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Topic: Power a big solenoid (Read 4381 times) previous topic - next topic

Yaro

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.

Grumpy_Mike

Yes you need a diode.

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

jackrae

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.

Yaro

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

jackrae

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

vffgaston

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)

Yaro

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

Here's datasheet: http://www.digchip.com/datasheets/parts/datasheet/456/BYW81-150-pdf.php

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.

jackrae

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.

Grumpy_Mike

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

vffgaston

Quote
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?)

michinyon

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.

michinyon

" (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.

Yaro

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.

Grumpy_Mike

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.

michinyon

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 ].



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