No, one pin (the Anode) must be taken High and the other (Cathode) pulled low, so that current flows through the diode.Do both pins go high to turn on the designated LED?
Since hammers are hand-held, their position when striking is not fixed, so (easy) detection by laser would not work.
Ahh yes the laser idea will not work, although I've got another (hopefully not covered yet). If the wires are conductive then attach some tinfoil to the hammer which is +5V then connect and pull down all the strings with a 10k resistor; the strings will also connect to one pin on the picaxe. The idea is when the hammer contacts any string it'll pull the picaxe pin high.We thought about lasers earlier, Or at least infrared sensors. How would you position them so the entire keyboard was covered?
Yes it would be much better if we did not have any hardware on the hammers.
If the strings are conductive, then an electric guitar pickup (or similar sensor) could be an option?The strings are surely conductive. . .
Piezos would probably work--like these: https://www.amazon.com/15Pcs-Trigger-Acoustic-Pickup-Guitar/dp/B07B8RJ8NXIf the strings are conductive, then an electric guitar pickup (or similar sensor) could be an option?
I don't think so. With a guitar, you only use one (I don't mean the kind of pickup which has 6 sensors). You don't need to distinguish notes, just determine that a strike has caused vibration which the piezo converts into an electrical impulse.The piezos would probably work, but would we need fifteen pins to spread them all over the instrument?
In general it will require two separate commands, though if one of the pins is already at the level required you would only have to alter the pin which isn't.Right i understand that but in this case we have 16 LEDS connected in an array going to 8 pins. To light an LED we must tell one pin to go high and another pin to go low. Must that be two separate commands?
' G3,-G#3,A3,-Bb3,B3,C4, C#4,D4,-Eb4, E4,F4, F#4,G4, G#4, A4,Bb4,B4, C5, C#5, D5,-Eb5,E5, F5,F#5, G5,-G#5,A5,-Bb5,B5, C6,-C#6,D6
'$00,$FF,$01,$FF,$02,$03,$10,$C0,$FF,$C1,$11,$12,$C2,$13,$C3,$20,$C4,$C5,$21,$C6,$FF,$C7,$22,$23,$C8,$FF,$30,$FF,$31,$32,$FF,$33
It's hard to say. As far as I can tell, all the "serial" LCDs and OLEDs on the first few pages of Amazon (which you might get quickly), or ebay are actually I2C (TWI) two-wire modules, and I don't think you have the pins available, especially the default port B pins (though you might be able to bit-bang it on other pins (which you still don't have available)).We still need to purchase an LCD screen. Which one should we get?
Alternatively, you could use an i2c i/0 expander like the MCP17008 (for 8 pins) or MCP17016 (for 16 pins). That would take away C.3 and C.4--hi2c scl and hi2c sda--but would enable you to use an i2c display without taking up additional pins. The 40X2 would make for cleaner code, though.... we'll get a 40 pin chip.
It is not clear exactly what you have done. If you have grounded the anodes then the only way current could flow through a LED to illuminate it would be if the output drive were at a negative voltage ...We grounded the Anodes instead of the cathodes! Pins B0 through B7 work the other way.
| ___ K A
X.Y |---|___|----|<|----.
| LED _|_ 0V
If you have all the anodes commoned together and going to a ground point, can you not cut the connection to ground and instead make one to +5V? Then the cathodes go individually to the appropriate C0-C7,A0 pins, and are activated by taking the pin low? (This actually follows a suggestion made, I believe, by Allycat many posts ago, because a picaxe pin can safely sink more current than it can source.)(Maybe I misunderstood--do you mean on the 16-LED matrix?)
Nope. As you suspected it is the 9 pins that connect to the eight diode pairs and one diode triple.
It appears the simplest thing to do will be to physically reverse the 19 diodes. (crying)
Manual 1 gives the needed download circuit--page 8 in the version I have.is there a list of needed parts to use a pic on a breadboard?