Mod for Direct Outs on a Yamaha M-508 vintage mixer?

GroupDIY Audio Forum

Help Support GroupDIY Audio Forum:

This site may earn a commission from merchant affiliate links, including eBay, Amazon, and others.

simoncastic

Member
Joined
Aug 29, 2012
Messages
16
Hey all, I recently picked up a nice old Yamaha M-508 (surprisingly nice pre's and EQ), but I want to modify the board for direct outs. There's plenty of room inside and each channel is on a seperate PCB. There's no inserts though, so where should I be looking to modify it for direct outs? Wouldn't mind the direct outs post-EQ as well, if there's a point that I can do that.

Cheers!
 
Looking at the schematics (you have downloaded the service manual, right?) the best spot is probably after the post-fader buffer amp (IC3) and before the pan pot.  This is the same spot where the post-fade FB and Echo sends are picked off.
 
Yep, downloaded the service manual. This is a first time mod kinda thing for me though - Is it simply a matter of soldering the positive and negative pin of the TRS jack to a part on the schematic between the IC3 and before the pan pot? Or do I need to put in resistors and capacitors, etc?

Thanks!
 
ground will stay ground  , easiest way would be to  " break " or interrupt the line
you could do this with a switching jack 1/4 " type  that breaks the connection when inserted
 
Well, if this were my mixer I would disconnect the wiper (center leg) from the FB control and run a wire to the tip of a 1/4" switching TS jack.  Return the other side of the switch to the spot where the wiper was connected (at the 33k resistor).  This way the FB will work as normal when nothing is plugged in.  When you plug into the jack it disconnects the FB path and the FB control becomes your direct out level control.  It looks like the FB can be jumpered for pre EQ, post EQ, or post fader so make sure it is set how you want it.

The sleeve of the jack is ground, which may be OK just attached to the chassis of the mixer or may need a wire to a ground trace on the PCB.  This is an unbalanced direct out so there is no Ring or (-) connection.
 
There are several considerations when doing this.

1) find the signal you want.
2) interface with outside world with appropriate drive capability
3) do not degrade the internal path quality.

Console designers often use different active components to interface with the outside world. Further, the signal when broken out as a DO, needs to be properly referenced to ground at that output jack (typically unbalanced).

Doing a DIY one-off has the luxury of knowing what you plan to drive with the DO. If you follow with a relatively high impedance line input, you don't need to use a high current driver. Further you can add a bunch of build out resistance to prevent loading from cable or input capacitance.

You may want to experiment with exactly where you grab signal for best result, and value of build out resistance. I would be inclined to use some R in series with low also, and send the signal 3 legged (+,-, and ground) to the DO, even though the - is just ground through a resistor.

JR
 
mjrippe said:
Well, if this were my mixer I would disconnect the wiper (center leg) from the FB control and run a wire to the tip of a 1/4" switching TS jack.  Return the other side of the switch to the spot where the wiper was connected (at the 33k resistor).  This way the FB will work as normal when nothing is plugged in.  When you plug into the jack it disconnects the FB path and the FB control becomes your direct out level control.  It looks like the FB can be jumpered for pre EQ, post EQ, or post fader so make sure it is set how you want it.

The sleeve of the jack is ground, which may be OK just attached to the chassis of the mixer or may need a wire to a ground trace on the PCB.  This is an unbalanced direct out so there is no Ring or (-) connection.

Thanks! This sounds like the way to go actually. I take it the wiper is a switching type connection on a pot?

There's no way to make this balanced? Doesn't matter too much but would be preferable.
 
The wiper is the variable leg of the pot, with the two ends being fixed.  You can make the output impedance balanced as John suggested above by using a resistor from (-) to ground.
 
Thanks for your help so far guys! I'm slowly working my way through this - going where the FB send is seems the best way so far.

Referencing this manual:

http://89.228.128.119/RemoteDownloader/files/20121124/31815a538f949b2962b6956e0d6cf759/M508YAMAHAServiceManual-EN.pdf

On page 7, that's the board I'm looking at. (Input C Board)

My current understanding is the following:
1. the pins of the pot, looking from the view of the board on the schematics is: right is ground... and I'm unsure which is the incoming signal and which is the signal coming out.

I'll want to get the signal from where it is coming out of the pot, correct?

To be honest with John - a lot of that went over my head, but I'm here to learn from experienced DIYers like yourselves! Are you saying that I would have to ground the direct O/P jack from the ground at the pot rather than anywhere else in the desk? I think I'll do that anyway.

-

I'm thinking I could solder some clips (not sure what they're called - but they're on this board so they can be plugged in and out) with 3 pins, and solder a wire coming from the clip onto each of the pins on the FB pot. this way, I could keep a balancing resistor and all my nasty soldering / connections clear from the board, and have it so there aren't too many messy wires leaving the channel cards.

I can draw up a diagram for you (well, for me) if that would help, but is this what I'd need per channel?

switching TRS (5 pins?)
male and female 3 pin clip things

Thanks!
 
Ah, think I got way too confused before... I've been tracing the PCB and see a point in which I could do the mod. The wiper / centre leg ('output' from the pot?) goes back to a 33k resistor, which then goes to the output of the card (which would go to the FB buss).

So would I solder the tip of the switched jack after the 33k resistor, so it's uninterrupted when nothing's plugged in, and a ground off the PCB to the sleeve?

And then, to make it balanced, wire a resistor between the ground and the ring (which can be done directly on the jack?)?

Is this a true balanced connection? I was always under the impression that there was a negative polarity version of the single on the ring, which would go through a balancing transformer on the other end.

Thanks for all your help (and sorry for the questions!). Looking forward to getting this underway once I get my head around it all. I'll post some (uninteresting) pictures once it's underway!
Simon

EDIT: Would I have any issues using this type of connector things? Signal degridation?
http://img380.imageshack.us/img380/9981/120mmleadml4.jpg (bad example... but I'm not sure what these connectors/leads are called)
 
You will have better results taking the signal BEFORE the 33k resistor.  You could lift the leg of the resistor closest to the pot, solder your "send" wire (to the tip of the TRS) into the hole it leaves, and your "return" wire (from the switch contact of the TRS) onto the lifted end of the 33k.  Then a resistor (say 50 ohms) from Ring to Ground will give you an impedance balanced output - not true balanced.
 
mjrippe said:
You will have better results taking the signal BEFORE the 33k resistor.  You could lift the leg of the resistor closest to the pot, solder your "send" wire (to the tip of the TRS) into the hole it leaves, and your "return" wire (from the switch contact of the TRS) onto the lifted end of the 33k.  Then a resistor (say 50 ohms) from Ring to Ground will give you an impedance balanced output - not true balanced.

Yeah, I was wondering about that, cheers. There is only one 33k resistor after the pot, and it'll be the easiest place for soldering. Just so happens that it goes straight to the large clip after the resistor.

Any issues with using those clip things I linked?

And what kind of O/P level do you think I'll get from the individual FB sends? around +4 line level kind of thing?
Cheers
 
Yes, 50 ohms can work (or the 49 or 51 ohm resistors you'll probably have an easier time finding), but this is one of the things with which John recommended experimentation, so maybe grab some 620 ohm and 1k resistors and see which provides the best results. You can install the different values in different channels and perform a double blind test if you can wrangle up some help.
 
Ego Tripper said:
Yes, 50 ohms can work (or the 49 or 51 ohm resistors you'll probably have an easier time finding), but this is one of the things with which John recommended experimentation, so maybe grab some 620 ohm and 1k resistors and see which provides the best results. You can install the different values in different channels and perform a double blind test if you can wrangle up some help.

Sure, will do, thanks. Is the impedance balancing going to affect the sound though? Or will it simply work or not work?
I've done a bit of reading about impedance balancing, I found this:

"connect the ring contact to ground through the same resistance as used for the build-out resistor on the tip contact... If you're building this into a piece of gear, then take the time to try to figure out what the value of the buildout resistor is from the signal source, then match that value from the ring side to ground. If there is a coupling capacitor, then add another of equal value in series with the resistor going to the ring contact."

Which is completely over my head. What is the build-out resistor (would this be the 33k resistor that is just before the output of the card to the FB buss?)? Would I just find out by trial-and-error?

Does the rest of my plan seem ok? What kind of level output will I be expecting?
Cheers!
 
The 33k is used to drop the level of the FB going into the summing buss.  You are taking the signal before that, so don't worry about it.  Try various resistors, but it shouldn't make a huge difference.  It is hard to say what output level you will have but I doubt it will be +4.  Possibly closer to 0 or even as low as -10.  If it is too low with the fader all the way up and the FB control maxed then you would have to add an output stage.  This can be one opamp which provides gain and also balances the signal, but it is a bit more involved.
 
mjrippe said:
The 33k is used to drop the level of the FB going into the summing buss.  You are taking the signal before that, so don't worry about it.  Try various resistors, but it shouldn't make a huge difference.  It is hard to say what output level you will have but I doubt it will be +4.  Possibly closer to 0 or even as low as -10.  If it is too low with the fader all the way up and the FB control maxed then you would have to add an output stage.  This can be one opamp which provides gain and also balances the signal, but it is a bit more involved.

Sure, thanks. I can start work on one channel soon as soon as I get the wiring clip things - what are they called? good idea? Will the gauge of the wire degrade the signal if it's too thin?

Stupid questions, I know haha. But thanks for all your help.
 
simoncastic said:
I can start work on one channel soon as soon as I get the wiring clip things - what are they called? good idea?
We generally refer to those as Molex connectors, although I'm sure there are other companies that make these. But, as mjrippe points out, utilizing this connector would put you after that 33k resistor, which would give you an output impedance higher than the source impedance you're driving... big no-no. You're gonna need to hardwire this one.

simoncastic said:
Will the gauge of the wire degrade the signal if it's too thin?
Yes, but you're gonna have to get pretty thin before it's thinner than the currently thinnest part of the circuit (i.e. the circuit traces). I usually use 24 awg for something like this.
 
mjrippe said:
If you use that clip, you are taking the signal AFTER the 33k resistor.  Better to desolder the leg of the 33k closest to the pot as I described earlier.

I'll be desoldering the leg of the resistor and taking the signal before the resistor, but I'll draw up a quick rough diagram to show you what I mean... done.

The squiggly things are supposed to be resistors (the one on the board is the one that one leg will be removed from the board so I can take the signal before the resistor - the other is the resistor to make the circuit impedance balanced.

The switching TRS jacks I purchased were the most solid but most fairly priced ones, but as you can see (hopefully haha) the switching is independent to the tip and ring or sleeve, so I'll have to solder a few more wires per jack. I also purchased a few different metal film resistors, with the values being what you mentioned I should try out.

Does this all look ok?

Looking through all the projects on here makes me realise how basic this is... impressive stuff! One day I'll do the Gyraf 1176 clone. One day...
Thanks again!!
 

Attachments

  • mod diagram.jpg
    mod diagram.jpg
    27.6 KB · Views: 104

Latest posts

Back
Top