# Driving a 5A current through a coil

To many electronics engineers this might seem a stupid question but I was given by my boss the task to find a solution to drive approx. 5A through a coil. The coil's resistance (not considering inductance which would be ~15mH) is about 90 Ohms. The signal is sinusoidal with a frequency of about 3kHz. Since he made this sound pretty easy I am a little bit unsure of how to proceed.

Considering Ohm's law, driving 5A through a 90 Ohms would result in 450V which doesn't seem so trivial to me.

I have a feeling that I'm overlooking some fact but I cannot figure out which one.

A possible idea would be to use a OPA549 which gets at least roughly into that area (bandwidth, output capability) but again, I think that I'm overlooking something. All power amplifiers I can find are intended for audio applications and typically drive a 4 Ohm or 8 Ohm load.

I'd be really happy if you could tell me where I'm wrong or if this actually isn't such an easy task.

• I smell an xy problem, you tell what you want to do, but not why. Nobody is given just the task to run some current through a coil. That coil and the current that runs through it has a purpose. Knowing about it will make it possible to suggest better options. – PlasmaHH Mar 30 '17 at 8:33
• Your first instinct that this is not trivial is correct. In fact it gets worse. The inductive impedance of an inductor is given by 2πFL for you this will be 282 Ohms this adds to the resistance at right angles like a right angled triangle so the total impedance is sqrt(282²+90²)=296 Ohms. To get 5A you are going to have to generate about 1500V from the driver. – RoyC Mar 30 '17 at 8:43
• Sounds like you need a better boss. I'd be going back and questioning his motive for the request. If he insists this is what he needs then tell him he needs to contract a high-voltage specialist because this is patently outside your comfort zone. – Trevor_G Mar 30 '17 at 10:15
• Continued: Reading that back is sounded harsh. It was not intended to be. It was meant as advice. Never let your boss pressure you into trying to develop something that you are not comfortable with. Especially if it might just kill you, or others, if you mess up. – Trevor_G Mar 30 '17 at 10:30
• I'd love to tell the motives behind the question but am simply not allowed to. And, yes, it has a purpose altough it's a questionable one. The main idea is to drive a Helmholtz configuration and use the magnetic field created by it for further tests. There is not bad thing in being harsh, at least that's helpful advice. My ballpark estimations (not taking into account the inductive nature at first) already kind of told me that this is going to be out of my league but I wanted to confirm if I was overlooking something. There's no way I'm going that road. – Tom L. Mar 30 '17 at 16:36

The inductance of the coil at 15mH gives you a reactive impedance of about 280 johms at 3kHz. That's a larger impedance than your resistance, so will dominate the voltage you need across the coil, 5A * 280ohms = 1400 volts, before you add the extra voltage for the resistance.

Assuming the 5A is 5A rms, you will be dissipating $I^2R$ = 2250 watts in the resistance of the coil, no small amount, and way out of the league of anything like OPA549.

I would suggest that you reduce the VA you need to drive the coil by cancelling the series inductance with a series capacitor. To resonate 15mH at 3kHz needs a capacitor of 180nF. You will still need to supply the full 5A, but only supply the 450v needed to drive it through the resistance. The alternative parallel tuned circuit connection ideally needs a current drive (or an inductor) and must supply the full 1500v resonant voltage, but at a lower current. Obviously, the series connection is easier on two counts.

Finally, you need a voltage source. One option is to buy a 3kW audio amplifier, and use a transformer to match its output drive to the requirements of the tuned circuit. Obviously this transformer will need to handle 2.25kW, but at 3kHz, you will be able to use a much smaller core than an equivalent rating mains transformer. At 3kHz, it cannot be a conventionally iron-cored mains transformer, you will need to use ferrite. At that low frequency, ferrite heating losses will be low, so you will be able to run up near the saturation field.

Another option that's just within reach is to use a 450v H bridge. Here, although the voltage supplied to the resonant circuit will be a square wave, the current flowing, and the voltage across the coil, will look very sine-like. If you can tolerate the waveform distortion, and engineer the high voltage H bridge, then this will be cheaper than a 3kW amplifier and a transformer.

• For those interested in the details of the calculation, I get the value of $C=\frac{L}{\left(L\cdot\omega\right)^2+R^2}\approx 170.37\:\textrm{nF}$. Good idea. – jonk Mar 30 '17 at 11:55
• Thank you @Neil_UK. That confirms (even in numbers) what I actually feared. As this is so out of my league, I'll leave this to someone more suited to the task (I usually care about FPGAs, Microcontrollers and such ...). – Tom L. Mar 30 '17 at 16:38
• @jonk My plan was to cancel the inductance of the coil, requiring 180nF, so that the drive sees the resistance alone. Why does your forumla include the resistance? Is that so the undriven ring-down is at 3kHz? If, so, I think my value is more appropriate to the use case. – Neil_UK Mar 30 '17 at 18:17
• @Neil_UK No. I started out imagining as you did. But quickly realized that isn't the case. The series resistance shifts the resonant frequency. Do you see why? (The series part is $a+i b$, but to go parallel you need to invert the equation into conductance to add again -- it's gnarly on paper and took me a few minutes.) Then I just looked it up on google like I should have done before and that confirmed my results. The resistance shifts the resonance and cannot be ignored. You get a close value because R isn't as big as $Z_L$, but R is too close to fully ignore. Try in spice, I suppose. – jonk Mar 30 '17 at 19:35
• @Neil_UK If you do try it in spice, make sure to look at the RMS current required by the power supply (the RMS voltage obviously is the same in all cases) and so then take a look at the resulting power factor. You'll find the value I gave almost provides the optimal power factor (you can compute more digits if you want them) and the lowest RMS current from the supply rail. Any deviation away from it makes the RMS current from the supply worse. The computation is good, I think. – jonk Mar 30 '17 at 19:57

The voltage across the coil (driven DC, ignoring inductance) will be 450V, as you have calculated. The power dissipated in the coil will be $450\mathrm{V}*5\mathrm{A}=2.25\mathrm{kW}$.
When you add the inductance of the coil in, things get even worse. The impedance of the coil is $z=j\omega L \approx 280\Omega$. To drive 5A through it requires 1400V and 7kVA.