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I've been working on a model maglev train as my Grand Hobby Project; something around the H0 scale.

The initial idea was to use the Inductrack mechanism to keep it simple. To summarize, a Halbach array of rare earth magnets produces current in a conducting track, which in turn generates a repulsive force. Unfortunately the theory doesn't scale well downwards; either you have to go really fast (Mach 3) or need 3 cm of copper to drive the resistance down.

My current plan is to keep the array, but manage the currents in the track actively. Back-of-the-envelope calculations indicate I can get satisfactory lift with 2-3 amps, and I can even use PCB with 1 mm traces as the track without turning it into a hot plate.

However, that's still a respectable current where I come from, and the load is on the order of milliohms. I have no idea how to build a circuit that can drive those currents without blasting 99% of the energy in a sense resistor.

Other "requirements" are reproducibility and ease of setup. If I have to tweak a trimpot or solder 10 components for each trace in the track to get the forces to balance out, this thing will never get done. Optimal solution would be some off-the-shelf voltage-controlled IC.

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3 Answers 3

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So you can build a +/- 0.5V at several amperes power supply, using switching regulators, say.

Then build an output buffer stage for an op-amp fed from that supply, current feedback from a small sense resistor (say 10-20m\$\Omega\$) with Kelvin connection. Done.

If you don't care about crossover distortion, a complementary pair of BJTs will do the trick. Since the op-amp will be powered by a much higher voltage (say +/-5V) there will be plenty of voltage to drive the bases.

Edit: Something like the below circuit. The transistors need to have a beta of something like 100 at your 2-3A, so the base current will +/-20-30mA. For example 2SB1412/2SD2118.

schematic

simulate this circuit – Schematic created using CircuitLab

As @GeorgeHerold suggests, if you only need one polarity, you can leave out the transistor and associated supply (probably the PNP would be best to excise), which makes it much simpler.

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  • \$\begingroup\$ Unfortunately I've always found the BJTs a bit mysterious, so could you please provide a bit more details about the transistor solution? (Should have taken up the opportunity to learn in the university, but I only needed to pass the basic electronics course and knowing the op-amp was sufficient...) \$\endgroup\$
    – lrasinen
    Commented Jul 14, 2014 at 19:27
  • \$\begingroup\$ See edit above. You could use higher voltage for the supplies, (like +/-5v) but the transistors will get hotter (10x the dissipation) and would require heatsinks. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jul 14, 2014 at 19:56
  • \$\begingroup\$ Looks like I could pull that off. One more thing(TM): What's the rationale for the R5 value? \$\endgroup\$
    – lrasinen
    Commented Jul 14, 2014 at 20:08
  • \$\begingroup\$ R5 is not too small wrt the medium frequency output impedance of most op-amps (usually around 100 ohms) and prevents oscillation via C1/R6. If you make it much higher, then it won't allow enough current to get to the bases, much lower and it won't perform the desired function. \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jul 14, 2014 at 20:31
  • \$\begingroup\$ Interesting choice of power transistors with low Vce(sat) implies Rce~ 1/16 Ohm with high hFE but is completely unnecessary with negative feedback reducing the driver impedance and the mag power is far too low with <1W \$\endgroup\$
    – D.A.S.
    Commented Dec 8, 2020 at 18:58
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I think Spehro is talking about a push-pull output. A simple one is here. http://en.wikipedia.org/wiki/Push%E2%80%93pull_output

But I wonder if you need both polarities? If not then just an opamp current source with a transistor buffer. like fig 7 here, http://en.wikipedia.org/wiki/Current_source#Op-amp_current_sources

Should be fine...

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  • \$\begingroup\$ Welcome, George! \$\endgroup\$
    – Spehro 'speff' Pefhany
    Commented Jul 14, 2014 at 20:05
  • \$\begingroup\$ The magnetic field is sinusoidal, so I'll need to match the direction of the current with the sign of the field. In the first prototype I plan to do this by physically routing the same current in $\phi$ and $\phi+\pi$ conductors, but in opposite directions. But when/if I scale this to full-track size, it might make more sense to have independent control of the currents for propulsion. \$\endgroup\$
    – lrasinen
    Commented Jul 14, 2014 at 20:16
  • \$\begingroup\$ Hmm so you will have loops of copper that match the period of the Halbach array, with a flip in the sign for each loop? \$\endgroup\$
    – George Herold
    Commented Jul 14, 2014 at 20:43
  • \$\begingroup\$ The protoboard will have just transverse copper strips matching the period which I'll connect with wires. Reason being that I can then use the same boards to test non-Halbach approaches. \$\endgroup\$
    – lrasinen
    Commented Jul 15, 2014 at 8:39
  • \$\begingroup\$ Did some back-of-the-envelope calculations: the system is the very definition of unstable. But that solves the propulsion question nicely; I can just plug a moving setpoint in the stability control. I also realized that an H-bridge can also handle the polarity issues, and I suspect that'll be cheaper than a separate supply for negative. But that's for prototype II, "Make it stay put". \$\endgroup\$
    – lrasinen
    Commented Jul 15, 2014 at 17:28
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I'd go for a buck converter to give you low voltage high current with good efficiency. Then use a hall effect sensor to determine the current flow and use that as feedback to your buck converter to precisely control the current.

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