Choosing an Opto Isolator

[KrisBlueNZ]

Cool. You understand that this all relates only to the release of the solenoid, right? The zener allows the solenoid to release more quickly; it doesn't affect the solenoid pull-in at all.

 

[BodhiSci]

Yeah, if you mean release by when the power is cut, yes. That's what it seems like too when I compared them. Seemed slightly faster. Thx

 

[KrisBlueNZ]

Yes, cool.

 

[BodhiSci]

Hi, I know this is older, reviving it so the background is available for anyone who can help. Still working on a project for my school lab. I made a PCB, part of which acts as a switchboard for a solenoid. It is run off a large 4x5" 12v battery (don't have current on me). On the board there is an SSR which isolates the control side vs the 12V power side. (has to be isolated for various reasons)

-------12V------------|------------|---------------To Solenoid
| SSR |
| |
____5V_______|_ diode_|_________control line (active 5v TTL on until dropped)
(from devboard)


The SSR closes the circuit and activates the solenoid when the 5v TTL is dropped.

Here is my problem:

I connected the solenoid to the board on it's jumper, the TTL was not there I believe since I was still working on the software end (telling zbasic what devboard pin to use for the solenoid).

I don't know how long it took but it could have been as short as 5-10 seconds. I really wasn't paying attention and don't remember. one of the traces on the board between the SSR and the solenoid burnt out/melted like a fuse.

I can repair that with a wireor something, but how can I prevent this? It seems that if I ever have this board connected to the power and the solenoid connected without the devboard running and supplying the TTL to keep the SSR open, then this will keep happenening. The trace is pretty thick.

I'm thinking a resistor might work to reduce current, but that would reduce the voltage too wouldn't it? I don't have the solenoid model on me currently but we want it to work with as many different kinds as possible. I suppose a different resisitor could be used for different models, but I'd really rather not have later people using the board have to figure that out!

I think that the other option would be to remove the 5v line and connect the control line there , but I would have to be able to get the current to be in the 15mA range. I think someone said a transistor would work? Not sure what they do exactly. Post #8 has a site that describe a circuit but to be honest I don't undertand it all, If I would still use the SSR and how the resistor is connected and if it would still work with the SSR (and whether it boosts the control\logic to power the SSR diode)

Thank you for any help!

 

[(*steve*)]

Perhaps you can draw a circuit diagram using paint or one of the online tools. I can't make head or tail of that.

 

[BodhiSci]

Hi, I know this is older, reviving it so the background is available for anyone who can help. Still working on a project for my school lab. I made a PCB, part of which acts as a switchboard for a solenoid. It is run off a large 4x5" 12v battery (don't have current on me). On the board there is an SSR which isolates the control side vs the 12V power side. (has to be isolated for various reasons)

The 5v and control logic TTL line is from development board, 12v is from a battery

The SSR closes the circuit and activates the solenoid when the 5v TTL is dropped.

Here is my problem:

I connected the solenoid to the board on it's jumper, the TTL was not there I believe since I was still working on the software end (telling zbasic what devboard pin to use for the solenoid).

I don't know how long it took but it could have been as short as 5-10 seconds. I really wasn't paying attention and don't remember. one of the traces on the board between the SSR and the solenoid burnt out/melted like a fuse.

I can repair that with a wireor something, but how can I prevent this? It seems that if I ever have this board connected to the power and the solenoid connected without the devboard running and supplying the TTL to keep the SSR open, then this will keep happenening. The trace is pretty thick.

I'm thinking a resistor might work to reduce current, but that would reduce the voltage too wouldn't it? I don't have the solenoid model on me currently but we want it to work with as many different kinds as possible. I suppose a different resisitor could be used for different models, but I'd really rather not have later people using the board have to figure that out!

I think that the other option would be to remove the 5v line and connect the control line there , but I would have to be able to get the current to be in the 15mA range. I think someone said a transistor would work? Not sure what they do exactly. Post #8 has a site that describe a circuit but to be honest I don't undertand it all, If I would still use the SSR and how the resistor is connected and if it would still work with the SSR (and whether it boosts the control\logic to power the SSR diode)

Thank you for any help!

 

[(*steve*)]

If you melted traces then it's likely that you shorted your power supply.

SSRs are turned on when current flows through the control input. I assume you're using TTL to sink current rather than to source it.

 

[BodhiSci]

I don't think that it was shorted, all I did was plug in the 2 pin connector (don't see how that could short it). I think that the current was too high through the solenoid. It is the battery with the positive going through the board to the solenoid and then returning to the battery's negative. I doubt that the EM coil took that much out of it. Do you see this as likely?

The TTL is normally ON, so it stops the current from coming through the other side (5v). The TTL drops to 0 when the solenoid is to be activated, and the SSR turns it on by closing the other circuit. In this case, the TTL was not on at all so it was activated continuously and melted quickly. So I think the situation is clear now... what do you think about the resistor idea? and what is the other option?

Thank you

 

[(*steve*)]

Well, if you want to switch power to that solenoid, and presuming the voltage is correct, you need to have thicker traces.

What current is required to operate the solenoid and how large were the traces?

 

[KrisBlueNZ]

If one of the tracks from the "contacts" (output) of the SSR to the solenoid went up in smoke, that's because too much current was flowing through it. That means that either the current was higher than it should have been, or the track was thinner than it should have been.

Solenoids are normally able to be energised continuously. Their DC resistance limits the amount of current they draw while activated (assuming you are applying the right voltage across them). So if there was no short, there's no reason why a very high current would have been flowing.

As Steve asked, what is the DC resistance (or the operating current) of the solenoid, and how wide was the track that vaporised?


Here's a schematic for how you should have the CPC1708 SSR connected. This demonstrates several things that I've mentioned in earlier posts on this thread.



Most logic-level outputs cannot provide 15 mA to drive the LED in the CPC1708 so you should use a transistor buffer, Q1. The type is not critical but I've suggested a 2N3904 which is widely available. When the controller drives the input (marked CONTROL) high, current through R1 forward-biases Q1 and it conducts current in its collector circuit. This current flows from the +5V rail, through the LED inside the CPC1708, through R3, and through Q1.

R3 limits this current to around 16~17 mA. You could actually increase R3 from 220Ω as marked, to 270Ω or even 330Ω and still be within the guaranteed specifications for the CPC1708, while saving a bit of current.

When the LED in the CPC1708 illuminates, the MOSFET inside the CPC1708 conducts. This applies the 12V supply (which I've shown as an isolated battery) across the solenoid. The solenoid's DC resistance will limit this current to the value specified in the data sheet for the solenoid.

When the CONTROL signal goes low, Q1 turns OFF, the LED stops emitting infra-red light, the MOSFET turns OFF, and the current in the solenoid starts to collapse. This causes the solenoid to generate a reverse voltage called "back EMF" which is clamped by the combination of the 1N400x diode and the zener. The presence of the zener allows the solenoid current to collapse more quickly than it would if just the diode was present, because the back EMF voltage is clamped to a voltage much higher than zero.

Since the MOSFET in the CPC1708 is rated for 60V maximum and there is already 12V from the battery, the back EMF voltage must be limited to 48V by the zener diode; I would use a 39V or 33V zener to be safe. The zener will actually dissipate nearly all of the energy from the solenoid and may need to be rated at 3W, 5W, or even more.

I don't know how to calculate the required power rating; it depends on the inductance of the solenoid (including the motional inductance) and the surge rating of the zener. Perhaps you can post a link to the data sheet for the solenoid and Steve may be able to provide guidance on the zener's power rating.