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I'm fairly new to electronics and am stuck at how to get a circuit working. I have 3 components connected in the circuit:- a card reader, relay and an electronic door lock release, and all operates at 12v DC. The card reader is continually powered on and is a NC type, meaning when the reader has power an internal switch is always closed and it outputs a short to ground. When a registered card is presented the card reader's internal switch closes for a set time (approx. 15 secs) and the short to ground becomes an open circuit, and after the set time the internal switch returns to a NC position and the short to ground is restored.

The reason this card reader operates as a NC type is because it is primarily used for magnetic locks which are always energized to lock the door (hence a constant short to ground to power a component) and de-energized when the door is unlocked (short to ground becomes an open circuit and power removed from a component).

I will be using a conventional electronic lock release which requires the relay to be powered on only when the door is required to be unlocked and no power to the relay when the door is required to be locked (basically the opposite way a magnetic lock works). So when the door is required to be locked I need the relay to be powered off so the lock release (on the load of the relay) is not operating. When a registered card is presented to the card reader I need the relay to be powered on and in turn the load circuit will be complete and the door lock release will operate and unlock the door. After approx. 15 seconds the card reader will return back to the NC position and I need the power to the relay to be removed and hence the door lock release will not be operating and the door will be in the locked position again.

I have looked at NPN / PNP transistors but they require the output to be in the “on” position to operate the circuit, but my card reader had no output when I want the circuit to operate. I think I need a MOSFET transistor but can’t figure out how to make it work.

NOTE: I cannot do this with a relay with NC outputs, i.e when the relay isn’t powered to use the NC outputs of the relay to complete the load circuit because of fail safe. By using a NC relay if the card reader was to lose power, so will the relay and it will default to the NC position (as it won’t be powered on) and the door will open when it shouldn’t be.

Thanks in advance for any help.

enter image description here

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  • \$\begingroup\$ Is the lock solenoid fed from the card reader PSU? \$\endgroup\$
    – Transistor
    Jun 20 '20 at 16:29
  • \$\begingroup\$ Yes, the lock release is on a different feed and does not share with the card reader PSU. Also the lock is a strike lock release not a solenoid so when the lock release is energized it will open the door. \$\endgroup\$
    – DJKalz
    Jun 20 '20 at 21:19
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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. What you've got, want and can have.

If the reader and solenoid are on the same supply then you don't need RLY1 as the lock can't open without a supply.

Figure 1c shows a workable solution:

  • RLY1 proves the reader power is OK. If power is lost then its contact opens as shown on the right and the UNLOCK3 solenoid cannot operate.
  • RLY2 will be normally energised so its contact on the UNLOCK3 circuit is normally open and UNLOCK3 is de-energised.
  • If RLY1 is on and the card-reader gets a valid card CR1 will open, RLY2 will turn off. Now RLY1's contact will be closed, RLY2's NC (normally closed) contact will close and UNLOCK3 will be energised to unlock the door.
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  • \$\begingroup\$ Thanks for the answer but that won't work. I don't have a maglock, I have a strike lock release. Also, the lock has a separate PSU and can't share the same PSU as the card reader. \$\endgroup\$
    – DJKalz
    Jun 20 '20 at 21:21
  • \$\begingroup\$ I've relabeled the unlock solenoids for clarity. Your question doesn't mention a separate power supply but I assumed one as otherwise the solution is simple. (The lock can't be released if the common PSU fails.) Have another look at my circuit and tell me why Figure 1c won't work. \$\endgroup\$
    – Transistor
    Jun 20 '20 at 22:10
  • \$\begingroup\$ I'm not sure I understand figure 1c. Can you explain what RLY1 and RLY2 are? Also if RLY2 is the relay than controls the lock release, then when that relay is powered on it will unlock the lock, like I said it is not a maglock I have (which locks then powered) I have a lock release (which when powered releases or unlocks the door). Also, if you have a look at my original diagram I have also drawn the relay (far right) that needs powered and it shows the load on the relay (which operates the door lock release), which is a completely separate circuit from the card reader circuit. Thanks. \$\endgroup\$
    – DJKalz
    Jun 21 '20 at 10:52
  • \$\begingroup\$ As indicated on the schematic, I've used an NC contact of RLY2 so when it's powered on it disconnects the UNLOCK solenoid. (You can stop telling me it's not a maglock.) See the update. \$\endgroup\$
    – Transistor
    Jun 21 '20 at 12:03
  • \$\begingroup\$ Thanks for the clarification, I can see how it works if you use a proving relay. However, I was hoping for a solution which uses transistors for the switching, rather than having relays that need to be powered on 24/7, therefore have a higher chance of failure over the years. If there anyway it can be done with transistors / MOSFETs ? \$\endgroup\$
    – DJKalz
    Jun 21 '20 at 18:48

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