So I have this coil and I'm driving dc current through it. This coil has inductance and thus stores energy from the dc current.

I have to remove this energy before I can change the polarity of my h-bridge to prevent dangerous voltage rises. How can I efficiently remove this stored energy from the coil immediately after it has been disconnected from the current source? Would just using a bipolar capacitor in series with small resistor and paraller with the coil like this?


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Can I just ask how much energy are we actually talking about here? because remember even 1 Joule can be a million volts, or 1 volt, and various currents/discharge time. If the entire system is only "low" power, then you can get away with standard clamp diode configurations seen in H bridge circuit designs intended for DC motor driving, as these operate the same way (coils, discharging and changing direction etc) \$\endgroup\$ – KyranF Oct 2 '14 at 13:21
  • \$\begingroup\$ @KyranF We are talking about 16 volts and about ten amperes. \$\endgroup\$ – The amateur programmer Oct 2 '14 at 13:23
  • \$\begingroup\$ and what's the estimated switching time from one direction to another? Is your application low Hz or is it high Hz switching? \$\endgroup\$ – KyranF Oct 2 '14 at 13:24
  • \$\begingroup\$ you may benefit from reading this, the author's reference to "flyback diodes" is the important part here, which deals with inductive loads being switched and their resulting voltage spikes. Only the first few pages are actually useful to you! hades.mech.northwestern.edu/images/8/84/ProjectW2011.pdf \$\endgroup\$ – KyranF Oct 2 '14 at 13:27
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    \$\begingroup\$ People suggesting diodes haven't considered the supply polarity reversal .. what you want is a large snubber capacitor across the coil. See google.co.uk/… \$\endgroup\$ – pjc50 Oct 2 '14 at 14:19

Just parallel the inductor with back-to-back Zeners, or a TVS, like this:enter image description here

Or don't do anything at all if the MOSFETs you're using have parasitic diodes which can take the current hit from the inductor when you switch.

Or, if they don't, you could do this:

enter image description here

  • \$\begingroup\$ Is that just a single zener or two zeners with cathode cathode connection? \$\endgroup\$ – The amateur programmer Oct 2 '14 at 16:11
  • \$\begingroup\$ Oh I see what u did there... Only realized this after taking a closer look at how zener diodes work. Very clever :D \$\endgroup\$ – The amateur programmer Oct 2 '14 at 16:15
  • \$\begingroup\$ Would the schematic editor (CircuitLab) not suffice for this example? Just wondering. \$\endgroup\$ – JYelton Oct 2 '14 at 16:30
  • \$\begingroup\$ @JYelton: Yes, it would. Why? Was my intent not clear? \$\endgroup\$ – EM Fields Oct 2 '14 at 16:36
  • \$\begingroup\$ Not that I don't appreciate hand-drawn schematics (they are often quite nice) but I just thought it might look more professional here (since there are so few components). :) \$\endgroup\$ – JYelton Oct 2 '14 at 16:38

In general, you dissipate the energy in an inductor by allowing it to circulate it through a resistance. In the simplest (single-ended) form, you have a 'flywheel diode', which just circulates the current through the inductor. The dissipation occurs as Vf * I in the diode and Rl * I^2 in the inductor, where Rl is the resistance of the inductor.

The voltage of the 'bottom end' of the inductor rises to Vf above the supply rail during circulation, so doesn't impose much extra voltage stress on the rest of the circuit.

To cause the current to decay faster, you can add additional resistance in series with the flywheel diode. This adds R * I^2 to your dissipation, but increases the overvoltage by IR volts, which is the trade-off.

Alternatively you can add a zener diode in series with the flywheel diode (but anode to anode) which allows the voltage to rise higher, and then dissipates Vz * I in the diode, while adding Vz to the over-voltage.

Pretty much you're just trading-off voltage spike height against speed of dissipation.

  • \$\begingroup\$ If I put diode paraller with the inductor would it allow current to go through the diode instead of the inductor when polarity is changed? \$\endgroup\$ – The amateur programmer Oct 2 '14 at 14:06
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    \$\begingroup\$ A couple disadvantages of the resistor (vs the zener diode) are: 1) the peak voltage varies with changes in inductor (resistance of inductor) or power supply voltage, and 2) the voltage decays as the current decays making it less effective. Item 1 can be important if switching electronically so as to avoid burning out the switching device. \$\endgroup\$ – Tut Oct 2 '14 at 14:06
  • \$\begingroup\$ We use the Zener and series diode, and then a Resistor in parallel with that and the coil. The zener sucks up the big spike and then you get an L/R time for the rest. You are always wasting a bit of current through the resistor. \$\endgroup\$ – George Herold Oct 2 '14 at 14:46
  • \$\begingroup\$ @GeorgeHerold So u mean like having zener and "normal" diode in series and that circuit in parallel with coil and resistor? \$\endgroup\$ – The amateur programmer Oct 2 '14 at 14:56
  • \$\begingroup\$ @GeorgeHerold What about diode polarities? Anode to anode or cathode to cathode? Or cathode to anode? \$\endgroup\$ – The amateur programmer Oct 2 '14 at 14:57

OK this also may work (in response to the parallel cap idea.)


simulate this circuit – Schematic created using CircuitLab

So this might work, depending on L and R.
L is the coil inductance and R the coil resistance.
You choose C such that RC = L/R or C = L/R^2.
This then makes it a low Q resonant circuit. (Search for Zobel network)
And it will decay with a time constant of RC = L/R.
If you have to voltage head room you can add more series R to the coil and get it to switch faster.

(Is there some way to make the schematic smaller?)

  • \$\begingroup\$ Why u put another resistor in series with the inductor? To represent the resistance of the inductor? \$\endgroup\$ – The amateur programmer Oct 2 '14 at 15:15
  • \$\begingroup\$ Yup, it's the coil resistance. But you can also add more resistance in series to make the circuit snappier. (at the cost of more voltage to drive it, and more power dissipation in the resistor.) \$\endgroup\$ – George Herold Oct 2 '14 at 15:19
  • \$\begingroup\$ Is there a way to calculate how long does it take to dissipate the energy then? \$\endgroup\$ – The amateur programmer Oct 2 '14 at 15:21
  • \$\begingroup\$ I'm going to do this with zener diodes though just realized how to :D \$\endgroup\$ – The amateur programmer Oct 2 '14 at 16:22
  • \$\begingroup\$ @George: The brute force way: add dots on the left and right to give you the display width which will shrink the art to the size you want. There may be a nice way to do it with commands, but I looked and I couldn't find anything. \$\endgroup\$ – EM Fields Oct 2 '14 at 17:17

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