Time to add a footswitch to my Dean Markley K-20X

[joecool85]

Alright, I'd like to add a footswitch to my K-20X.  I love the built in overdrive, but hate having to run over and hit the button.  I'm thinking I could remove the stock switch and put in a relay.  Would this work?  Also, I'd need to add a button on the front of the amp to activate the relay when the footswitch isn't plugged in - that part is easy enough.

The big question, what relay should I use/where can I get it?  The amp runs on +/- 19.5v DC.

Here's the schematic: http://deanmarkley.com/Info/LegacyAmps/Schematics/D1515.pdf

[DJPhil]

I did something a little like this for a friend once. We used a small reed relay removed from an old modem and wired it in parallel. This sets it up like an OR gate, if either the footpedal or the front switch is on then OD is on, and they both have to be off (or the footswitch disconnected) for clean.

In my case we made a small pcb daughterboard and crammed it in as close to the switch as possible to keep the leads carrying signal short. I soldered small leads from the relay pins directly to the switch pins on the solder side of the amplifiers board. It's a bit tricky and kind of kludgy but it keeps the modification to the amp minimal.

Any DPST relay should work, and you can use a zener regulator to drop your rail voltage down to 5V or 12V for use with a small relay. The smaller relays tend to operate on lower voltages and currents, and this keeps you from having to run full rail (or both!) through a TRS jack in the back for switching (disaster in the making). The relay's datasheet should list the hold current, and you can plug that into a zener regulator calculator to see what power rating you'll need on your zener.

I recommend including a protection diode across the relay coil, any 1N400X series rectifier will work fine. This will keep your amp safe from voltage spikes when the coil switches off, and reduce the chance of the switch causing audible noise in the amp's output.

Hope that helps

[joecool85]

Hope that helps

That does help.  Still at a bit of a loss as for what relay to use.  Where can I even get one?  Also, from looking at the sheet it looks like I will need a DPDT not DPST relay.

[J M Fahey]

Here you go.
You'll need a 12V DPDT relay, because your DM uses a pushbutton DPDT to switch distortion in, and you need to replace it.
Pull it from the PCB and route its contacts to the corresponding contacts in the relay.
The series resistor is needed to bring those 19V down to needed 12V.
Value?: measure the DC resistance of the coil, and multiply it by 0.58 .
Example, if your relay coil has 1200 ohms, you'll need a (1200 * 0.58)=696 ohms.
680 ohms is close enough. 1/2W is enough dissipation.
Repeat calculations for *your* relay.
You can put a footswitch jack somewhere and a switch by it.

[joecool85]

Juan, I guess I'm a bit confused.  What is the "left switch" and "right switch" about?

**edit**

Also, wouldn't something like this make more sense for the jack/switch situation?



Wouldn't this mean that if the foot switch was plugged in it would control the relay and if not, the switch on the front of the amp would control it?

[DJPhil]

That does help.  Still at a bit of a loss as for what relay to use.  Where can I even get one?

For something like this I'd usually use All Electronics. Here's an example from them that should work fine, plenty in stock, and an easily found data sheet. They've got a wide selection, here's a 12V DPDT if you'd rather use 12V (and it's datasheet).

Also, from looking at the sheet it looks like I will need a DPDT not DPST relay.

Here you go.
You'll need a 12V DPDT relay, because your DM uses a pushbutton DPDT to switch distortion in, and you need to replace it.
Pull it from the PCB and route its contacts to the corresponding contacts in the relay. . .

That's doing it the right way.
It's true that you'll need a DPDT relay, that was my mistake. I should not post from memory, something made me think DPST would work.  :duh
Using a 12V relay is probably preferable over a 5V because it will not stress the zener regulator as much (less voltage drop). This may not be the case if you use very large 12V relays that require a lot of hold current though (automotive relays perhaps), as the higher current will also cause stress. The small signal type relays above are well suited to the job. The 12V version above uses 140mA 140mW nominal, 79mA 79mW minimum, so you'll probably want to design your regulator for 200-250mA 200-250mW (assuming 12V ~16mA, 20mA would be fine) just to be safe. The 5V relay uses roughly the same current, but you have to shed more voltage to use it. If you were designing for production you could balance the loss between the zener (voltage drop) and the resistor (current demand), but it really doesn't matter too much here so long as you use parts that won't burn up.

Note: Many of these relays are sized right to fit in DIP sockets, which can be useful, but you always have to check the sheets. The 12V version above, for example, would fit in a standard DIP socket (like this, but preferably with 10 pins), but the 5V version would require a skinny DIP socket (0.2"[5.08mm] wide as opposed to 0.3"[7.62mm]).

Sorry, I had a lot of coffee this morning so I'm kind of babbling.  :loco

Juan, I guess I'm a bit confused.  What is the "left switch" and "right switch" about?

Those represent the relay output, and you'd wire them in to the six pins where the old switch came from. They get wired in to the amp's mainboard where SW1A and SW1B are on the schematic.

Also, wouldn't something like this make more sense for the jack/switch situation?
Wouldn't this mean that if the foot switch was plugged in it would control the relay and if not, the switch on the front of the amp would control it?

I think you've got the right version. Switching always makes me a bit dizzy, but it looks like yours would behave as you describe.

Bleh, I hope that made some sense. Sorry about the mistakes in my first post.  

Edit: Corrected zener calc math above (blue text).

[joecool85]

Looks like this one would work: http://www.radioshack.com/product/index.jsp?productId=2062483

Am I right in thinking so?

[DJPhil]

Looks like this one would work: http://www.radioshack.com/product/index.jsp?productId=2062483

Am I right in thinking so?

Should work just fine (though I hate RS's prices). It'll use a bit more current (~60mA), but that's no big deal.

Also, corrected my math above. I read mW and wrote mA, sheesh.

[joecool85]

Should work just fine (though I hate RS's prices). It'll use a bit more current (~60mA), but that's no big deal.

I hear you about the price, but I can get it locally and it will be $2.50 cheaper than that other switch since I would have to pay shipping on it.  So far I haven't been able to find one online store that has an NC 1/4 Mono jack, 12v non-latched dpdt relay and a spst (or dpdt for that matter) stomp switch.  Looks like I'll get the relay from RS and the jack and stomp switch from smallbearelec.com.  I'll end up using a DPDT switch, but for $3.75 you can't beat it.  I can't even find a good stomp switch that is SPST, never mind for that price.  That and Steve at smallbear is a great guy and I like supporting his business.

I'll probably use a raco box like this one that I can get locally for the enclosure: http://electrical.hardwarestore.com/14-45-electrical-boxes-metal/octagon-box-with-cable-clamps-601951.aspx

Nothing like over-doing it strength wise.  No way am I going to be able to break this one!

**edit**
I'll probably add a LED indicator to this box as well.  Since I'll be running a DPDT switch anyway, and the wire I want to use is 2 wire plus shield, I think I will end up using one rail on the PSU to run the relay, and the other to run the LED.  I will just need to use a stereo NC 1/4" jack and plug.  The relay power will come through the wire at ~13.2v DC and the LED I will have a resistor in the amp to bring the voltage down to whatever is appropriate for the LED I pick out.  Should be pretty slick.

[J M Fahey]

Hi joecool, let's answer one by one.
1) In your K20 you use a dpdt switch, equivalent to two spdy but mechanically joined.
They are called Sw1a (on the left) and SW1b (on the right).
You'll have to wire each of them to the corresponding half of the relay
2) you don't need to get 12V regulated for it (even less 5V); the resistor I suggested turns your 12V relay into a 19V one.
That 19V rail can supply way over the current you need, since it's meant to  feed the power amp itself.
3) The way I wired the foot and hand switches (in parallel) is equivalent to the other (in series), but in practice is safer and will last longer, because jack contacts are much more likely to become worn/open/corroded, etc. , rather than shorted.
Your system is "automatic", plugging the footswitch kills the switch; with mine you must remember to leave the switch off , which isn't that difficult because in normal use you use the amp clean and switch the OD on when needed.
Of course, if you feel more confortable with the other way, by all means use it.
4) If you buy a regular relay, you do not need a board at all, just contact-cement it to any convenient place and wire it as needed.
I said contact-cement and not epoxy because rigid adhesives tend to unglue on impact, and semi-flexible ones do not.

[joecool85]

Ok, it's been a while but I'm finally going to do this.

One question though, the protection diode across the relay, how important is that really?

I plan on using a 5v low signal relay that only draws 140mw - I can't imagine this would affect the PSU a whole lot on my amp.

[J M Fahey]

The diode is to protect the switch which activates the relay.
A mechanical one will arc over; a transistor one will die.
Besides, you lower clicks and thumps.

[Enzo]

A diode costs one or two pennies, not worth trying to be cheap.

The coil in the relay is an inductor, in addition to its duties as an electromagnet.  WHen you try to turn off the relay, that inductance tries to keep the current going - that is what inductors do.  It generartes a voltage spike.   It can be a pretty large voltage spike.  That spike is shunted off by the diode across the coil.  WIthout the diode, as JM said, it can arc across your switch, ir if a transistor was controlling the relay, it will destroy the transistor.

I used to work on pinball machines.  SOlenoid coils ran the moving things like flippers and bumpers.  24v solenoids.  They came in AC and DC versions.  The DC coils had a diode right on the coil and in parallel with it.  The AC coils were the same, but no diode.  I remember one time I was replacing a bad coil in a DC circuit, but only had AC coils.  No problem, we carried diodes, easy to slap one on there.  But I decided to test the circuit just to see if it was working, before I installed the diode.  Power up.  I tapped the switch blades together to fire the solenoid, and instantly a bright blue-green arc formed between the contact blades, and burned steadily until the switch blade burned in half.   That is what the diodes prevent.

[joecool85]

Great explanation guys, I had no idea.  Glad I asked :-) 

Now, I've used relays in cars and motorcycles before, do those have built in diodes or something?  I've certainly never added one and never had any issues either.  I use ones like this normally: http://www.parts-express.com/pe/showdetl.cfm?Partnumber=330-073

[J M Fahey]

Not used in car/motorcycles because activating switches are large and robust, not delicate as those used in Electronics.
Anyway, look at the contacts with good light and a loupe and you'll see the pitted contacts.
Looks like the craters of the Moon.
And in the dark you see the blue sparks.
Not to mention that in an "explosive atmosphere" (nothing more unusual that some gasoline spill) they trigger fires or explosions.