We're designing a circuit which essentially has the topology shown below. We're driving an inductive load (24V, 3.8Ohm solenoid - max avg current = 2A, but peak current can be 6.3A) using PWM. Unfortunately, the distance between the controller and the solenoid is quite large (8m).


simulate this circuit – Schematic created using CircuitLab

I'm quite concerned about the EMC implications of driving a high current PWM signal over such a long cable. I was thinking of mitigating it by doing the following:

  • Using a twisted pair cable to cancel out magnetic fields and eliminate electric fields
  • Adding large tank capacitors in parallel with the 24V supply and ground to reduce sudden current spikes going through the power cable. (Thanks to this post).
  • I can control the PCB layout at the microcontroller to make sure that the current loop for the PWM signal on the PCB is minimised.
  • I'll be adding unpopulated footprints for a series resistor and parallel capacitor on the MOSFET gate drive signal to experiment with reducing the edge rate of the PWM signal to reduce any high frequency components.

What other steps can I take to help avoid any EMC problems with this?

  • \$\begingroup\$ You have already gotten some good answers. In addition, physically put D1 right at L1. \$\endgroup\$ – rioraxe Mar 12 '15 at 20:23

The solenoid doesn't care if you supply it PWM or a clean analogue voltage so you could use an LC low pass filter to "convert" PWM to a smoother DC value. Values depend on dynamic expectations of your solenoid and PWM frequency. Fit it at the driving end of the cable to reduce EM emissions on the cable.

Note that using a twisted pair won't eliminate e-fields being generated by the cable because it is not a balanced output drive due to one wire being 0V. Add a screen or use coax for better results.

  • \$\begingroup\$ Balanced has nothing to do with 0V or not - it has to do with current. Unless there's another path the current can take, it's balanced. In the case of RF signals and COAX, it CAN be unbalanced because of the skin effect since the inner-shield and outer-shield are effectively two separate conductors. \$\endgroup\$ – iAdjunct Mar 12 '15 at 14:15
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    \$\begingroup\$ @iAdjunct You are considering magnetic fields - I'm considering E-fields - if you have a driving signal and 0V wire closely twisted - the e-field will still be present. If a balanced drive were used, theoretically there would be no significant emanating e-field at a short distance. Maybe I've misunderstood your point - your comment is a little confusing i.e. "Balanced has nothing to do with 0V or not" - what do you mean "or not" - this seems to be self-contradictory. \$\endgroup\$ – Andy aka Mar 12 '15 at 14:21
  • \$\begingroup\$ @iAdjunct - I suggest you read sigcon.com/Pubs/edn/ReducingEMI.htm and please note that what you are saying appears to discount the whole reason why balanced drivers are used. This is nonsense so please rethink what you are trying to say. \$\endgroup\$ – Andy aka Mar 12 '15 at 14:27
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    \$\begingroup\$ @iAdjunct If I used the words "differential driver" instead of "balanced output drive" would you still believe me to be wrong? \$\endgroup\$ – Andy aka Mar 12 '15 at 14:56
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    \$\begingroup\$ @iAdjunct the magnetic fields cancel, yes. But for electric fields, the pair of wires is at an average of half the supplied voltage with respect to ground. Thus the long wire to the solenoid becomes an antenna, effectively driven at the edge of the PCB with half the actual pwm voltage. It will start to radiate when it gets anywhere near 1/4 wavelengths, a few MHz, which could easily be created by the PWM driver if not filtered. \$\endgroup\$ – tomnexus Mar 12 '15 at 20:05

A two-conductor wire is basically a ladder-line. As long as there's no other path the current can take, the EMF from these is extremely minimal in the far field. In the near field (1x or 2x the spacing between the wires) is where you might EMF and be sensitive to EMF because of the tidal effects of the waves (i.e. the two wires are close enough to you that they don't appear as a single wire any more).

Twisted-pair can make this "near-field" effect smaller since you have to be much closer to it for your signals to have tidal effects. This, again, reduces both emitted EMF and differential-mode interference on your line. Generally, with a solenoid, the latter isn't your problem.

Generally speaking, protecting power supplies connected to inductive loads with a capacitor is a very, very good idea. Always.

One more note: solenoids are typically not modeled as purely-inductive devices. They have inductance, but at steady-state they're resistors. A quick Google search yielded fewer results than I expected, but I did find this: some random circuit diagram

He used a BJT, but I've personally always liked MOSFETs for this sort of thing.

I do agree with Andy's idea of putting a low-pass-filter at the end to turn it into a DC signal. This'll help you control it better since it is fairly inductive. If you can characterize its inductance, you can tune the capacitance to be optimal as well.

  • \$\begingroup\$ If I'm already using a protection diode at the solenoid, why do I need a protection capacitor? \$\endgroup\$ – Amr Bekhit Mar 16 '15 at 15:45

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