lylehaze

I regret that I don't have time to get deeper into this.. I'll be going offline in about an hour, and will probably not see the net again for most of a week. SO if it seems like I'm ignoring you, I am. :) Pins 1 and 3 need to be analog inputs OR +5 volt supplies. To make this work in MIOS8, you HAVE to use analog inputs for these. You can also attach PNP transistors that will turn these into +5 volt suppliers when either transistor is turned on.. The transistors will have the emitter connected to +5 volts, the collector connected to the AIN pin in question, and the Base connected, through a resistor to a spare DOUT... when the DOUT is HIGH, the input should act like a analog input, when the DOUT is LOW, the analog input will read 5 volts, and it can supply power to your XY pad. For pins 2 and 4, which must be ground or (no connection), you can use NPN transistors,in the same way. The emitter will be ground, the collector will be your Pin1 or Pin4, and the base will connect to a spare DOUT through a resistor. When the DOUT is HIGH, pin 2 or 4 will be GROUNDED. when the dout is LOW, there will be no connection to pin 2 or 4. Yes, yes, this is all very messy. Working with individual transistors, scary stuff, all that. Radio Shack sells bags of 10 transistors for about 2 bucks each. You'll need a bag of PNP and a bag of NPN.. the resistors I mentioned can be.. Oh hell, I don't know. 1K will work, 10 K probably will too. Build it, test it, have fun, that's what counts. What's nice about this method is that you do NOT have to re-configure MIOS. You're doing the switching in hardware outside the core. You only need control of a few DOUTS to switch the transistors, and of course expect to see 5 volts on the analog inputs when they are being used as power supplies (as expected) I'll be offline for a few days, but there are others here that can help with this if you ask nicely. :-) Radio Shack: PNP transistors, bag of 15 276-1604 $2.59 NPN, bag of 15 276-1617 $2.59 Have Fun, LyleHaze

Just Guessing.. What is the PIC device ID? This was burned in with the bootloader, and is used to identify some important options in your core.

To spot the expert, pick the one who predicts the job will take the longest and cost the most. -- Warren's Rule

I'm going to try a slightly different approach. It is common to draw a full-wave bridge as a square, and everyone seems to have a preference as to how it should be rotated. Fair enough, whatever helps you to see how it works. To make my point, I'm going to show you a three phase bridge.. one that has three AC legs coming in, instead of the two that you have already seen. As you might guess, it's no longer a perfect square, but if you look closely, it is THE SAME CIRCUIT. Now, once you get adjusted to seeing it this way, you'll realize that ANY TWO of R,S, and T make the exact same bridge you've already seen.. So if EACH incoming AC leg could connect to R OR S OR T, what is the difference between them? No difference at all. I hope seeing the "different" drawing of the same circuit helps. Have Fun, LyleHaze

If you'd like, I can explain further why the AC terminals of a bridge are not polarized. HOWEVER I suspect that at this point such details will serve to confuse you further instead of help to clarify things. SO I will stand by my original assessment that they may be swapped without concern, and I'll offer a friendly invite to discuss it further only if you ask for it. :-) Have Fun!, LyleHaze

The + and - matter, the AC leads do not. Good Luck, LyleHaze

No need to be that complex. It's still very likely that they will work if they are just connected to the DIN inputs and ground. If op-amps are needed, most of the parts from the last drawing can be eliminated. Something like this: Aaargh! No picture upload.. We can do 4 inputs per chip, each input also needs ONE resistor, and a single pot to adjust sensitivity for the whole set. So worst case, 4 14 pin chips, 16 resistors and a pot and you'll be connected. SIP resistors would reduce the footprint further. But really, try connecting straight to the DIN! According to what you have posted so far, that should work fine. Have Fun, LyleHaze

http://hackaday.com/2009/08/22/maple-beats-up-arduino-takes-its-shields/

I thought the MidiBox 64E was for encoders instead of pots. I might be wrong.

I see.. There are many examples of using a PIC to receive remote control signals, but I have not seen any here with the MidiBox. Most of what we do here is controlled by Midi. One of my "open projects" is a IR remote receiver that sends MIDI messages, but it's not yet ready. There is a discussion of a MIDI controlled audio switcher, but I don't think it's done yet. http://www.midibox.org/forum/index.php/topic,11262.0.html Except the DOCMatrix, which does appear complete... there's a picture of it in there somewhere. Well, that's a starting point, anyway. Have Fun, LyleHaze

If you need to adapt them, I would be happy to provide a circuit. Have Fun, LyleHaze

MIOS by itself is an operating system, not an application. There are many different applications, depending on what you want your core to do. Like Midibox 64, Midibox 64E, Midio 128,Midibox Seq, Midibox Sid, Midibox FM, Midibox LC, Midibox Mon, Midibox CV, MidiMerger, Midibox Mixer. So, what do you want to do with your Midibox? If the LCD says "Ready", then it's time to send it an application. Try this page: http://www.ucapps.de/mios_bootstrap_newbies.html Available in English, French, and Spanish, for your enjoyment. Have Fun, LyleHaze

OK, I'll take a stab at this, but it comes with a "disclaimer": Digital (TTL) circuits expect all signals to be "high" or "low", with as little time in between as possible. Digital(CMOS) circuits are more tolerant of analog voltages, but I don't know how well they will like being permanently in analog mode. at a minimum, I would expect higher current draw, and possibly erratic behavior near the boundaries of high and low.. Now, what all that said, I'll proceed. It looks like the DIN modules use 74HC165 chips. these are CMOS, so far, so good. Each input is already "pulled up" to +5 volts by a 10K resistor. When nothing else is acting on it, you'll "see" 5 volts at the input, meaning "button not pushed". The datasheet says the guaranteed maximum voltage for a "low" input is about 1/3 VCC. If we already have 10K pullup, then we need a resistance LOWER than 5K pulling to GROUND to get the input that low. If your sensors reliably go below 5K when pressed, and above 20K when released, then hook them up just like a push button and see what happens! Worst case is that the 74HC165 chips eventually overheat, and can be replaced for under a buck each. I can get deeper into the math of voltage dividers if you want.. but only if I am asked to. Have Fun, LyleHaze

I'm having a bit of trouble understanding the problem. You say that Pin 25 RX has no outgoing data.. Pin 25 is TX, pin 26 is RX.. You show in a pdf that you uploaded MIOS in smart mode. For this to happen, both TX and RX MUST be working, or the upload would never have completed. SO the problem is not TX or RX... or you never would have made it this far. You have demonstrated so far that you have working MIDI IN, MIDI OUT, and LCD display. Sounds like you are doing quite well. Next is to upload your application. Good Luck, LyleHaze

Depends on the relay, and which chip is driving it. Most PIC outputs can sink or source about 20 to 25 ma. That is not much current to work with. The 74HC595 chips used in the DOUT modules must stay below 35ma per pin, 70ma per chip. Other driver chips may be available, I'll let you look them up. If it's an electromagnetic relay, you also have to protect the output from the voltage spike that you'll get when the output turns OFF. Many relays have an optocoupled input (LED). These are a much better choice for driving from an output. Have Fun, LyleHaze

The schematics are available in PDF format. I don't think the Eagle sch files are available online, but you could probably create your own if you wanted to. Have Fun, LyleHaze

Also, the information given will give each module 5 inputs and 5 outputs. If you need 10 inputs separate from 10 outputs, you'll need 2 modules on each side. Have Fun, LyleHaze

"Last week, I lost an Electron" I said to the bartender... "Are you sure?" he asked, seeing the despair in my face.. "Sure?", I said.. " I'm POSITIVE!"

I'll have to wait until tonight to be sure, but I think that last post glowed in the dark, just a little bit. :-)

Just a guess, try increasing the delay between MIDI packets. Good Luck, LyleHaze

It sounds like you are well along. I'll add a few comments, but please don't take anything I say as discouraging.. I'm just point out some details you are probably already aware of. Sometimes a second train of thought is useful, or sometimes it just makes a great train wreck. :-) For the DINx4 side, You only need to short the input to ground. There is a pullup resistor in the DIN module. debouncing is done in software. Also, if you actually create a "negative-going" pulse the DIN may be damaged, as input voltage levels must stay between 0 and +5 volts. (you probably didn't mean that it would go below ground, but I'm covering all bases here!) Logically, LEDs that display button action are cool, but often with MIDI controls there are multiple control origins (your panel, a window on the PC, a stompbox on stage, etc..) If MIDI ties them all together, sometimes it's nice to see if some OTHER source is pressing the same button.. This would be an argument for controlled LEDs.. I have not played with that, perhaps someone else has. Sounds like a great experiment.. Being a hardware type, if I wanted to use those pots, I'd just add op-amp buffers if there was a problem, but that may not be necessary. I seem to fix everything with Op-Amps.. But that's just me. :-) Have Fun, LyleHaze

Our hearing sensitivity follows a log curve, so most pots that handle audio directly are (an approximation) of log curve. If you were to use a linear pot as an audio volume control, you would find most of the range of rotation wasted, and a small range where most of the useful settings are. On the other hand, when we make "control circuits", like the MB64, we usually want the data to be a fair description of the position of the knob, so we use linear pots, and the voltage is a straight description of the knob position. If you used a log pot with an MB64, you'll find that most of the number range comes from one end of the pot, with MUCH slower response at the other end. When I made the MBMixer, I scaled the knob data directly into dB data for the PGA chips. The result was the same as if I had used linear pots for audio material. It was necessary to add functions to scale from Linear to Log (and back again). The source in PIC Assembler is posted with the project, but it's NOT a great way to adapt the "wrong" types of pots, as resolution is lost in each conversion. If you're the visual type, you can see Excel charts that I used to create the log conversion tables. They are in the MBMIxer wiki, or http://www.midibox.org/dokuwiki/lib/exe/fetch.php?media=pga:logcurve.zip Have Fun, LyleHaze

Make sure the optocoupler is facing the right way. Hope it helps, Lyle

depends on what "trick" you are trying to accomplish.. If you're trying to make the minimum current, then an LED and a 100 ohm resistor won't be enough. Getting back to ohms law: assuming 5 volts, - 2 for the LED, with the 100 ohm resistor you specified.. you'll get about 30 ma of current.. 30 ma is more than most LEDs are rated for, so it will be bright, but will probably not last too long. Your power supply "minimum" is 0.5A, or 500ma. so you're not drawing the minimum it needs either. So the power supply will probably just shut off, saving your LED. :-) Are you trying to light an LED, or are you trying to set up a dummy load for your switching power supply? We have given you instructions to do each, but you seem to be somewhere in between. Have Fun, LyleHaze Hmm. maybe you want to use an LED circuit AND make the minimum for your switching power supply.. Assuming 20 ma per LED, and a 500 ma minimum for your power supply, you'll need at least 25 leds all lit up at once, and all connected in parallell, to make 0.5 amps

OK, my last response was regarding your text, not your math.. You said: SO that's 64 analog inputs and 41 * 2 + 43 = 125 digital inputs Yes, I think that will work.. My last post was just pointing out that you cannot get those 8 "extra" inputs from J5 if you are using any analog inputs (as you are). Sorry if my hasty response came across badly. Some days I really shouldn't be allowed into public places. Have Fun, LyleHaze