Many of you who make audio circuits are likely familiar with the following simple design:

(source: synthesizers.com)

It takes the gate output (typically a +5V pulse) from one piece of analog electronic audio gear, and uses it to close a switch on another piece analog electronic audio gear (using a transistor).

Recently I've used this little circuit to take the gate signal from one piece of gear (I'm using a LepLoop) and trigger (or close) the "push" of the One-Key-Play button on my old Casio PT50. This allows the Casio to be sequenced in time. It works great, for the most part. More specifically it works while playing sound out of the Casio's speaker, and with headphones plugged in. The problem arises when I plug the Casio's audio output into a mixer which also receives audio output from the LepLoop. In other words, the LepLoop sends gate-out to the Casio and sends it's own audio to the mixer. In this setup the Casio is no longer triggered.

Some strange things to note:

  • The actual gate pulse can be faintly heard on the Casio channel on the mixer. I'm guessing this means that the gate pulse found an easier path to ground through the Casio rather than returning to the LepLoop through the gate cable?

  • I can also get the S-trigger to stop working by simply touching the grounds of the audio out cables from the LepLoop and Casio together (mixer aside).

  • Checking the transistor emitter and collector with a multimeter, I get beeps in rhythm with the gate (but notes are not triggered). I think this means that the switch is still closing when the devices share ground, although I'm not sure how accurate this test is given that the test leads may be affecting the circuit.

  • Using a multimeter on the LepLoop, I tested between the ground from the gate out and audio out, and found that I get beeps when the gate fires a pulse, so this means they share ground during gate pulses.

I've tried modifying the circuit in a hundred ways using different arrangements and values of resistance and diodes with no luck. At this point I'm pretty stumped so any help would be greatly appreciated. If I can't figure this out I might be forced to explore solid state relays for this application. Any suggestions there would also be appreciated. Thanks!


1 Answer 1


The circuit you used to convert from Voltage trigger to Switch trigger connects 'ground' from the V-trigger source to 'ground' on the device being triggered. This works fine when both devices do have their grounds on these ports.

But the Casio PT-50 probably has a keyboard matrix with neither side of the button grounded. Your hack connects one side of the One-Key-Play button to the LepLoop's ground. You may get away with this if you don't make any other connections between the PT-50 and the LepLoop (or any other equipment that shares a common ground with them), but as soon as you do a short circuit is created between the switch matrix and ground in the PT-50.

To prevent a short circuit you need to break the electrical connection between input and output sides of the trigger circuit, while still passing the switch closure signal through. This can be done with an opto-isolator, which uses an LED and phototransitor to transfer the signal via a beam of light. Here is an example:-

enter image description here

The transistor in the 4N25 works the same as in your original circuit, except that it is turned on by infrared light falling on its Collector-Base junction rather than a direct connection. As in the other circuit the transistor only conducts in one direction, so you should connect the Collector and Emitter to the same points.

The 1kΩ resistor sets the LED current, and the diode between pins 1 and 2 prevents damage to the LED from reverse voltage (it can be left out if you ensure that the V-trigger polarity is correct).

  • \$\begingroup\$ Thanks this answer is perfect! I'm currently working to implement an optocoupler. In another forum, I found that these can be found in old PC switched power supplies, old TVs, and old monitors, although there are many cheap options online. \$\endgroup\$
    – Jim Slice
    Commented Jan 22, 2018 at 18:28

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