# Power a big solenoid

Hi, i need to power a big solenoid, about 7A 24v. How i can do that? i don't want to use transistors but relays, the problem i have is how i can discharge solenoid without problems. I need also to use a diode?

Thank you.

Yes you need a diode.

I don't understand your question about discharging. You simply remove the power to discharge it.

With a 7A solenoid coil you will get one mighty fat spark when you de-energise the coil so the diode needs to be rated at least 15A and of the schotty type. General rule is that the diode is required to pass the same current as the coil draws when powered. A simple 1n400x type diode will not suffice.

Ok thank you, so if i connect diode to solenoid connectors and connect all to gnd it will discharge?

Draw us a picture of how you intend to connect your circuit. Then we will be able to advise.

When you power off the solenoid (by opening the relay), the energy stored on the solenoid creates an inverse voltage (the solenoid itself behaves as an -almost- infinite voltage instantaneous power supply) that, if not prevented, creates a big spark in between the relay contacts destroying them after a few manoeuvers.

The diode evacuates all this energy trhough it, so it has to be a very big one (I'm not sure if the energy to dissipate is 7 A x 24 V = 168 W; i think it is)

I’ve found this type of diode BYW81-150 of ST.

I have a power supply with + - 24v and a GND.

I need to power solenoid(7A, 24V) with it, and control his ON/OFF with a relay (common songle relay https://encrypted-tbn1.gstatic.com/images?q=tbn:ANd9GcRjn_UGYIeMUcFeQcJMIz6p6TLRgUTwsoq2waiw8M65Jht_gNf_gQ ).

I tought to connect it like this:

I think i’ve connect wrong diode pins.

That should work OK. Ensure the diode is placed as close as possible to the solenoid and that the connection leads are capable of carrying the current. You might even want to consider glueing the diode to the solenoid body and connecting directly to the solenoid wires.

I would not connect it to GND with the other contact just disconnect it when you want it off.

I've found this type of diode BYW81-150 of ST.

I DO insist on the energy the diode has to absorb (sketch is correct).

The current the diode has to drive is MUCH HIGHER than 7 A.

What is the duty cycle?. (i.e. How many times per minute are you going to connect/disconnect?)

You can shout if you like, but it doesn't make it true.

If 7 A is flowing though the solenoid when you interrupt the supply circuit, then the solenoid current will be 7 A at that instant, and less than 7 A at all subsequent instants, as the current is dissipated by resistive and magnetic field losses.

" (I'm not sure if the energy to dissipate is 7 A x 24 V = 168 W; i think it is)"

... this is wrong, too.

This diode is rated of 15-35 A.

But if i won't connect it to gnd, it'll discharge fast? i need to discharge it faster as possible, i saw that someone connect solenoid to gnd to discharge it faster otherwise the plug remains attached to the walls of the solenoid.

Connecting it to ground will not make it discharge any faster as there is no current path to ground. It could infact give you problems with ground bounce. Do not do it.

Assuming that your solenoid does indeed draw 7A at steady state when powered from a 24 volt source, then it has a resistance of about 3.5 ohms. As soon as the current in the power circuit is interrupted, the inductance of the solenoid coil will cause a negative voltage at the switched end of the solenoid, relative to the ground, according to the formula V = L ( di/dt ). For this inductance voltage to be negative, di/dt must be negative, in other words, the current through the inductor is reducing.

Without the diode, di/dt becomes arbitrarily large, and the voltage at the switched end of the inductor can go negative several hundred volts, which can cause the switch ( in your case, the relay contact ) to flash over if it hasn't openned fast enough.

With the diode, as soon as the induced voltage at the switched end of the inductor falls to about -0.7 volts, then the diode will start to conduct. The current occuring in the diode loop, reduces the negative value of di/dt, which reduces the tendency for the voltage at the switched end of the inductor to become large and negative. The current in the loop consisting of the inductor and the diode is then limited by the resistance of that circuit path, which is at least 3 ohms.

At all material times in this switching off process, di/dt is negative, and the current flowing through the inductor from either of the possible loop paths will be 7 amps, and falling.

If 7 amps is flowing through your inductor when you interrupt the power supply loop, then the peak current that will flow through the diode will not be more than 7 amps, at any instance in the turn-off process. People are uninformed, confused and delusional if they think otherwise. The opportunity for a potential of several hundred volts to appear at the contacts of the switch is correct, but that does not mean that a high current will occur there, either. You can have a spark with a very small current if the voltage is high enough. It is the negative potential caused by the inductance which causes the problem, and this negative potential only exists while di/dt is negative, that is, the current is falling.

You need to understand the difference between power and energy dissipation. The energy which existed in the magnetic field has to go somewhere, and it gets removed by the resistive losses in the inductor-diode loop. This loss of energy occurs mainly in the series resistance of the coil. If the solenoid coil is capable of dissipating the resistive heating losses from a constant current of 7 amps flowing through it, then it is not suddenly going to overheat from the resistive losses of less than 7 amps flowing through it.

From the point of view of components overheating, the heating depends on the power of the heating and the duration of the heating. From the point of view of accumulating heat, a heating effect that lasts a few milliseconds is inconsequential [ although you might have to consider other effects, like insulation breakdown ].

If your relay connects the switched end of the solenoid to ground, then the switched end of the solenoid won’t be able to reach the -0.7 volt potential which would make the diode do anything for you.

What will happen, if you connect the switched end of the solenoid to ground using the relay, is that the persistence of the solenoid current will attempt to suck current out of one part of the ground and feed it back into the other part of the ground. This is unhelpful.

In any case, by the time the relay had switched over from the power contact to the grounded contact, the discharge would have finished.

If your solenoid is rated for 7 A, does it have a time duration specification on that ? Putting 7A continuous though a coil with a resistance of around 3 ohms is going to be generating 150 watts of heat - that’s quite a bit.

The resistance of the solenoid coil is larger than the forward resistance of the diode, so most of the heat during the turn-off process is going to appear in the solenoid coil, not in the diode. And this amount of heating, is going to be less than the amount of steady-state heating while the solenoid is actuated.

If you want the solenoid current to collapse as quickly as possible, you need to increase the resistance in the solenoid-diode loop. The increased resistance will dissipate more heat. Looking at this the other way, it would take a larger amount of inductor potential to drive the same current, which requires di/dt to be greater, which means that i reduces to 0, sooner. You can cause the solenoid current to decline faster by putting a 1 or 2 ohm resistor with a suitable heat rating in series with the diode. Note however, if you do this, you will have a slightly higher unwanted negative potential at the switch contact when you interrupt the circuit.

The rate at which the armature of the solenoid moves, is also going to depend on its inertia and the return spring, as much as the electrical behaviour.

There is a quite useful explanation of this phenomenon, with a good diagram, here

http://en.wikipedia.org/wiki/Flyback_diode

You'll notice at the end, they recommend schottky diodes for this - but note, that is for the application of switching power supplies, where you are energising and de-energising the inductor continuously at a rate of several tens of kiloHertz, switching at a very high rate.

That is a quite different scenario to your solenoid mechanical actuator, which will not, and cannot, be actuated at such a rate.

I have another question. If i want to shield the solenoid, i need to connect shield to gnd or to - of the solenoid?

What michinyon says IS ABSOLUTELLY CORRECT. Excuse me for giving mislead (altough not dangerous :cold_sweat:) observations.

I used to choose a forward current higher rated diodes to supress spikes form relay coils than the coil current (And the wiki article you refer to says "In an ideal flyback diode selection, one would seek a diode which has very large peak forward current capacity (to handle voltage transients without burning out the diode)".

Any way, power concern stands correct (about duty cycle, I mean). Does it?

Regards