Does an inductor act like a simple resistor in a DC circuit?

[chipwitch]

I'm designing a simple circuit that uses a small PCB, through-hole mounted relay.  I need 5V for the relay coil.  The voltage source is 10VDC.  I'm using a 5.1V Zener to regulate the voltage for the relay coil.  I know how to size the series resistor for a resistive load but I'm just not certain that an inductor qualifies.  It seems to me that without a change in voltage, there's no dI/dt therefore I can simply measure the coils resistance with an ohmmeter and calculate my zener's series resistor like usual.  No?

I'm going to use a flyback diode across the coil.  Am I also correct in assuming that the brief dI/dt occurring at switch-on and switch-off can also be disregarded as inconsequential?

[Zero999]

Of course an inductor acts as a resistor at DC. Connect the relay in series with a resistor with a similar value to the relay's coil resistance.

[chipwitch]

Thanks.  I knew I was making this more complicated than it needed to be.

[richard.cs]

The best option here is to skip the zener entirely and put a resistor equal in value to the relay coil in series with the coil. The relay will then see 5V steady state and work as normal (there is a small improvement in switching time as more of the 10V can appear accross the relay inductance during the switchon transient but it usually down't matter). Use a flyback diode, but put it across the relay-resistor pair, the extra resistance in the current loop will help the stored energy in the inductance decay more quickly and the relay to open faster (faster opening is better for the contacts if they're switching any kind of load). Note: this does allow the bottom of the relay to rise to 10.6Vish rather than 5.6Vish when switched, if that's a problem for your application put the diode in the normal place.

[cybermaus]

You could also consider to add that zener to the flyback diode.

The flyback avoids a spike upon opening of the relay. But it does so by shorting the coil and keeping voltage low, so it takes a while before the coils energy is dissipated. (energy is volt times amps times time) If you add a zener of reasonable voltage, voltage will be higher, more energy is dissipated faster.

The relay will open faster, not only improving response time, but also increasing the relays lifetime.

See this simulator (wait a sec to start, allow Java)
You can switch on normal diode, zener pair, or nothing.



Added considerations:

You need to select your zener not too high: while it will avoid a negative high spike, but will still cause a negative smaller 'bumb' on the supply side. So consider what the driving components can handle.

You need to select your zener not too low. No diode at all will switch immediately, a diode means the relay contacts only slowly switch, the higher the voltage, the faster the switch. If there is no electronics involved, and you are connected to some lead-acid battery, consider not using a diode at all. (fastest switch time)

For industrial/automotive/marine mounting, I often use 2 zeners (so polarization does not matter and its easier to screw under relais terminals) a bit above max supply voltage. So 32V for a 24V battery system (seeing as how a charge can be up to 28.8V. (but as stated this is not needed at all if directly connected to the battery bank)

But on a PCB, where no mistakes in polarization can be made, and negative voltages can be a problem, I guess a zener plus normal diode would be better, as you can select any voltage that you think is acceptable.


PS: you still need that resistor! The zener addresses a different item. Best place the resistor between the (zener)diode and the coil, so it helps with the power dissipation.