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I'm building this:

enter image description here

Albeit with a single pulse adding a bit to the shift register when the circuit is powered and another 555 timer (I'll probably use a 556 for both) to provide the clock for the shift register.

I want to avoid using the relays I have integrated in favor of this:

enter image description here

Which avoids the relays (or analog switches or w/e), but which I also do not understand. That being said, my question is,

How do I integrate the style of the latter circuit in making a shift-register based 555 sequencer without relays/analog switches?

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  • \$\begingroup\$ I can't see the top schematic very well but I think I can make it out enough to try to make a suggestion. If I understand the question correctly, you want to replace the physical switches with ones that can be controlled electronically (but not electromechanical a la relays). Why wouldn't you be able to replace the switches with transistors? \$\endgroup\$ – Michael Choi Dec 13 '13 at 3:01
  • \$\begingroup\$ You can click on the images to blow them up. They're pretty high-res. I want to replace the relays in my diagram with... well, either something cheaper, or I want to eliminate them. \$\endgroup\$ – Fruity_Grebbles Dec 13 '13 at 3:46
  • \$\begingroup\$ Sorry, right click and "Open image in new tab." That'll give you high-res. \$\endgroup\$ – Fruity_Grebbles Dec 13 '13 at 4:07
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    \$\begingroup\$ I understand the relay part, but why do you want to rule out analog switches? 4066's cost 10 cents or less. \$\endgroup\$ – tcrosley Dec 13 '13 at 5:42
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Assuming you want your frequency generation to be "analog", I would suggest that circuit performance will generally be much more predictable if you use the "selector" to produce a voltage (which would not be required to source or sink significant current) rather than having it be an active part of an oscillating circuit. For example, you could use an op-amp as a non-inverting buffer and have that either vary the control-voltage pin on the 555 oscillator of modulate a constant-current source.

Alternatively, there are a variety of ways you could generate your frequencies digitally. If you use divide-down circuits, the notes of an ascending diatonic scale have approximate frequencies (60 53 50 48 45 36 32). Use a presettable counter to divide a fixed-frequency clock by one of those values, feed that into a counter, and select one of the taps of that counter to select an octave.

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  • \$\begingroup\$ for an analog signal, would this work as the op-amp? I don't really know a lot about them. ebay.com/itm/… \$\endgroup\$ – Fruity_Grebbles Dec 14 '13 at 19:45
  • \$\begingroup\$ For an analog signal, an op amp whose inverting input is connected to its output and whose non-inverting input is connected to a voltage source, will source or sink as much current to/from its output as is required to make its voltage match the input voltage, within certain limits [in your case, the most important limits are the amunt of current an op amp can source or sink; this will often falls off as an op amp's output nears the supply rails]. \$\endgroup\$ – supercat Dec 14 '13 at 21:08
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As I see it, you want to select one of (how many? 8?) adjustable resistors into a timing circuit to control pitch. They are shown as 100R in one schematic and 100K in the other; IMO 100R is far too low (given that the other timing resistor is 10K.)

Now diode switching is possible but it's tricky to get it working, especially with large amplitude LF signals. Essentially you need enough excess voltage to keep all the "off" channels firmly off - their diodes reverse biassed, - and the "on" channel should approximate a constant current through the diode whatever the signal voltage.

That's because you don't want the switching current to affect the voltage you are transmitting. So the switching current must be (a) greater than the maximum possible negative signal current (or the diode turns off) and (b) be independent of the signal voltage (or you add some distortion to the signal).

Which either means additional transistors acting as constant current circuits, or relatively high switching voltages and high value series resistors. It's going to be difficult to achieve for this application.

An alternative is an 8:1 analog switch - like the CD4051 or faster lower voltage 74HCT4051 connecting 8 inputs to one output. It's controlled by a 3 bit binary counter (counting from 0 to 7). Internally the switches are like the 4066 analog switches, but in a more convenient package. If you need more than 8 channels, an "Enable" input turns the whole chip off or on so you can connect several to your input (the 555 "Discharge" pin)

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  • \$\begingroup\$ I may go with this, given that I understand it better, but it's not quite what I was asking for. Thanks nonetheless. \$\endgroup\$ – Fruity_Grebbles Dec 13 '13 at 19:46

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