# Way to discharge solenoid without (heated) resistors?

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.

• I have little or no experience with solenoids, but I have never seen it with resistors -just the flyback diode direct across the solenoid. Can anyone explain why you'd have resistors in there for me? – Majenko Nov 4 '11 at 10:35
• @Matt - As I understand it the 100$\Omega$ is the internal resistance of the solenoid; the 1H coil is a perfect coil with zero resistance – stevenvh Nov 4 '11 at 10:59
• Ah, so the whole question is invalid then, as those resistors don't really exist then and their dissipated heat is irrelevant? – Majenko Nov 4 '11 at 11:11
• @stevenvh, I think the user is using them to limit his current and may just have a missunderstanding about the solenoid resistance. we will find out. – Kortuk Nov 4 '11 at 11:28

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.

• I was just playing with some pinball-machine solenoids and observed that when one the solenoid is energized (with a PTC fuse on the input), the applied voltage goes up briefly to several times the supply voltage. My guess is that the solenoid imparts a magnetic field to the slug, and when the slug moves it generates back EMF, but I find it surprising that the effect would be that strong. Have you noticed back EMF with solenoids? Is it possible to exploit it (e.g. switch in a holding resistor at the moment the back EMF exists, since the solenoid takes about 30-35V to energize and 6V to hold? – supercat Apr 2 '12 at 18:50
• @supercat - Is this viewed with a meter or an oscilloscope or ...? The viewed rise in voltage is possibly due to switch bounce causing inductive ringing. You COULD operate a solenoid by pulsing it, rectifying the ringing voltages and storing them and then applying them to the solenoid. What you are doing is using the inductor as a boost converter, saving the energy and then utilising it. The switching required and pulse drive adds complexity. – Russell McMahon Apr 3 '12 at 0:02

When you see the rating of a resistor it is for peak dissipation, often if you are just going to be discharging resistors on an irregular basis you do not need to have high wattage resistor as your resistors are really not getting many joules of energy. It is high wattage, but a very short time.

Do some testing with lower wattage resistors for discharge, you will find they can dissipate much more then needed. The issue in your current circuit is that your resistors are forming the limiting for your solenoid all the time, meaning a lot of heat. The major point that is invalidating your simulation is the lack of a resistance for your solenoid. You should look up the resistance of your solenoid as it will significantly resist current flow on its own.

• Alright, I appreciate the answer. I was picturing a solenoid having somewhat resistance, however on another question on here about a solenoid the extra resistors were hinted to reduce the current during discharge, otherwise the solenoid would throw current in a loop a hundred times longer (if you know what I mean) until it can eat it up. – Kim N. Nov 4 '11 at 11:46
• Yes, seems to be my misconception. The simulator's solenoid has a ceiling of 400 or so mA, no need to limit both ways. The looping discharge however takes ~30ms, however that in real time perspective it just silly. – Kim N. Nov 4 '11 at 11:49
• @kimN. give a set voltage for a long time to the real one and see how much current their is. This will tell you what its resistance is. Then you can model with that. Most solenoids are pretty resistive and are designed to operate at a certain voltage on their own. – Kortuk Nov 4 '11 at 11:51
• The power rating on a resistor is for continuous dissipation, not peak dissipation. They can handle peaks higher than the rating as long as the average is lower. – endolith Nov 4 '11 at 13:46
• While what the others say about a resistors peak vs. average dissipation is correct, you MUST use resistors that are pulse rated. All resistors are spec'd for continuous dissipation, but only some resistors will be spec'd for a pulse, which is normally 3-5 times the normal wattage and for a specific duration. If you use a non-pulse-rated part in a pulsed circuit, the value of the resistor can change. I have had non-pulse-rated resistors change from 50 ohms to 500 ohms after only a pulse or two. – user3624 Nov 4 '11 at 14:07