Ok, back to adding this reverb circuit to my Firefly amp. I wanted to add a footswitch to turn the reverb on and off. I looked at the remote bypass switching at GEOFEX and I think that might be overkill for what I'm looking to do. I looked at my footswitch for my Peavey Classic 30 and I'm wondering if it couldn't be done with a simple stereo jack in the amp and one in the footswitch and use a stereo (TRS) cable to connect. I've attached a diagram of what I think will work. Can anybody verify this or tell me no way, and I'm crazy? If this will work, is there anyway to wire it so that if nothing is plugged into the jack at the amp, the reverb will stay on? Here's the diagram:
(http://i865.photobucket.com/albums/ab217/twt_fx/REVERB_zps1e179e8f.jpg)
Sending the audio signals through all the wiring may cause some trouble/degradation.
You would need to isolate the sleeve from the enclosures so you don't ground the signal.
The way the switch is set up, when you throw the reverb on, the input goes to both the reverb in and the amp in. The out of the reverb then connects to the isolated sleeve which connects to the amp input again? I think your switch wiring is a bit messed up. You are going to want to send the output of reverb section to the amp in, and just the input to the amp when the reverb is off, not both the input and the reverb output at the same time. The reverb will have a level control for you to get the dry signal.
I think using the Tone God's wicked switches or GEOFEX's switching are much better suited. Then the only signal that needs to go over the long wire to the amp is just 5v or whatever that is being connected with the switch is thrown, not your entire audio signal.
(http://www.geofex.com/article_folders/cd4053/cd4053-4.gif)
Use the part on the right hand side "SPDT control"
(http://www.geofex.com/article_folders/cd4053/cd4053-2.gif)
Then use this circuit, wired up as indicated. Bear in mind that the third section of the CD4053 relay does nothing in this application so it's left open apart from C is grounded.
With this simple little set up, your wire from the switch only needs to have the control signal voltage on it, and this will keep your signal as clean as possible, minimal attenuation and completely contained in your Firefly. This is a lot less wiring than it looks - 9 components and a relay inside your amp, and 3 components and a switch in your pedal ;)
All credit to R.G. Keen of course!
just use remote switching and remote power over a regular guitar cable and a latching SPST or whatever footswitch. you can have the control line pulled up by the LED resistor. then when you engage the footswitch, the LED is pulled down to ground and your control line is also pulled down to ground. LED current limiting resistor or logic level button pull-up, it is both! what you do with that two state control logic is totally up to you. could be some simple relay driver with a transistor buffer, could be something fancy like FET switching. it all works.
Thanks guys for the replies. I gain so much from this board.
So, as I guessed, simplifying it is not going to work. I was trying to get away with not having to add another board inside the chassis since space is already limited. Looking at the RG's diagrams posted by gcme93, let me see if I have this straight. So, in my footswitch I can use an SPDT with +9V going to the pole and the throw is going to send the 9V to the CD4053 which I can do using a mono jack and regular guitar cable. I understand that the 4.7K is the limiting resistor for the LED, but what does the 10K to ground do? At the switching circuit, the +9V control signal is fed to 11 & 10 on the CD4053. I also need to supply +9V to 16 on the CD4053. Input goes to 14 on the CD4053 with a 1M pull down resistor to ground. Output (which will be the amp board in my case) is connected to 15 with a 1M pull down resistor to ground. 13 connects to the reverb in and 1 to the reverb out. 12 and 2 are connected to each other. 6, 7, & 8 are connected to each other and to ground and 9 is connected to ground. So far so good?
Now it looks like there is a 2.2uf cap between the In and the CD4053 and the Out and the CD4053. Are these just in and out caps? Also, each line has a 1M resistor to Vref. What is this for and where does Vref come from? I see the diagram for Vref. So does this mean I need +9V to Vref, +9V to 16 on CD4053 and +9V to my SPDT for Control signal? I planned on powering the reverb board with a heater from one of the tubes on the amp board. Can I power this circuit with that as well?
To further my knowledge, if I'm getting this straight, if > 6V = Effect and < 3V = Bypass, if I'm powering the Control Signal with a 9V battery, once it drops below 3V, I'll have no reverb? Also, if I don't power the footswitch at all (passive, No LED), it will stay bypassed? Also, if I unplug the footswitch from the amp, the reverb will be bypassed, correct?
The switching circuit seems pretty simple and after checking the pin-out of the CD4053 I'm guessing I should be able to do this on vero. Should I shield the switching circuit for interference like I plan on doing with the reverb?
Thanks for all the help. I'm learning tons.
> adding this reverb circuit to my Firefly amp
Which reverb circuit?
Anything with a spring-pan, just short the pan's OUT jack.
It's a digital reverb using a Belton Brick.
I'll do my best to answer this clearly one step at a time:
Quote from: blues_mang on February 11, 2013, 10:39:05 PM
Thanks guys for the replies. I gain so much from this board.
So, as I guessed, simplifying it is not going to work. I was trying to get away with not having to add another board inside the chassis since space is already limited. Looking at the RG's diagrams posted by gcme93, let me see if I have this straight. So, in my footswitch I can use an SPDT with +9V going to the pole and the throw is going to send the 9V to the CD4053 which I can do using a mono jack and regular guitar cable. I understand that the 4.7K is the limiting resistor for the LED, but what does the 10K to ground do? At the switching circuit, the +9V control signal is fed to 11 & 10 on the CD4053. I also need to supply +9V to 16 on the CD4053. Input goes to 14 on the CD4053 with a 1M pull down resistor to ground. Output (which will be the amp board in my case) is connected to 15 with a 1M pull down resistor to ground. 13 connects to the reverb in and 1 to the reverb out. 12 and 2 are connected to each other. 6, 7, & 8 are connected to each other and to ground and 9 is connected to ground. So far so good?
That all sounds right. It'll be useful to print of the diagram as you're building / designing your vero.
the 10K resistor is there to create the voltage difference between your 9V and your ground. If it was just wire (with very small resistance) you'd get a very high current and a lot of heat which could burn out the wire (or track). From power = (V^2) / R , the 10K resistor dissipates 0.0081W which is suitable for a 0.25W resistor (standard) to not burn out.
QuoteNow it looks like there is a 2.2uf cap between the In and the CD4053 and the Out and the CD4053. Are these just in and out caps? Also, each line has a 1M resistor to Vref. What is this for and where does Vref come from? I see the diagram for Vref. So does this mean I need +9V to Vref, +9V to 16 on CD4053 and +9V to my SPDT for Control signal? I planned on powering the reverb board with a heater from one of the tubes on the amp board. Can I power this circuit with that as well?
Yepp, standard filter caps that will let all your audible guitar frequencies through. You'll see that the section with all the weird Vref voltages added in (Vref = 4.5V roughly) is completely contained within the input caps and the output caps for that circuit (some of them being in your effects board). All the varying current (small AC signal) passes straight through these capacitors, but the caps block the positive voltage caused by the Vref in this part of the circuit. I'm not very familiar with the chip, but I guess it just needs to operate with this bias voltage of about 4.5V
As for powering it all, the circuit is quite nice in that it doesn't need exact or particularly stable voltage supply. It is dealing in "pretty much 9V to swing the switch one way" and "pretty much 0V to swing it back". So in this case, I think you should be okay borrowing some voltage from elsewhere.
QuoteTo further my knowledge, if I'm getting this straight, if > 6V = Effect and < 3V = Bypass, if I'm powering the Control Signal with a 9V battery, once it drops below 3V, I'll have no reverb? Also, if I don't power the footswitch at all (passive, No LED), it will stay bypassed? Also, if I unplug the footswitch from the amp, the reverb will be bypassed, correct?
Correct. I'd personally put a little optional switch on the amp which you could have as "reverb on, unless the pedal says its off" which would be simple enough to do
QuoteThe switching circuit seems pretty simple and after checking the pin-out of the CD4053 I'm guessing I should be able to do this on vero. Should I shield the switching circuit for interference like I plan on doing with the reverb?
As I said before, the switch circuit really doesn't mind about a bit of frequency interference. It will be switching with 9V or 0V, so the noise it picks up will only vary this by a couple of mV, definitely not enough to change the switch. Therefore you can even put it in a plastic box if you like! No shielding needed.
Hope this all helps, and congrats on pushing yourself with a design - going out of your immediate comfort zone is how we all learn ;)
George
George, thanks for the info. and the encouragement. I've built an amp, many pedals, and power supplies, but they have all been pretty much follow-by-number. This is my first delve into thinking outside of the box, electronically speaking. I'm doing my best to understand schematics and hash it out myself rather than be lazy and ask someone to spoon feed me the info.
Looking at the pin-out of the CD4053, I'm thinking stripboard is going to be the best way to do this. I'm still a little fuzzy on Vref though. I'm guessing I just take a wire off the 9V+ strip to a 47uf cap with the negative side to ground and bring that to its own strip. Then all the 1M resistors on each of the lines will connect to that strip. Is my assumption right?
I think you're right that a switch on the amp would be good to allow the reverb to be turned on if the footswitch is connected. I'm not positive on how to go about this. Should I use the switch to somehow bypass the relay switch circuit altogether kind of like a bypass switch in an effects pedal?
Thanks again,
Brian
Ok, here's my first stripboard layout. I think this is corresponds with the GEOFEX diagrams above, but I'm still unsure of the Vref area. That part may be incorrect. Can someone verify if my layout matches the schem.? Also, am I right in thinking that I can send the 9V+ and the control signal to and from the remote switch with stereo jacks and a TSR cable?
Final thoughts is how to wire a switch so that if the remote footswitch isn't plugged in, I can still have the reverb on. Is it a matter of putting a switch between the 9V+ source and the footswitch jack going out of the amp. One throw would connect it to the jack and the opposite side would connect to the same strip as the control signal?
(http://i865.photobucket.com/albums/ab217/twt_fx/BYPASS_SWITCH_zps61ed72c8.gif)
Anyone verify that my layout is right? At least the stripboard part. This is my first attempt at doing a layout from a schematic.
Hey Brian,
Here's what I've come up with. I believe you don't want to be skimping out on adding all those 1M resistors because they do a certain job of absorbing stray voltages around the chip that we don't want interfering with one another (someone please correct me if I'm wrong!)
This circuit should do the trick:
(http://s7.postimage.org/z5757zgaz/Remote_Effect_switching.png)
Let me know if you're confused about anything!
As I said in the personal message, you were pretty much there (you had an idea of what voltages should be on each pin and where the connections needed to go) but there's a reason R.G added the extra details in. I think you might have missed the mini Vref circuit as well.
Good luck, and definitely keep designing those stripboard layouts! It becomes easier with practice for certain
George
George, thanks again. You've been a huge help. I saw the mini Vref circuit on RG's schematic but the little rectangles that I now know are resistors weren't labeled with a value, so at the time, I wasn't sure what they were. That's why those were missing. I see now how you added more cuts to the board to make the layout more compact. I knew looking at mine and after seeing others' layouts that mine seemed to be all over the place. That looks to be something that will come with practice.
2 final questions.
1.) Instead of using a battery in the footswitch, can I feed the 9V from the amp using 2 stereo jacks and a TRS cable. Run the 9V to the footswitch on the tip and run the control signal back on the sleeve. Of course, the ground would be provided by the ring.
2.) Using the section from question 1 above, in order to make a switch so that the reverb stays on when no footswitch is plugged in, I thought of using a DPDT an run the 9V to it before running to the stereo jack out of the amp. So that in position 1, the throw will send the 9V to the footswitch and the control signal will be provided by the footswitch and in position 2, the 9V will connect to the 10K resistor to ground and send the signal directly to the amp. Sound right?
I opened up a marshall footswitch yesterday and they actually did something exactly as I described. the only thing in the box was a resistor and an LED. you could also mount the resistor in the amp so I have no idea why they didn't.
Brian,
Those two ideas sound perfectly good. I'm bored at the moment so I had a go at it on paint:
(http://s24.postimage.org/yytwbxcv9/Circuit_wiring_CD4053.png)
The stereo jack is one of those plastic ones where one side of the connections get broken when you plug the jack in.
In this layout:
DPDT switch on (yellow LED always on too):
- Reverb always on unless footswitch plugged in. Then the reverb is controlled by the footswitch (footswitch is exactly the same as before)
DPDT switch off
- Reverb always off, regardless of whether footswitch is plugged in. Footswitch led and yellow led on amp don't light at all.
As ever, let me know if you have any questions or want it to work in a different way? Hopefully you can see what voltages are where when the switch is on or off, and when the jack has a plug in or not :)
George
That's about what I was thinking, although using a switching jack will make things a little more convenient wiring wise. Now to just find one of those jacks with solder tabs. I'll use my multi-meter to trace which tabs are linked when the plug is inserted, etc. Ok, it looks like everything is in line. Time to gather the parts.
Thanks again George. I've learned a lot.
Brian
Quote from: blues_mang on February 12, 2013, 08:05:53 AM
It's a digital reverb using a Belton Brick.
I added one to two of my tube amps. It's not remotely switchable, but I love it in any case. Makes it quite simple to get a reverb on an amp that is not setup for it or has too many physical constraints to make a reverb pan fit etc.
Quote from: gcme93 on March 05, 2013, 07:06:54 AM
(http://s24.postimage.org/yytwbxcv9/Circuit_wiring_CD4053.png)
this is way more complicated than it needs to be but at least you tried. you do not need a stereo jack or stereo cable. you do not need a switched jack with normals. you do not need DPDT. you only need a SPST footswitch and a mono jack. obviously you need an LED but the LED resistor can be on the other end of the cable. the LED resistor can be part of the amp. if you still need help I would be happy to draw everything and explain it.
Quote from: EATyourGuitar on March 11, 2013, 01:11:51 AM
Quote from: gcme93 on March 05, 2013, 07:06:54 AM
(http://s24.postimage.org/yytwbxcv9/Circuit_wiring_CD4053.png)
this is way more complicated than it needs to be but at least you tried. you do not need a stereo jack or stereo cable. you do not need a switched jack with normals. you do not need DPDT. you only need a SPST footswitch and a mono jack. obviously you need an LED but the LED resistor can be on the other end of the cable. the LED resistor can be part of the amp. if you still need help I would be happy to draw everything and explain it.
I'm afraid I think you've got a bit confused. This set up is the wiring for inside the amp so that we have a switch to turn the reverb on if the pedal isnt in use. Therefore it has it's own LED on the amp to indicate whether the reverb is in use. The stereo jack set up is so that when the pedal IS plugged in, the reverb is only on if the pedal is switched on (again with its own LED but using the circuit a few posts ago or other)
The stereo jack at the bottom is meant to indicate what goes to the pedal, and it supplies the pedal with a +9V, ground and control voltage out
George
I'm sure you could refine this some more with two relays and a cap to debounce the switch. I think half of the people know what this is already. it is a comparator and a current switch in saturation mode turning on a relay. I took the marshall footswitch and moved the resistor inside the amp. you can use your own method of bypass instead of a relay. heck you could even loose the opamp but then you need to be concerned with your LED Vf being greater than your logic level high. for example the top of your LED could be 2.2v when ON and 9v when off. therefor it might not get below some logic level low that you need. cmos is about 1.7v or less to trigger logic high in most cases. not including hysteresis so add that to your tolerance. could be another 0.1v+/-
(http://i50.tinypic.com/1jnml0.gif)
(http://i47.tinypic.com/2ymei5l.gif)
and you said you want this too. the only down side to this simple system is the two LED's drawing current from one resistor might make it a little dim when both switches are closed.
(http://i45.tinypic.com/bfgqk8.gif)
Thanks EAT, that seems like quite a pretty way of doing this. I was working from the R.G Keen schem and making sure that the rest of the circuit fitted with that, but yours is definitely a lot cleaner. I have to say I'm a little confused as to where the LED switch fits in to your first schematic post, but I'm going to gracefully bow out of this thread I think - I've provided a slightly messy method but I hope it's quite easy to follow theoretically.
Hope this has given you some good ideas for how to work in your control switch Brian! Of course there's always going to be many ways of solving the problem ;)
George
the third schematic is the same as the second schematic. the only difference is that it is mounted in the amp internally so I did not draw the mono jack. they are actually all connected. the red wire is the tip in the first schematic and the black wire is the sleeve in the first schematic. if you use a plastic mono jack on the amp, you can not possibly wire it backwards cause it will still work either way. same thing for the amp panel switch. mixing the red and black wires up is no problem. the first schematic only cares is its closed or open circuit to ground. you can use a metal mono jack on the amp if you follow my schematic exactly.
Well, I know it has been over a year since I first posted this, but I finally got around to trying this. Everything is finally wired up and everything is looking like it's working like it should... but I'm not passing any signal. When I turn the up the volume, all I get is super loud static. So, I've checked all my wiring then went to a signal tester. When the amp is off, I get signal through the probe on the wire that is going from the input jack to the IN of the CD4053 board as shown above. When I turn on the power switch (just the power, not the standby) of the amp and probe the same wire, I get the same super loud static. There is a hint of the signal, but it's pretty much drowned out by the static. Now this is just one wire going from the input jack to the board, so it's not passing through any circuit yet. 2 things I'm thinking and correct me if I'm wrong: 1.) A bad ground? 2.) The wire goes from the input jack on one side of the chassis, underneath the amp circuit board to the other side where the CD4053 board is. Is it possibly picking up static from somewhere on the board like the tubes or the power wiring?
Anyone have ideas, I'd appreciate. I think I'm close to getting this to work... just need a little guidance if you will. Thanks.
Ok, I took the input wire off the CD4053 board and I get signal with the audio probe. If just touch the end of the input wire to hole on the stripboard (where it's supposed to be soldered) I get the loud hum and the signal is gone. So I think I've eliminated that the power wiring is contributing to it. Next, I guess I'll eliminate the CD4053 board and put the input and output wires directly to the reverb board and see what that does. That would at least narrow it down to the CD4053 board. Anyone have any ideas in the meantime?