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Some products in my company are now required to use an auto shut-off switch to be compliant with new regulations in European space (European Eco-Design Directive ErP).

I got word that the device being integrated into the current products will be the TIPPMATIC iF by Johnson Motor.

Can anyone please explain how these devices are supposed to operate?

The only datasheet I found for this is more of the mechanical kind, and has no guidelines whatsoever towards its use case.

The datasheet has a graph that displays a \$V_{coil}\$ level during a certain \$T_{(on)}\$ and then displays 0V during a \$T_{(off)}\$. It also provides the following values:

  • \$I_{coil\,peak}: 1.0 +0.2/-0.3A\$
  • \$I_{coil\,steady}: 20mA \,DC\$
  • \$R_{coil}: 20 \pm1.5\Omega\$
  • \$T_{(on)}: 3 \times 10^{-3}s\$ (coil command pulse length)
  • \$T_{(off)}: 2s\$ mininum (dead time for next disconnection)
  • \$V_{coil}: 20VDC\$

This is the representation it provides:

schematic

simulate this circuit – Schematic created using CircuitLab

From what I could make of it, I should connect the mains' line between pins 1/1. and neutral to N. Then, should I pulse 20VDC for at least 3ms, and the switch should be "armed" (enabling operation on the load). Then I can't "arm" it again for at least 2s (even if it triggers meanwhile, disconnecting from the load). Can someone attest if this is correct?

Simply put, it's almost as the switch requires a "keep-alive" signal of sorts, that each 2s, requires a 20VDC for 3ms in order to remain closed, right?

EDIT: Still can't make much sense of what I read here - but I know the problem is me above anything else! :) Anyway, the manufacturer's FAE, told me this:

you have to connect the switch in this way.

N: neutral

  1. main voltage

1 to the load (heater, etc...)

On the drawing the switch is shown in OFF position the puls 5vdc should be done during 1-2ms You have to wait at least 2s to do again the operation otherwise a shorter time will heat >up the coil.

This validates @SpehroPefhany suggestion (+1'd), although, I'm still missing the whole auto shut-off feature they claim. The way I figure it, if we enable (turn on, close) the switch manually, it should operate as if no switch was there. Then, should we want to disable (turn off, open) the switch we have to pulse it at 20V for 3ms, is that it? If so, at what point does it shut-off automatically?

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  • \$\begingroup\$ Where's the data sheet dude? \$\endgroup\$
    – Andy aka
    May 12 '14 at 17:06
  • \$\begingroup\$ @Andyaka I believe you would have to register at johnsonmotor.com to get it. It was forwarded to me by a distributor. I'll try to attach it here asap. \$\endgroup\$
    – Joum
    May 12 '14 at 17:09
  • \$\begingroup\$ Actually, the datasheet isn't available from their website. At least not directly and/or it's not easy to retrieve. Also, apparently the component being sampled here is a custom version of a solution they offer. \$\endgroup\$
    – Joum
    May 12 '14 at 17:15
  • \$\begingroup\$ Ask yourself how this question can be most easily answered without a data sheet \$\endgroup\$
    – Andy aka
    May 12 '14 at 17:16
  • \$\begingroup\$ This sounds like a situation where you should be in contact with the manufacturer, not the internet. \$\endgroup\$
    – Matt Young
    May 12 '14 at 17:17
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My interpretation (which could be completely wrong) is that you pulse the switch coil with ~1A at 20V for 3msec to turn it off.

It can only be switched on manually but it can be remotely switched off. If the mains between 1. and N disappears it "drops out" (but maybe the switch rocker does not move?).

You need to wait 2s after switching it "off" before you command it "off" again, or the coil may overheat.

But that guess and $3 will about get you a cup of coffee.

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  • \$\begingroup\$ Thank you for your help! But maybe something's missing: the idea of this rocker-switch (and the euro regulation I mentioned) is that it should turn off automatically. This is the way to produce a real shutdown instead of a standby mode. Your explanation seems to contradict this, right? \$\endgroup\$
    – Joum
    May 13 '14 at 14:31
  • \$\begingroup\$ It says in the datasheet that the shut-off coil is activated by an external circuit, so I think for this model it is provided by the appliance microcontroller (except power loss). Other models have a built-in timer microcontroller that automatically shuts it down. Again, I'm guessing to some extent here. \$\endgroup\$ May 13 '14 at 14:45
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So I finally was able to validate @SpehroPefhany's opinion.

When connected as indicated by the manufacturer, the switch operates correctly.

The logic behind the component is the following:

  • The switch is normally off. Turning it on is a manual process.
  • The only scenario in which the switch turns off automatically, is when the mains' power fails. This way, when it comes back up, the appliance using the switch isn't automatically turned on.
  • To turn it off, it's only necessary to pulse a certain \$V_{coil}\$ (varies with version of the component) for at most \$T_{(on)}\$. This will make the coil release (open) the switch.
  • In case the switch is turned on manually after pulsing the coil (powering on the circuit/appliance), the coil can only be pulsed again 2 seconds after this operation - otherwise it heats up, damaging the component.

As for connection of the switch, the instructions I added in the question edit are 100% correct and straightforward.

Hope this helps someone else who might have a similar problem.

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