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I am using 0.010" nitinol shape memory alloy wire (Datasheet) to actuate a lock. The wire needs to be resistively heated to actuate, and I have empirically measured the cool resistance to be 0.83 Ω (~3").

When using a lab power supply with a constant current limit of 700 mA, the system works wonderfully, but for production we are looking to use a single 12VDC 2A supply for the system. As this supply is expected to heat the wire and while still powering the control circuitry I need something to limit current. The device does not have forced cooling or room for large heatsinks, so we do need to stay fairly efficient.

How can I embed a simple ~700 mA supply? Would a buck converter intended for driving LEDs work? (1, 2, 3)

These are niche devices – if it is not possible to do this efficiently with a low part count, a <$30 prebuilt module is within budget.

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  • \$\begingroup\$ You could use a buck convertor, but it's probably overkill. A simple 1:20 PWM driver is probably all you need. \$\endgroup\$ Commented Apr 4, 2017 at 3:17
  • \$\begingroup\$ @Trevor, I understand how I can control the heating of the wire by modulating the full 2A supply, but how can I do that without knocking out my control circuitry? I only have one supply. \$\endgroup\$ Commented Apr 4, 2017 at 3:22
  • \$\begingroup\$ You would need to add an inductor in the line so the current averages at 700ma. (plus a flyback diode of course) \$\endgroup\$ Commented Apr 4, 2017 at 3:29
  • \$\begingroup\$ You can make is smarter by adding a sense resistor of course. \$\endgroup\$ Commented Apr 4, 2017 at 3:31
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    \$\begingroup\$ @TonyStewart.EEsince'75, I am happy to use a prebuilt assembly, but all of the CC regulators I can find are labeled as "LED drivers", hence the question "can I use an LED driver for nitinol?" \$\endgroup\$ Commented Apr 4, 2017 at 4:32

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I'd suggest LED CC drivers would be unsuitable for your application since your voltage requirement is so low.
You could use a simple PWM solution but this is completely unacceptable due to the high pulse current. (0.83 Ohms will result in 14 Amp pulses, which exceeds your wire current capability (may act like a fuse) so you need to have a switching solution with a step down inductor which increases complexity)

Looking at the Nitinol datasheet, you should have a maximum operating current of 1 Amp for your wire.
With your off resistance of 0.83 Ohms that requires a voltage of about 830 mV.

Note: The datasheet says your wire is 0.47 Ohms per inch ...so I'm not sure your question has the right values. Why is the resistance not 1.42 Ohms??

enter image description here

You could design a bespoke constant current switching regulator for the purpose, but I'd suggest a quite reasonable solution would be to use a TI LM2596 based Buck regulator which supplies a minimum voltage output of 1.235 V. You could simply use a 0.5 Ohm (1 W) series resistor so you pass 1 Amp when the Nitinol wire is cold. It's resistance goes down as it heats up so you can adjust the series resistor to ensure you don't overcurrent the wire when hot.

There are many LM2596 based Buck regulators on Ebay (I use them all the time as replacements for LM78XX regulators) at very cheap prices (around $1.50). There are also plenty of constant voltage/current modules, but they are really based on a foldback current design which I doubt you actually need.
You could design you own of course ...but I'd suggest the constant voltage modules would be fine for the task.

Be aware that no matter what approach you take, there will be pulse currents of at least 1 Amp drawn from your power supply. You may need to ensure you have sufficiently large output capacitors on your 12 V supply since it is limited to 2A.

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    \$\begingroup\$ Hysteric Continuous Conduction Mode (CCM) Buck regulators ought to have triangular currents over DC but well under rated (sat) current of choke. Ripple voltage is irrelevant here as you very well pointed out. Cap, FET & L [mΩ] affects efficiency losses but reducing hysteresis range of CCM reduces peak/avg current with some limitations. It should work down to 0V out since the 50~100mV Vense is what controls FB with ampilifer to internal Vref. \$\endgroup\$ Commented Apr 4, 2017 at 14:12
  • \$\begingroup\$ Agreed @TonyStewart.EEsince'75 and if the OP want to design his own solution using CCM: mouser.com/pdfdocs/BuckConverterDesignNote.pdf It adds complexity of course, with a multiphase solution having the least ripple current. \$\endgroup\$ Commented Apr 4, 2017 at 15:38
  • \$\begingroup\$ for PWM RMS current is only 700mA, that's not going to need a fat wire. \$\endgroup\$ Commented Apr 8, 2017 at 2:12
  • \$\begingroup\$ Yes, this is good. There are even buck regulators with minimum output voltage as low as 0.8V. \$\endgroup\$ Commented Apr 8, 2017 at 5:17
  • \$\begingroup\$ Yes PWM with LC filter aka Buck regulator is the way to go with suitable feedback for temp Resistance or current along with Hall sense for contraction sensing. I would expect R to rise sharply with Temp as a better feedback to T and thus force but Op has not tested/reported this. \$\endgroup\$ Commented Apr 8, 2017 at 6:00
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For small loads PWM is best

PWM is best. resistors don't respond to average voltage or current, they respond to average power. so the greatest heating efficiency if got by PWMing the load.

PWM will cause ripple on the supply, but the higher the frequency the smaller the ripple amplitude

so about 400mW into the load (when cold) from 12V requires a PWM ratio of about 300:1 the peak current will be about 15A but the RMS current only 700mA ripple current in the filter caps slightly less than that figure.

driving the mosfet fast enough and limiting EMI are going to be the worst challenges,

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  • \$\begingroup\$ I think you mean PWM with an inductor for stored energy is actually more efficient than non-inductive PWM because the peak current is lower \$\endgroup\$ Commented Apr 4, 2017 at 13:57
  • \$\begingroup\$ it's a trade-off between heating the mosfet (my plan) and heating the inductor and the diode, or second mosfet (your plan) The load is a resistor so a high current at a low duty cycle is no less efficient than a low current at 100% duty cycle. \$\endgroup\$ Commented Apr 6, 2017 at 6:37
  • \$\begingroup\$ try again , PWM 12V/0.7A = 17.1Ω , load starts at 0.83Ω, so passive R ratio= 0.83/17.1= 5% which means ripple current in supply cap can be up to 12V/0.8Ω= 15A peak stressing ESR in source cap ( overheat) . I agree your efficiency is same as a linear CC (100% duty cycle) and only 5% efficient due to R ratio, thats why SMPS or Class E Audio uses chokes to store energy. So load of I²R= 0.7A * 0.83 (cold) = 0.5W and loss is 10 W \$\endgroup\$ Commented Apr 6, 2017 at 16:01
  • \$\begingroup\$ as far as heating from the ripple current is concerned it comes out about 3.2A RMS, so yeah. that could stress the filter caps a bit. \$\endgroup\$ Commented Apr 7, 2017 at 22:31
  • \$\begingroup\$ heat in caps and MOSFET, very low efficiency, more EMI and more issues with 15A pulses unless controlled, which is why LC filter is preferred to reduce Peak/avg current near 1.1 \$\endgroup\$ Commented Apr 7, 2017 at 22:37

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