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A while ago, I bought a wireless switch (similar to this) for a wall outlet. It has recently failed so I opened it to see if there was a simple fix. It is a two part system (the transmitter which sticks to the wall, and the receiver/relay which plugs into the outlet) - but since the receiver gets knocked around a lot and has the relay, I figured I'd start there.

I don't have much experience working with relays, but the circuit I found seems to defy logic.

First, an image of the board - note, the image of the bottom was taken against the white screen of my monitor, so that the profile of the relay is visible (making its pinout clear).

bottom view

On the left edge, you'll see prongs for an outlet. So, this module plugs into an outlet and is suppose to switch the outlet power to the black socket (bottom-left) where some other appliance would be plugged in (a desk lamp, in my case).

The module receives signals from an RF transmitter which acts as a switch on a wall - the RF receiver circuitry is clearly the mess on the bottom right of the board. When a signal is received, the board is supposed to activate the relay, and connect power to the output socket.

Now to the point: What's got me confused is that, when tracing out the front-end, I find that the circuit is (or at least, "was") somehow activating the 24 VDC relay coil via the mains. Here's what I see: simple trace of front-end

(Note: I realize that I drew the relay a little unorthodox. I did it that way to mimic the symbol as it is presented in the datasheet.)

This must be possible, since it was working for a couple of years, but how? From the image you can clearly see that the top of the HRS4H-S-DC24V relay coil (bottom-left pin of the relay, as shown in the image) is connected to a mains prong. The bottom of the coil then goes to the collector of a C945 NPN transistor - which could never drop 120 VAC, so I assume there must be some weird biasing going on.

Q: How is it possible that the circuit is activating the switch, using the mains as a driver for the coil, when the coil is clearly defined for use with 24 VDC?

I feel like I've missed something very obvious.

Here's the top view: top view

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  • \$\begingroup\$ Before anyone votes to close as repair, OP is asking how a DC relay with a NPN transistor can be used with mains AC voltage. This is a design question. \$\endgroup\$
    – Passerby
    Dec 17, 2017 at 4:32
  • \$\begingroup\$ @Passerby Yes, exactly. Thanks for pointing that out. \$\endgroup\$ Dec 17, 2017 at 4:33
  • \$\begingroup\$ That said, this would benefit from a full schematic. It looks simple enough you can crank one out with some time. \$\endgroup\$
    – Passerby
    Dec 17, 2017 at 4:33
  • \$\begingroup\$ @Passerby I'm actually working on that. It's a bit of a task, so I figured I'd make the post in the meantime. The only caveat is that the chip in my image labeled "Unmarked IC", is in fact completely unmarked. \$\endgroup\$ Dec 17, 2017 at 4:35
  • \$\begingroup\$ That's okay. We can assume it provides simple switching of a gpio. But how all the passive interact is important too. \$\endgroup\$
    – Passerby
    Dec 17, 2017 at 4:36

2 Answers 2

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Q: How is it possible that the circuit is activating the switch, using the mains as a driver for the coil, when the coil is clearly defined for use with 24 VDC?

It's done like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The coil only ever sees 24VDC here. DC shares a common potential with AC but as the circuit is encapsuled into a casing with no other outlets than the mains one, it doesn't matter.

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  • \$\begingroup\$ I see what you're getting at, though this differs a bit from my drawing. I should have pointed out that the drawing is not a guess - it's precise (at least, for the few traces I drew). The relay has direct connections to each of the two main-prongs: one at the coil and one at the switch, whereas yours only connects one of the prongs (to two places on relay). And the other side of the coil is at the npn collector. I may be able to extrapolate your answer to come to an understanding, but I'm still having a little trouble realizing exactly how. \$\endgroup\$ Dec 17, 2017 at 5:28
  • \$\begingroup\$ It's unlikely that a NPN is used like that in this circuit \$\endgroup\$
    – Passerby
    Dec 17, 2017 at 5:29
  • \$\begingroup\$ Like it better that way? The relay contacts do not meddle with the function of the DC supply or relay coil. \$\endgroup\$
    – Janka
    Dec 17, 2017 at 5:34
  • \$\begingroup\$ Yes actually. That's much more representative. Other than the NPN discrepancy, I think this is making sense. \$\endgroup\$ Dec 17, 2017 at 5:46
  • \$\begingroup\$ I’m just waiting to verify this with a DMM, before I +1/accept. Sorry for the delay. \$\endgroup\$ Dec 18, 2017 at 1:47
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This uses capacitive reactance to drop the AC voltage, the brown capacitor (C13) is likely a X2 type that is used for this. The thick resistor next to it acts a overload fuse. From the pictures it looks like the circuit only uses a half bridge rectifier with a Zener diode and capacitors to regulate and smooth the voltage to ~24V DC. The unmarked IC is likely to be a microcontroller get the signal from the RF part of the circuit and toggles the relay off and on. This site shows a circuit for a transformer-less power supply and how to calculate the value. https://www.nomad.ee/micros/transformerless/index.shtml

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