Behold the MT-10,000
June 30, 2008 6:48 PM

Messin with electronics.. filter. I'd like to slightly circuit-bend my Casiotone MT-100 - creating a switch that cuts one circuit when power is applied.

Okay, so here's the deal. When I play a chord, and switch the Casio Chord mode, the chord is essentially triggered again - the notes start a anew just as I had played them right then.

The plan is pretty simple in my mind. The drums are controlled separately (fairly simple to locate the trimpots for each drum). There's also a LED which blinks each measure. I'd like to pilfer a tiny bit of this voltage in order to possibly run into a transistor that cuts the power to the Casio Chord selector. (Note: the selector is a slide switch).

Okay, so questions: Is a transistor the right thing to use for this purpose (essentially I want an electric switch activated by a current?) What would this effect be called?

Heres a spec on a similar keyboard if that helps
410V
Sound Club Article
posted by tmcw to Technology (3 answers total) 1 user marked this as a favorite
It sounds like you're probably on the right track, but it depends on a number of things (which are usually true, but sometimes not).

The Casio Chord switch probably has two pins, one of which is the "output" (typically connected by a resistor to the +V supply, and also to the microcontroller in the keyboard. You're going to use the transistor to let current flow between those two pins even when the button isn't pressed. You need to know which of these pins is ground, as the cheap general-purpose NPN transistor you're going to want to use cares about polarity.

If that's true, then this is easy. Trace the path to the blinky LED, and cut it right before it gets back to ground. Connect the positive side of what you just cut to the base (B) of the transistor. Connect the ground side of what you cut to the emitter (E) of the transistor. Connect the collector (C) of the emitter to the high side of the chord switch, and the low side of the switch to E along with that LED path you connected before.

So: When the Casio turns on the LED, current will flow through it and then into B, out E, and on to ground. This base-emitter current will cause the transistor to start conducting, essentially closing a switch between C and E, which will mimic pressing the chord switch (which does the same thing).

Some caveats:

1) If the LED is just barely working already, you're introducing another .6V diode drop into its circuit, and it may get dimmer or shut off entirely (in which case none of this works). This is pretty unlikely.

2) If the switch is somehow floating in the circuit (instead of being a standard pull-up resistor to ground setup) then this single-transistor setup might not work, or you might have to try something a little sneakier to get it to work completely.

3) If the LED is only on for a short time, it's possible that the chord switch is de-bounced enough that the microcontroller will ignore the signal entirely (de-bouncing is a normal hardware or software trick where you ignore blips in a switch's signal that are too short for a human to have produced; without it what you think is a single button press looks, to a speedy computer, to be 50 presses as the contacts chatter against each other for 1/1000th of a second).

There are probably a bunch of other possible ways this could go wrong, but why take the fun out of it? If you're feeling cautious, fire up your multimeter and confirm that all the assumptions above are correct. If not, start soldering.
posted by range at 8:14 PM on June 30, 2008


A slider switch might be a double-throw switch, meaning it disconnects the chord selector input from one terminal and connects it to another when you flick the switch. It might also be a single-throw switch, meaning it is either connected, or disconnected. You can determine this by examining the switch; it should have 3 connected terminals in the former case, and 2 in the latter.

If this is a double-throw switch, I recommend getting a solid-state SPDT (single-pole double-throw) relay, which essentially a prepackaged circuit consisting of several transistors that has the functionality you need. If it's a single-throw switch, you could get away with a single pMOS transistor. These conduct when the input is low (i.e. the LED is off) and do not conduct when the input is high (i.e. the LED is on). But first, you'll need to measure the voltage level generated at the node you wish to tap off, and the voltage levels used by the slider switch, using a multimeter or scope. For instance, if the voltage across the opened slider switch is 10V, and the voltage generated at the LED is 2V, your PMOS will not be fully off when the LED is on. The LED might have a biasing resistor in series with it -- if so, tap off the end of the resistor that is not connected to the LED.

Note that, for both the relay and the pMOS, the switch will only change state while the LED remains on (i.e. it won't "switch on" at one blink, and "switch off" at the next blink). Note also that a pMOS has polarity. The "source" terminal should always be connected to the higher voltage.
posted by Krrrlson at 8:25 PM on June 30, 2008


A one-two punch to my electrical knownothingness. I'll be heading to HacDC perhaps to do this up soon.
posted by tmcw at 10:03 AM on July 2, 2008


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