triggering a switch with audio?

[Gainzilla]

Hi all,

I'm looking for a way to change my amp channel with my existing pedal looper. My thought was that I could put a box in one loop that would flip a connection to ground (connected to a stereo switching Jack, essentially replacing my Footswitch) when audio passes through. I would engage that loop whenever I'm using clean channel settings. So basically I'm looking for ways to trigger a circuit with an audio signal.

Any ideas?

Alternately, I might be able to install a relay in the looper attached to one of the switch circuits, but I'd like to explore looking at less invasive options first.

Cheers!


- Bryce

[Gainzilla]

Yeesh... The more I think about this the more I feel like I may need to just mod my looper. Audio travels on an ac signal in tube amps, so it must be the same in a pedal (albeit with much lower voltage). I started to wonder if I could just rectify the signal and then use that DC signal to trigger the relay/switch/jfet... But then I realized that just using audio to trigger a switch isnt going work for my purposes. The amp would flip-flop channel switch every time I stop or start playing.

Still, if anyone has any ideas, I'd love to hear them.

[JazzGuitarGimp]

If the looper has an LED that changes states when you want one channel or the other, you could pick the signal off that LED and use it to turn on a transistor. Then connect the transistor's collector (open collector) to to your footswitch input on your amp.

[Gainzilla]

Thanks Lou!

There is indeed an LED. Something like this?

[JazzGuitarGimp]

Put a 10K resistore from the base to the emitter, and a 4.7K resiator from the LED to the base. It would also help to measure the DC voltage, with respect to ground, on bothe leads of the LED in both the on and off states

[Gainzilla]

You just preemptively answered my next question.

I've been reading up on transistor switching circuit design, and it looks like the 4.7k to base is needed to prevent the transistor from committing suicide, and the 10k provides stability when the base to emitter connection is made (I think).

I do need to measure the actual voltage provided by the LED, but my working assumption in the meantime is that it's a 5mm red diffused, so I think it should be around 2v. I'll get actual readings tomorrow.

It looks like my amp footswitch circuit is fed directly off the end of the filament supply through a 10r resistor. Again, I've yet to measure this, but I guess it would be in the neighborhood of 5-6v. This is fed right to the jack tip connection. On the proposed switch jack (to be added to the looper), it would see that same voltage at the jack tip, supplying voltage to the transistor collector? Is that sufficient, or would I need to provide a dedicated voltage connection?

Thanks a bunch!

[JazzGuitarGimp]

Depending on how the LED is being driven (hence the need to measure it), you may be able to take the signal from the other end of its current-limiting resistor (assuming you can track it down) - that would likely give you a full 0 to 5V swing to work with - probably not a big deal as you only need about a volt to drive the base of the transistor.

The 4.7K resistor limits the amount of current that flows into the base. The base will clamp at about +0.7V, so an unlimited voltage higher than that will kill the transistor.

The 10K resistor probably isn't really needed in this case. It keeps the base from floating up and turning on. The source impedance of the driving circuit should be low in both the high and the low states in this case. So the chance of the base floating around isn't likely. In general, the inclusion of the 10K resistor is considered good engineering practice.

You said your footswitch jack is driven by the heater supply? Is that to say they are using an AC voltage to power the relay coil? If this is the case, a BJT might not work, and a MOSFET might be the better choice. In any case, make sure the transistor you choose, whether it be a BJT or a MOSFET, has a high enough collector to emitter (or drain to source) breakdown voltage to accommodate the relay coil voltage - I would shoot for at least two times the relay coil's working voltage. It will probably be easier to implement a BJT than a MOSFET, so I am hoping for DC heaters (or a misunderstanding of your statement about the the heater supply on my end).

[Gainzilla]

Hi Lou,

Apologies for the miscommunication. The switching voltage does come from the heater supply, but it's rectified to DC first (see attached schem). I measured 2.5v from tip to sleeve from a cord plugged into the Jack.

I just pulled apart my looper and the voltage across the LED when lit is 1.781v. There appears to be .01v when unlit.

Given the voltages, and the 60v Max from collector to emitter of the 2n3904, does this look good?

Thanks!

[JazzGuitarGimp]

2N3904 is a good choice.

You might want to add "K" after the 4.7 resistor value.

The only other thing is, you may need to experiment with how you connect ground for the new circuit. You'll nneed to connect the emitter of the 2N3904 to the looper's ground in order for the the transistor to turn on. You will wind having an additional ground connection (besides signal ground) between the amp and the looper, which could cause ground loop hum. If this happens, the first thing I would try is floating the new footswitch output jack you'll add to the looper, and let sugnal ground be the return path for the footswitch signal.

[JazzGuitarGimp]

If you haven't already done so, you should consider the polarity of the footswitch circuit: When the LED is on, the footswitch will be closed (shorted). I don't know if that state puts the amp into clean mode or od mode, but if it is not the desired mode, you'll need to another 2N3904 to flip the polarity.

[JazzGuitarGimp]

Upon closer examination, 2N3904 may not be a good choice. If I am reading the schematic correctly, it looks like the current through the footswitch circuit will be nearly 300mA. The 2N3904 has an absolute Ic max of 200mA. It also looks like the DC current gain will be only about 30 at 300mA of collector current, which means you will need 10mA of base current to get the job done. Can you measure the current required by using your DMM, set to DC Amps, to complete the curcuit at the footswitch jack? Once we know for sure how much current the transistor needs to sink, we will be better equipped to specify the transistor.

[Gainzilla]

Hi Lou,

Thanks for giving this so much thought!

Ok, so it looks like the lead channel is on when the signal shorts to ground, so that looks backwards from what I was intending. If you look at the attached schematic, you'll see a connection right before the Jack that goes off screen. This connection goes to the lead channels indicator LED. Since this is a switching Jack, it looks like the LED turns on when the Jack is shorted. I checked this by simply unplugging the switch, and sure enough, the amp is on the lead channel when no Footswitch is plugged in. There is no front panel switch either, so the only way to change channels is with the Footswitch Jack. In any event, I suppose I could plan this so that I program the looper to turn on the channel switch triggering LED when I'm using the lead channel rather than the clean.

It looks like the amp puts out about 102ma Max current at the Footswitch. I was expecting a solid reading but I was getting a fluctuating reading that went from 95 up to 102 and back down. To be clear, I set my DMM (a pretty nice Fluke I was given by my father in law) to ma and measured across the tip and sleeve. This amp changed channels instantly upon touching the positive lead to the tip.

Assuming I measured correctly, it looks like the 2N3904 works? Max 200ma, and the amp reads 102?

Cheers!

[JazzGuitarGimp]

Okay, we're back to the 2N3904!

If you don't want to mess with reprogramming the LED on the looper, you could use the circuit attached. However, this requires another transistor and two more resistors. It also requires you find a constant source of +5V in the looper ( it can actually be any voltage from +5V up to +20-ish V although R1 will need adjusting).

According to the Fairchild 2N3904 datasheet, the saturation voltage at 100mA is about 0.15V, so that should get the amp's footswitch circuit sufficiently close to GND to work properly.

[Gainzilla]

Holy cow, thanks for taking time out of your life on a Sunday to help. That was very cool of you, and I greatly appreciate all your help with this. Prior to this little thread I felt like I didn't really understand transistors much. Thanks to your help, and the necessity of needing to do some research I feel like I'm starting to get it.

Anyway, I'm not sure how soon I'll get around to actually doing this but it won't be until the weekend at a minimum. I'll follow up in this thread and show the actual looper, etc.

Quick question, what is the etiquette around changing the title of the post? I don't think the title will be very useful for anyone looking to solve similar problems.

Cheers!

- Bryce

[JazzGuitarGimp]

Etiquette? What's that? I'm not much on etiquette. But I think, and especially since you and I are the only ones who have contributed to this thread, that you should change it to whatever you like. Send me a PM with the new title so I can continue to follow along.

Cheers,
Lou