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Topic: Power a big solenoid (Read 4383 times)previous topic - next topic

michinyon

#15
Jun 11, 2014, 11:09 am
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.

michinyon

#16
Jun 11, 2014, 11:10 am
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.

michinyon

#17
Jun 11, 2014, 11:19 am
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.

Yaro

#18
Jun 11, 2014, 11:49 am
I have another question. If i want to shield the solenoid, i need to connect shield to gnd or to - of the solenoid?

vffgaston

#19
Jun 11, 2014, 12:15 pm
What michinyon says IS ABSOLUTELLY CORRECT. Excuse me for giving mislead (altough not dangerous  ) 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

Grumpy_Mike

#20
Jun 11, 2014, 03:37 pm

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

Connect the shield to ground. Any energy picked up from EM ( electro magnetic radiation ) will not be referenced to any of your signal voltages so it is best to ground them.

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