airgap.

• more current

You can't get more current from your amplifier. 160 W at 4Ω comes out to about 6 A rms, and bobflux has found the datasheet gives 10 A, which is consistent with that. The amplifier recommended load resistance is simply a way of limiting the output current, given that the voltage is automatically limited by the power supply.

But, at a particular frequency, you can resonate the coil with a capacitor, to increase the coil current to many times that of the amplifier output. If you had a selection of specific frequencies, and a box of capacitors with a switch, the changeover could be very quick.

• lower duty cycle

How can you tolerate the extra coil current? It doesn't matter if a coil of copper gets hot, it only matters if it gets too hot. Let it cool down between pulses, and you can safely use a single massive pulse that heats it to maximum temperature in milliseconds. It's quite easy to calibrate the I²t of a load, which is the adiabatic heating it can tolerate.

Answer 4

4 Ohms = 20kHz @ 30uH

For 2kHz use 300uH but wire resistance should be capable of dissipating x% of total power output such as 1% of 100W=1W to keep cool.

Since you used 0.5mm wire = 4.4 Ohms you lost about half the power in heat, so to achieve 1% of 4 OHms the radius must increase 10x for 1% of 4.4 Ohms and about same length of wire. L will vary due to geometry of radius and height as well.

The L/R =T ratio also defines the upper frequency limit or rise time to T=64%

R is a choice of heat dissipation and ability to cool it affect temperature rise by thermal conduction and not insulation.