I'm trying to build a simple circuit to control a church lampadary for an escape room game. I don't know if you've seen these but they're fairly common in catholic churches where I live: You input donations and a fake candle lights up for each coin.

Here's a basic blueprint for it: Schematics

I bought this coin acceptor: 12V coin acceptor

And powered it up with a 12V, 1A adapter. At the same time I'm using a car usb phone charger as converter (12V to 5V) to power my logic.

This acceptor has a NC signal configurable to output 100ms, 50ms or 22ms square pulses whenever a coin is inserted. To adapt this signal I used a couple of 10M resistors in series and a parallel 5V Zener diode in reverse connected to my 7474's ¬PRESET pin. Two ANDed flip-flops catch this signal and send a "clock pulse" to a 74164 which puts one bit up and lights a candle.

You can see the mounted thing in: enter image description here

(Sorry for this, newbies can't post more than two links) All ICs in the picture are pointing west. Pin #1 is on the south west corner.


simulate this circuit – Schematic created using CircuitLab

Still, when I input a coin in my acceptor, nothing happens. I've also tried to simulate acceptor output with a switch, and I saw two consecutive bits light up at the same time at each switch push. Maybe switch bouncing, which I thought I could solve with my flip-flops? Or maybe it isn't a good idea to use gate output as a clock signal for another IC?

Acceptor output gives 2.56V in my multimeter, which according to the datasheet should be enough to activate the latch.

This being the case, my question is: Is this a/the proper way to interface my 12V acceptor with my circuit? Am I missing something?

Thank you for reading. Any help would be appreciated!


I implemented the voltage converter using the IRF530 as proposed in this answer. Still the circuit wasn't responsive. I suspected the coin acceptors so I tried to test them connecting a couple of LEDs to N.O. and N.C. outputs through a 10K resistor. None of the LEDs lights up but they aren't burnt. Are the acceptors definately broken or is there something I'm missing?

Testing the acceptors

  • \$\begingroup\$ Voltage divider is the worst option to step down a voltage for power. Normaly you either use a linear or switch power supply. Also , D1 isn't doing anything here. because the minimal current of 1mA to make the diode isn't there. \$\endgroup\$ – MathieuL Jun 2 '16 at 19:55
  • \$\begingroup\$ What is the power consumption of the coin acceptor? If you don't have any reference, you gonna need to power it and check with an amp-meter the current consumption. \$\endgroup\$ – MathieuL Jun 2 '16 at 19:57
  • \$\begingroup\$ I didn't check the queue, I just edited the OP around the same time you did. Sorry if I missed it -- I checked the queue just now and don't see anything in there. \$\endgroup\$ – Krunal Desai Jun 2 '16 at 20:13
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    \$\begingroup\$ Are you using the original bipolar 74xx TTL family parts, or the more modern CMOS version? The inputs of the bipolar parts (74xx, 74LSxx) source current, and must be actively pulled low to be considered Low. CMOS parts (74HC, 74AC, or anything else with a "C" in the middle) have a very high input impedance, and will appear as Low with a fairly high (10K) pull-down resistor and will havew an unknown state if not pulled either up or down. Your 20 MOhm resistors are much too high for any logic family to reliably recognize as a high. \$\endgroup\$ – Peter Bennett Jun 2 '16 at 22:46
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    \$\begingroup\$ Why not just change the 20 M ohm to something around 7 to 10 K ohm. That way you will get a better signal. \$\endgroup\$ – David Drysdale Jun 3 '16 at 0:42

20Meg will only allow about 600nA of current to flow, which may not be enough to bias the input of your logic gate (check its input bias current requirements).

A quick and dirty way to do this could be to use an optocoupler, or a simple NFET that can tolerate a VGS that high. It would act as a simple inverter if your logic can tolerate that.


simulate this circuit – Schematic created using CircuitLab

For the above, I would add a protection diode to the input of the MOSFET (to clamp voltage below VGS(max)), or choose a MOSFET with a higher VGS(max). An opto has the benefit of isolating you from the higher voltage as well.

  • \$\begingroup\$ I tried this but to no avail. Then I tested the acceptors independently. I published an edit with subsequent changes. Thank you very much for your help. \$\endgroup\$ – Nirro Aug 27 '16 at 19:23

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