solid state replacement for contactor (relay) with 40VDC inductive load

[ballen]

Hopefully this is the right forum, if not please advise me and I'll move it.

I'm helping someone to repair the brake circuit in a 50-year-old Schaublin 135 lathe.  An annotated snippet of the circuit is below.  The brake is effectively a solenoid with about a 25 ohm coil resistance.



The contactor (relay) in the middle is no longer working correctly, probably because some of the contacts got welded together from the current generated when the contactor opened. I'm guessing that the capacitor and MOV have aged enough that they didn't do their job properly.

I'd like to suggest that he replace this with an off-the-shelf solid state part.  This needs to handle (say) 60VDC/10A for the load, and if needed a snubber diode or MOV can be added to handle the inductive switching spike.  The hard part is that the control voltage should be 230VAC/50Hz, and we don't want it switching on and off every 20 msec.

So my question: can anyone point me to an off-the-shelf BJT or Mosfet solid state relay which can handle 60VDC/10A for an inductive load AND has a 230VAC control signal?

(I know that it's easy enough to generate a low-voltage filtered DC control signal from the high voltage AC control input, but for this type of industrial equipment it's best to stick to 100% off-the-shelf replacement parts, if at all possible.)

Thanks!
   Bruce

[thm_w]

https://industrial.omron.eu/en/products/G3NA-D210B-UTU-100-240VAC
https://assets.omron.eu/downloads/datasheet/en/v9/j166_g3na_solid_state_relays_datasheet_en.pdf


If you want to copy the image link to insert it, the other ways dont work:

[ballen]

Thank you, that's incredibly helpful, I've suggested that he purchase a G3NA-D210B-UTU-AC100-240 along with a suitable MOV.

Design question: should we keep the 250uF capacitor or drop it?  If we keep it, should it stay where it is on the load side or move it to the power supply/source side?

I still don't understand how you got my attachment to show inline.  Could you provide a couple of sentences of instruction?  I tried using the "Attachments and other options" feature below the composition box, but it didn't work as I had expected and I could not figure out how to use it.

Edit: did you mean the following:
(1) Upload attachment
(2) Click on attachment to display it
(3) Copy image URL from browser
(4) Paste URL into post, inside IMAGE tags?

Trying this now...



It works, thank you a second or third time!  I've now gone back and fixed this in my original post.

[thm_w]

Yeah you figured it out, I wasn't very clear.

I'm still not entirely sure why they have so many contacts here. And how they even work timing wise (break before make, or whatever).

But, capacitor would probably reduce the load on the MOV, and help it last longer? I would be tempted to leave a capacitor on the load side, assuming its still functional and not failed.

[nctnico]

How about putting a big diode across the solenoid to keep the current flowing? Or is solenoid operated using two polarities?

[f4eru]

It seems to use the reverse polarity to discharge the inductive current fast enough.
II would either :
1) rebuild the original schematic to spec with all new components, as relay MOV, capacitor all age badly.
2) build a similar performing solid state circuit in the forma of a "3/4" H-bridge (the 4th switch being only a diode)

[nctnico]

It doesn’t seem to reverse polarity. When the relay is activated, the solenoid is powered and the capacitor is charged. When the relay is released the solenoid is powered from the capacitor (likely to make the solenoid hold longer).

[langwadt]

It doesn’t seem to reverse polarity. When the relay is activated, the solenoid is powered and the capacitor is charged. When the relay is released the solenoid is powered from the capacitor (likely to make the solenoid hold longer).

the charge on the cap is opposite polarity

[nctnico]

It doesn’t seem to reverse polarity. When the relay is activated, the solenoid is powered and the capacitor is charged. When the relay is released the solenoid is powered from the capacitor (likely to make the solenoid hold longer).

the charge on the cap is opposite polarity

You are right! It seems like it is very important to get the solenoid to 'stop' when deactivated. Likely quite a bit of time went into getting the timing of this circuit right. As simple as it may seem at first sight. I don't think it is wise to replace with a solid state solution or blindly replace any of the parts other than the ones that are broken. I'd measure the capacitor and if the value is within spec and the insulation of the wires is still good (not brittle) I'd leave it as is. It might be a special type of capacitor. If it ain't broken don't ruin it.

[DavidAlfa]

Sometimes the simple way is the better.
If that solenoid has worked for 40 years straight, I'd justo put a new one.
Electronics will hardly match the toughness of a beefy solenoid.