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 Forum index » DIY Hardware and Software » Lunettas - circuits inspired by Stanley Lunetta
Random Chimes
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synaesthesia



Joined: May 27, 2014
Posts: 291
Location: Germany
Audio files: 85

PostPosted: Thu May 12, 2022 12:57 am    Post subject: Reply with quote  Mark this post and the followings unread

For manual triggers, you could leave away U1C, U1D, U2, U3A, U3D and the resistors/capacitors around them. You will still need a logic level at the points s1 and s2. Your triggers (positive) would go to the diodes. Look at the diagram in the lower right corner to see the sequence of triggers and the states of s1 and s2.

You can leave away the LM368 and get the output signal at the capacitor behind R12 and R13. They form a passive mixer with a rather strong signal. It might need some buffering, but not really amplification to go to a mixer. You could also send the individual signals from R12 and R13 to two mixer channels through individual capacitors.
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R_Gol



Joined: Oct 17, 2019
Posts: 33
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PostPosted: Thu May 12, 2022 3:09 am    Post subject: Reply with quote  Mark this post and the followings unread

If I omit U2 what will cause the change of pitch? Is every new trigger with the manual button will change the pitch?
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synaesthesia



Joined: May 27, 2014
Posts: 291
Location: Germany
Audio files: 85

PostPosted: Thu May 12, 2022 5:17 am    Post subject: Reply with quote  Mark this post and the followings unread

The logic levels at points s1 and s2 determine the pitch and timbre. To be more precise, those and the actual content of the shift registers in the very moment that their level changes. In the current implementation the counter U2 presents a static sequence of outputs, and their changes produce the triggers. The trigger sequence is static, but the actual sound they trigger may vary because the clock for the shift registers and the clock for the triggers are not synced.
Code:
Example:
content is 0011 and s1 changes to 1 => the shift register will cycle through 0011,0111,1111,1110,1100,1000,0000,0001,0011,...
content is 0011 and s1 changes to 0 => the shift register will cycle through 0011,0110,1100,1001,0011,...
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R_Gol



Joined: Oct 17, 2019
Posts: 33
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PostPosted: Thu May 12, 2022 12:33 pm    Post subject: Reply with quote  Mark this post and the followings unread

Did I get it right?
On each new trigger of KEY1 a new note will generate based on PITCH1
On each new trigger of KEY2 a new note will generate based on PITCH2


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synaesthesia



Joined: May 27, 2014
Posts: 291
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PostPosted: Fri May 13, 2022 1:36 am    Post subject: Reply with quote  Mark this post and the followings unread

What you say is right, but the circuit will not work because input A of U3.3 is left floating. How do you intend to use the circuit and what are your inputs?
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R_Gol



Joined: Oct 17, 2019
Posts: 33
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PostPosted: Fri May 13, 2022 6:18 am    Post subject: Reply with quote  Mark this post and the followings unread

synaesthesia wrote:
What you say is right, but the circuit will not work because input A of U3.3 is left floating. How do you intend to use the circuit and what are your inputs?

I meant to connect input A of U3.3 after the button, just before the diode anode (same as in KEY1)

the original circuit generating a new note for pitch1 and pitch2 on every new clock of the TEMPO part of hte circuit. My intended is that on each manual pressing of KEY (either 1 or 2) a new tone will be generated.
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synaesthesia



Joined: May 27, 2014
Posts: 291
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PostPosted: Fri May 13, 2022 6:52 am    Post subject: Reply with quote  Mark this post and the followings unread

You may want to try this on a breadboard. The good thing is that it requires less parts as well.

There is only one oscillator built from spare XOR gates. Because they have no Schmitt-Trigger in the inputs, we need two gates for the oscillator. The same clock now feeds both shift-registers. However, the frequencies will be different, because their feedback path is different. Even better, the two tone frequencies will always stay in a harmonic ratio. I haven't tested the circuit, but it does simulate ok.


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R_Gol



Joined: Oct 17, 2019
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PostPosted: Fri May 13, 2022 7:59 am    Post subject: Reply with quote  Mark this post and the followings unread

Thank you very much for your suggestions. Can't wait to try it on breadboard asap.
The two xor gates I marked in blue are forming an oscillator at a fixed frequency? that is determined the fundamental frequency for both the two tones? if so - is it possible to add a potentiometer to change it fundamental frequency?

Why the bottom tone generator is feeding the xor gate from Q2 while the upper tone generator is feeding the xor gate from Q3?

btw -

What software are you using for those schematics?


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Steveg



Joined: Apr 23, 2015
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PostPosted: Sat May 14, 2022 2:34 am    Post subject: Reply with quote  Mark this post and the followings unread

The two shift register taps cause differing frequencies. The one fed from Q3 is 1/8 the frequency of the oscillator. The other fed from Q2 is 1/6 of the oscillator.

Looped shift registers form a divider depending on which step is selected Q2 is the third step and Q3 is the fourth. The two XOR gates that are not part of the oscillator invert the output from the selected shift register stage diving the frequency by another two. hence 1/6 and 1/8. Don't be tempted to remove those XORs or you will have no output.
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R_Gol



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PostPosted: Sat May 14, 2022 2:45 am    Post subject: Reply with quote  Mark this post and the followings unread

What is the functionality of the two xor gates marked in blue?
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Steveg



Joined: Apr 23, 2015
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PostPosted: Sat May 14, 2022 3:20 am    Post subject: Reply with quote  Mark this post and the followings unread

They act as inverters to make an oscillator. By tying one input to +5v the XOR gates become inverters. Two inverters can form an astable multivibrator. There are plenty of explanations of the circuit on the web. Here is one: https://www.eleccircuit.com/the-experiment-use-inverter-gate-as-oscillator-circuit/
The explanation only has one resistor but two is usual.
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R_Gol



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PostPosted: Sat May 14, 2022 3:32 am    Post subject: Reply with quote  Mark this post and the followings unread

Can I replace the 22k resistor with a 20k potentiometer? For controlling the frequency of the oscillator?

will that circuit will work with 9V supply?


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Steveg



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PostPosted: Sat May 14, 2022 4:25 am    Post subject: Reply with quote  Mark this post and the followings unread

Yes and yes. However the oscillator won't work at all if the pot resistance is too low. I don't have a good idea of where the cutoff will be but I would suggest putting a 10K resister in series with the pot.
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R_Gol



Joined: Oct 17, 2019
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PostPosted: Sat May 14, 2022 4:28 am    Post subject: Reply with quote  Mark this post and the followings unread

Steveg wrote:
Yes and yes. However the oscillator won't work at all if the pot resistance is too low. I don't have a good idea of where the cutoff will be but I would suggest putting a 10K resister in series with the pot.

Thanks! will do so. Can't wait and try it tonight.
So changing the potentiometer shaft will change both tone generator fundamental frequency right?
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Steveg



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PostPosted: Sat May 14, 2022 4:34 am    Post subject: Reply with quote  Mark this post and the followings unread

yep!
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R_Gol



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PostPosted: Sat May 14, 2022 4:40 am    Post subject: Reply with quote  Mark this post and the followings unread

Thanks.
What difference is it make if using 100k resistor from the shift registers Q outs instead of 47k as in the original schematics?

What will happen when using different values for each Q out?
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Steveg



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PostPosted: Sat May 14, 2022 5:03 am    Post subject: Reply with quote  Mark this post and the followings unread

100k vs 47k shouldn't matter much although 100k will pass lower frequencies on the emitter of the transistor. I think synaesthesia tried differing values elsewhere but found no value add. All it would do is make the output wave shape less regular.
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R_Gol



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PostPosted: Sat May 14, 2022 5:37 am    Post subject: Reply with quote  Mark this post and the followings unread

How can I control the decay time of the tone generated?
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synaesthesia



Joined: May 27, 2014
Posts: 291
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PostPosted: Sat May 14, 2022 5:57 am    Post subject: Reply with quote  Mark this post and the followings unread

Increasing the 10uF cap will increase the decay time. You can try 22uF or 100uF. However, it will take longer to load it through the resistor and switch to Vdd as well. So maybe it is better to increase the two 100K resistors to ground to 220K or 470K or even 1M.
Mind that the voltage at the cap in this circuit also is sensed by the XOR gates. So after the voltage falls to about Vdd/2 the frequence and/or timbre of the generated tone changes because the feeback path changes from inverting to non-inverting.
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Steveg



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PostPosted: Sat May 14, 2022 6:00 am    Post subject: Reply with quote  Mark this post and the followings unread

Ah! I missed that.
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R_Gol



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PostPosted: Sat May 14, 2022 6:04 am    Post subject: Reply with quote  Mark this post and the followings unread

synaesthesia wrote:
Increasing the 10uF cap will increase the decay time. You can try 22uF or 100uF. However, it will take longer to load it through the resistor and switch to Vdd as well. So maybe it is better to increase the two 100K resistors to ground to 220K or 470K or even 1M.
Mind that the voltage at the cap in this circuit also is sensed by the XOR gates. So after the voltage falls to about Vdd/2 the frequence and/or timbre of the generated tone changes because the feeback path changes from inverting to non-inverting.


Can I add a potentiometer in series with the 100k resistor to ground so the decay time could be change?

Will that work?
R6 changed to 10k instead of 100k and it will keep the path to ground at minimum of 10k (?). The P2 B1M will make the decay time variable when turning the shaft(?)


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R_Gol



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PostPosted: Sat May 14, 2022 12:48 pm    Post subject: Reply with quote  Mark this post and the followings unread

synaesthesia wrote:
You may want to try this on a breadboard.

So I finally had the chance to try it. when I press on button 1 a certain pitch is generated and pressing on button 2 another different pitch is generated. so far so goof. When I pressing on those buttons once again, instead of generating a difference pitch, same pitch is generated.
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Steveg



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PostPosted: Sat May 14, 2022 7:04 pm    Post subject: Reply with quote  Mark this post and the followings unread

It seems likely ... the pitch change should happen towards the end of the note not each note having a different pitch.
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synaesthesia



Joined: May 27, 2014
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PostPosted: Sun May 15, 2022 1:52 am    Post subject: Reply with quote  Mark this post and the followings unread

Add a potentiometer in series so the decay time could be changed?
The capacitor discharges through the transistor and the resistor to ground. Rather than exchanging the resistor to ground with a pot, try to replace the resistor to ground with a 10K and the resistor to the base with a 10K plus a 100K pot in series.

Why the same tone when I press the button?
While you press the button, the capacitor charges and the XOR gate acts as an inverter. The generated frequency will be low (twice the normal cycle length). Once you release the button and the capacitor voltage goes below the logic level throeshold for a HIGH, then the XOR will not invert any more and you have the chance of either doubling the frequency or getting silence.

If you want to have a different tone at the start, you could insert an inverter between the XOR and the input to the shift register. However, you then run the risk that you get no tone at the start and a high tone after some delay.
A better approach to create more variety in the tones is to double the clock speed and make the shift registers longer with a feedback after 6 or 8 bits.

The software I used for the circuit diagram is actually a circuit simulator: http://www.falstad.com/circuit/circuitjs.html
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Steveg



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PostPosted: Mon May 16, 2022 3:06 am    Post subject: Reply with quote  Mark this post and the followings unread

synaesthesia wrote:
Add a potentiometer in series so the decay time could be changed?
The capacitor discharges through the transistor and the resistor to ground. Rather than exchanging the resistor to ground with a pot, try to replace the resistor to ground with a 10K and the resistor to the base with a 10K plus a 100K pot in series.


I was going to say something similar and then I realised R_Gol has put the pot on the emitter of the transistor. In that position it probably will slow the discharge of the capacitor.
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