/tilted/

OK, so the AD8113 datasheet suggests (among other things) that "The input and output signals will have minimum crosstalk if they are located between ground planes on layers above and below, and separated by ground in between" This suggests a minimum 3-layer board. Is this OK?

AD75019 layout - done. Input buffers - not done yet. Output buffers - not done yet. AD75019 is fully cascadable in two axes using dual-row 2x8 pin headers. BAM.

So the interface with the 75019 looks to be a doddle. I've as good as done the layout for that. I do need to know... This sucker would connect to J10 on the core, right? Thus leaving J8/J9 free for UI stuff? The interface to this chip is very simple hardware-wise. SCLK - Sync for upload of switchpoint data in serial form. SIN - The input pin for aforementioned switchpoint data. SOUT - Output pin (allows for cascading of chips- nice!) PCLK - Tells chip that all switchpoints are loaded into serial buffer. This all reads as very similar to the way a 595 works. So my thinking is: SCLK - SC SIN - SO SOUT - to cascade output... PCLK - RC Any problems?

Pardon the pun- Shit yeah! If you can organise a bulk dump order of those ass-wash buttons MTE, I'm in!!!

Jesus! This thread has caught fire! I have started a layout for the 75019. I did a layout already for the MT-8816, but I think it is not worth it.

The AD8113 does look very nice. It also seems to be one of those classics from the audio realm, of "for something a little better, insert all of your money here" Something like $30 US per chip... Also it is only in a 100pin LQFP. Are we OK with that?

that looks like the one you want. I'm not sure where the "pierce oscillator is unreliable" bit comes from, I'm interested to find this out. In fact, the 4-pin oscillator is AFAIK a crystal and pierce oscillator in a sheilded case. This explains why it is a little bigger than just the crystal. About enough to house a resistor, a couple of caps and a couple of transistors...

...so, which chip? do we have a concensus?

Sorry, didn't mean to be rude. I guess there's ramming home the point, then there's being a {expletive deleted}. Damn I miss the chat. :'( Hear Hear! A large number of op-amp designs use same pin outs. I think for a time there was a bit of "me too", "mine's the best for everything" going on in the op-amp world, and the manufacturers used same pinouts to ease prototyping headaches. I think the idea of using a seperate buffer board is a good call, that way if some crazy audiophile or guitar hero wants to use discrete OTAs, germanium transistors, NOS radio valves from a soviet submarine pre-amp, they can. Possibly, we could also modular-ise the matrix board, say putting a 16x8, or 2 16x8s on one board, with input thrus and output nodes to connect more matrixes as needed. Make the board to get around the one-set-of-inputs, one-set-of-outputs thing.

I think you're missing my point. Doing a 4x4 matrix would be close-to useless, but drawing more than a 4x4 would have made my point less clear. The point again: This was in response to: What I'm saying is that whatever the size of your matrix, you are still routing one-to-N, not N-to-one *. To do N-to-one would still require a summing mixer, or horrible gain problems will follow. This means that your output is derived from only one crosspoint, on one chip. You don't gain anything by chaining chips. Noise will increase as you add chips, because you are connecting outputs together. But not distortion. You are not running through one chip, into another. Doing this does not increase the size of your crosspoint matrix. It simply makes your path more convoluted. ;) *Where N is the set of numbers >=0... This could include 16, 256 or 512 gujillion. :P

That is fantastic! I would've loved one of these as a youngster. Very nice.

But surely to expand a crosspoint matrix, you would parallel the chips, not chain them? Ie (using 2x4x4 for clarity) [tt] Y1 Y2 Y3 Y4 | | | | X8 -+---+---+---+- | | | | X7 -+---+---+---+- | | | | X6 -+---+---+---+- | | | | X5 -+---+---+---+- | | | | Y1 Y2 Y3 Y4 | | | | X4 -+---+---+---+- | | | | X3 -+---+---+---+- | | | | X2 -+---+---+---+- | | | | X1 -+---+---+---+- | | | | [/tt] I don't see why a chip manufacturer would bother to include bus outputs as well as inputs, as these would simply carry the same signal as the input, plus a wee bit of distortion and noise. By adding chips, you are adding to your noise floor, but not neccessarily adding to distortion, as your crosspoint is only happening on one chip.

Yes, these will fit the smash encoders. The shafts are 6mm round, with a milled face for metric D knobs, such as these wheels. I'm in for 10. If that's ok.

I'm happy to do a layout.

but surely with a patchbay, the entire point is to connect inputs to outputs... you could still connect inputs to inputs, or outputs to outputs, but you'd need at least one input and one output for a patch to be scanned correctly.

As maddox says... You can't spell PIRATE without IRATE!!

You could use the scan matrix code with cables. The trick is to use the cables themselves to act as the switches, rather than placing switches on the sockets. Assuming you have designated ins and outs (which you do, as this is a patchbay...) For each patch input, assign a connector (banana sockets would work well) and a DIN pin 0 -x. For each patch output, assign a connector and a DOUT pin 0-y. Implement the scan matrix for an [x,y] matrix. Now you use the patch cables to connect ins to outs as needed If you use banana sockets, you can connect one-to-many or many-to-one simply by piggybacking. You might need to use series diodes in the cable if you wanted to combine one-to-many and many-to-one, ie: In1 -> Out1 + Out2 In2 -> Out1 + Out3 Without diodes, this would register a connection between In1 and Out3 and In2 and Out2 because of links from other cables. The cables complete the connection between ins and outs, as a switch would ordinarily.

First up, no, you should not expect that the SR will read your button press. The VIL for the 74HC family of chips is 0.8V. The VIH for these is 2.0V. The WHAT? VIL = The maximum voltage :P allowed as a 'low' (logic 0) at an input. VIH = The minimum voltage allowed as a 'high' (logic 1) at an input. This means that a switch intended to pull the input low, must present to that input a voltage of less than 0.8V. Anything higher than this and results are not guaranteed. 1.6V falls right into the undetermined place between VIL and VIH The 10k resistors are a pull-up device, intended to give the PIC / SR chips a steady high state (logic 1) when the button is not pressed. Changing these for LEDs with resistors might lead to the SR reading incorrectly. Now, you may find that results are better if you keep what you have, but connect the SR pin to the junction of button+LED, not the LED+220R. ie: [tt]Vdd---220R---LED---+---BUTTON---VssGND | To DIN SR pin[/tt] Try that, take some measurements... If it doesn't look good, how about this? [tt] Vdd-------10k-------+ | Vdd----220R----LED--+----button/SR pin-----Vss(ground) [/tt] P.S. ** Edit: the 1.6V you are reading is the voltage drop across your LED ** (Don't forget to answer the question...) :) Edit:typo

Oscillators have 4 pins. Crystals have 2 pins. The clock divider (which I'm assuming is what the flip-flop is for) needs an oscillator input. You can get a crystal to run as an oscillator, but you'll also need two ceramic caps, a resistor and an inverter. Check these links. http://en.wikipedia.org/wiki/Pierce_oscillator http://www.midibox.org/dokuwiki/crystals_and_oscillators

What would help the most, would be a model number for the transformer. From this, we can determine the voltage :P taps, from them, knowing the 560R parallel value, we can determine the current limiting/low pass filtering R(x) value. Once we have done this, we (I) will suggest that you put in a higher power resistor, or do all your shorting out after a 78xx series regulator, as they have thermal cutout.... ::)

Paying Nebula truly is a rewarding experience. Mmmmmmmm 401s..

Seconded. I once sunburned the sh*t out of myself in about a 5 hour dive session at a Reef on Rottnest Island. Worth every disfiguring moment. I once broke the muffler of my old Laser hatchback, it sounded like a Cessna...

Happy birthday TwinX!

The resistor in question forms part of a low pass filtger, in conjunction with the two electrolytic capacitors below it. Most likely what has happened here is that when you shorted the supply, you exceeded the power rating (wattage :P) for the resistor. As this resistor is in series between the rectifier and the output, it took a whack. Do like the rabbit says. Feed us more info.

That's hot. I thought my 25cm long MTC was massive. Now you should build a 5 meter long 3 x MBLC, with 500mm motor faders, and industrial E-stop buttons for Mute/Solo/Record. Very handy when the Incredible Hulk is mixing...