I have a project which is proving to be more challenging due to my limited knowledge of electronics.

I am installing a homemade central vacuum system. It uses 12 V signal from the outlets to operate a 230 V vacuum cleaner. Plug the hose in, make the circuit, start the vacuum. Unplug the hose, the motor stops.

I have a solid state relay which switches it perfectly.

I wondered if there was a way to keep the vacuum running for 20-30 secs after the outlet has been unplugged, to clear any debris in the pipework?

I was thinking of possibly using a capacitor on the 12v circuit but have no idea how to calculate the size I would need!

  • \$\begingroup\$ You might edit your question to explain why you want a solid state relay. For relatively infrequent operation as in your application an off-delay relay module would be fine. \$\endgroup\$ – Transistor Jan 25 '19 at 10:19
  • \$\begingroup\$ I'm not sure what you're asking here. It sounds like you just want someone to find you a commercial timer module, which is not what this exchange is for. \$\endgroup\$ – pipe Jan 25 '19 at 10:21
  • \$\begingroup\$ It may be possible to modify your existing circuit quite simply but you'd need to add in the make and model of the SSR you're using and a link to its datasheet. \$\endgroup\$ – Transistor Jan 25 '19 at 10:31
  • \$\begingroup\$ A schematic might help. Don't use a capacitor on 12V line, it might burn your relay. \$\endgroup\$ – Dorian Jan 25 '19 at 11:39
  • \$\begingroup\$ Having worked with many central vacuum systems previously, I can tell you that keeping the vacuum motor on for an extended period after the hose is pulled out is NOT effective. That is because the flap that covers the hose connection point stops any air flow from occurring. If you have only a single hose connection, you can remove the flap / door and eliminate that problem. But most central vac systems have multiple hose connection points. \$\endgroup\$ – Dwayne Reid Jan 29 '19 at 5:23

You may be able to find commercial, off the shelf (COTS) devices. This site isn't for that purpose. But to answer your question, yes. The following schematic shows one basic approach (there are always others):


simulate this circuit – Schematic created using CircuitLab

\$Q_1\$ can be any of a number of small-signal PNP BJTs capable of supporting the relay current (I assumed \$250\:\text{mA}\$ here.) You didn't specify your DC relay, so I may have over-specified its requirements a bit (many require less than \$3\:\text{W}\$ to operate them.) \$M_1\$ is shown as a BSS145. That's because I'm depending upon its threshold voltage for the timing (along with the values for \$C_1\$ and \$R_1\$, which collectively set the hold time you wanted.) However, again, any of a list of small-signal NFETs can do the job. But the timing may need to be adjusted by changing the value \$C_1\$ and/or \$R_1\$ to get the time you want if you choose a different one. The point is that pretty much nothing here is all that critical.

As already noted, the timing of the circuit depends upon \$R_1\$, \$C_1\$, and the threshold voltage of the NFET you choose.

The circuit requires an NFET instead of a BJT because there's no gate current to speak of. This allows the RC timing to be selected over a very wide dynamic range of possibilities and for extended periods of time. (A BJT in its place would not allow such a dynamic range and would greatly complicate extended periods of time.)

\$D_1\$ is important, but \$D_2\$ can probably be eliminated if you don't want to add it (it helps protect the NFET gate, though.) \$D_5\$ is the usual free-wheeling diode for the relay inductance. You could use other methods (zener+diode, RC snubber, etc.) But you probably don't care how fast the magnetic energy collapses. So a simple diode is probably fine. \$D_3\$ and \$D_4\$ are required to hold \$C_1\$ while your switch is engaged (your hose is plugged into the wall.) Once you pull the hose out, the timing circuit will then be allowed to do its job and time out the period you wanted.

Be aware that there are SOT23 diode pairs, such as the BAV99 and BAT54A, BAT54C, and BAT54S, to check out. You can replace several pairs of diodes shown in the above diagram with one or two of those things. Again, this is your call.

The basic idea is that when the switch is engaged, it pulls upward on the relay load and supplies current to activate it. However, that's not the important detail (it's doing that just for a slight moment through \$D_3\$.) The important detail is that it also pulls up on the gate of \$M_1\$ via \$D_4\$ and also via discharged \$C_1\$. In doing so, the NFET \$M_1\$ turns on and pulls downward on the base of \$Q_1\$ via \$R_2\$. \$Q_1\$ takes over and bypasses the switch and \$D_3\$ providing a voltage still closer to \$12\:\text{V}\$ (with only a slight \$V_{\text{CE}_\text{SAT}}\$ voltage drop of \$Q_1\$) to the relay itself. (Most relays only require about 70% of their rated voltage to engage, so there's plenty of safety margin here.)

Meanwhile, the top side of \$C_1\$ has also been pulled up by both \$D_3\$ and then also by the collector of \$Q_1\$ and the bottom side of \$C_1\$ has been pulled up by \$D_4\$. Together, these keep \$C_1\$ from charging (and therefore keep it from performing its timing function) so long as the circuit and your switch remain active.

Once your switch disengages, this releases the "grab" across \$C_1\$ that had been held there by the switch (through \$D_3\$ and \$D_4\$) and now \$C_1\$ can start to charge via \$R_1\$, gradually getting to the point where it forces the gate of \$M_1\$ below its threshold voltage. When that happens, the base current supply for \$Q_1\$ expires and \$Q_1\$ rapidly turns off. Now, the top side of \$C_1\$ finds itself slammed down to ground (and still more negative, limited by \$D_5\$) via your relay inductance/load. The bottom side of \$C_1\$ rapidly discharges via \$D_1\$, almost immediately resetting the entire circuit for another use.

As Hearth notes in a comment below, \$D_5\$ is only needed if a mechanical relay is used for switching your power on and off. Given an SSR, the above circuit is actually over-built, since I've assumed that I need to support driving up to \$3\:\text{W}\$. A typical SSR requires only a modest amount of current through an opto-isolator and the circuit could be re-designed to reduce the base current in \$Q_1\$. For example, \$R_2\$ could easily be increased to reduce that base drive.

Or you can just buy something commercial. I'm sure there's something out there.

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    \$\begingroup\$ Since the asker is using a solid-state relay, D5 is likely unnecessary. \$\endgroup\$ – Hearth Jan 25 '19 at 15:40
  • \$\begingroup\$ @Hearth I take that point. I'll add that note to improve the answer. Thanks. \$\endgroup\$ – jonk Jan 25 '19 at 15:41
  • \$\begingroup\$ Very well-written answer otherwise! I don't mean to only point out the negative. \$\endgroup\$ – Hearth Jan 25 '19 at 15:42
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    \$\begingroup\$ @Hearth I think it's a good catch, though. And it gives me an opportunity to add a thought to improve the general usefulness of the answer. I really appreciated the comment. \$\endgroup\$ – jonk Jan 25 '19 at 16:00
  • \$\begingroup\$ @Hearth Also thanks for correcting your name!! hehe. \$\endgroup\$ – jonk Jan 25 '19 at 16:03

Potter & Broomfield (AMF) may still manufacture:

CUB-51-70060 'Time Delay on Operate, Adjustable 1 to 60 sec, contacts 10 amp. 1/3 HP at 120 VAC Not 100 % sure, but I believe the relay coil is also 120VAC

enter image description here

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    \$\begingroup\$ This doesn't look like a solid-state relay to me. \$\endgroup\$ – Hearth Jan 25 '19 at 15:39

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