Here is a simple simulation of a schematic for my solenoid:
Once the switch is closed, the solenoid will be activated with ~80mA charge. When it is open (where the screenshot is at) it will discharge via the resisters in a looped path, due to the diode.
Now those resistors are cumbersome. A 100 Ω resistor would reach ~800mW of power during discharge, so for this specific test I split it in to four 25's in series.
I would like to provide a lot more power to my solenoid however (200-800mA), how would I be able to discharge it safely, with minimal current resistance when the switch is closed?
I could possibly use some wire-wound resistor (~5W) as a last resort, however those I do not have currently.
Best Answer
It IS valid to use discharge resistors, but they would usually be in series with the diode and NOT the solenoid. You DO need to model the solenoid resistance by a resistors in series with the solenoid where you have the resistors now.
The resistors in series with the solenoid are a potentially* valid way to reduce the solenoid on current BUT it means your solenoid is not really the right one for the power supply you are using.
If you want extra resistance (see below) put it in series with the diode.
The reason to put extra resistance in the circuit is to cause your solenoid to relase FASTER when the power is turned off !!! This is not intuitive :-).
Thereason is that the time constant of an inductive loop with current flowing in it is
T = L/R, amd so by adding extra R we reduce the time cinstant ! :-).
This is STILL not intuitive!
What makes this so is:
When the current feed to an inductor happens there is NO instantaneous change in curremt. The SAME current keeps on flowing.
This is in fact part of the fundamental definition of what an inductor is.
Not many people know that.
The still flowing current MUSt be allowed to "go somewhere" and in the real world it will in fact just do that. Always. The diode in the circuit provided provides a current path. If there is no formal current path the inductor will "find" one. Always. If necessary the inductyor will deliver the current into its own parasitic capacitance. As this capacitance is very small it need a lot of voltage to store energy as E = 0.5 x C x V^2. Small C = big V squared.This is why you get an inductive kick and a spark and other interesting results.
With just an inductor plus a diode plus (we'll assume) negligible coil resistance the losses in the circuit are due to te diode voltage drop x the current plus I^2R losses from the (assumed neglible resistance. The diode voltage drop is typically in the 0.4 to 1.0 Volt range. Energy dissipated in the diode will slowly dissipate the circulating current. The losses are roughly linear with current (Power = Vdiode x I)
now assume that the coil resistance is not negligible. say we have a 12V , 100 mA relay. If the current when turned on is limited only by its resistance then R = V/I = 12/ 0.1 = 120 ohms. When we turn off the power the current will be as before so the coil resistance will have Vin across it and will start to disipate Iin^2 x Rcoil. Also the diode will dissipate Vdiode x Iin_initial. This will decay ~= exponentionally.
Now add a resistance of say 10 x Rcoil in series with the diode.
Let's call this Rdiss
wGEN vIN IS removed Iin will flow.
V_rdiss = Iin x Rdiss = Iin x 10 x Vcoil.
Dissipation in Rdiss = 10 x as much as in Rcoil before so ~ exponential decay will start at 10x the rate before. (Actually 11x as Rcoil is still present.)
The solenoid will hold in until Ihold is reached - and this is being approached at aout 10x the rate without an R.
SO adding a series resistor did in fact reduce hold time or increase release delay. FWIW.