Trying to solve a stuck state pushbutton gated oscillator.

[Musica_En_Fuego]

This issue has been driving me nuts. So, I have a push button gated 4093 connected to a 4040 IC. The issue i've been having is that whenever I release the button.

The output of the 4040 randomly stays in one state or the other. This wouldn't be much of an issue if I didn't have any LED connected.

I've tried to create an inverter using one of the other gates of the 4093 so it would output a positive state to the reset pin of the 4040 which should ground all the outputs.

But, well I kinda figured that would just create a standstill state.
I just want to get this damn project finished...

[JovianPyx]

[PHOBoS]

How fast is the oscillator running ? Can you see the status of the LED before you release the button or does it run too fast for that ?
If you can't see it it makes sense it's 'random' otherwise it could indeed have something to do with switch bounce.

Also which output(s) of the 4040 are you using,. does is change or is it a fixed output ?

[Musica_En_Fuego]

Here is a schematic.

https://drive.google.com/file/d/1-8pulU4UqYqbNQoP0eBOTpOb6IFxKRHv/view?usp=drivesdk

[PHOBoS]

I think this should work:

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[Musica_En_Fuego]

You got damn genius. It finally works the way I wanted to work for the longest time. Now I can add the rest of the circuit. And I'll post the completed circuit once I'm finished with it.

[PHOBoS]

great to hear that works

btw you can probably leave out U1c. I forgot that the 4040 counts on a high to low clock transition and I added U1c
so that when you push the button you immediately get a low to high transition. If it's for high frequencies
it doesn't make much of a difference though.

[Musica_En_Fuego]

[PHOBoS]

Yes you were close I was actually surprised you didn't have the NAND for the reset connected to the button but instead to the oscillator.
I have no idea what you were trying to achieve with the capacitor connected to the first NAND gate though.

[PHOBoS]

let my dissect the circuit a bit so you know exactly why it works although you will probably already understand it.

When the button isn't pressed the capacitor is charged through the resistor so both inputs of U1a are high and as a result the ouput is low.
When you press the button one input immediately goes low which results in the output going high. Because of the capacitor (which is now discharged)
and the resistor the pin can't go immediately high if there would be any switch bounce. This gets even improved because of the schmitt trigger
inputs of the NAND gate. (If you would add a much larger capacitor the ouput will stay high for a while after you release the button, which can be useful.
You can also replace the resistor with a potentiometer to make the time adjustable, but you do need to add a resistor in series with the pot)

When the gate input of the oscillator (U1b) is low (button not pressed) the output will be high and thus the capacitor is charged. When the gate input
goes high the output immediately goes low and then (after a short delay) it starts oscillating. When the button is released the output will return back
to a high state. With U1c it's the otherway around so it is in a low state when not oscillating.

U1d inverts the output of U1a so that the 4040 is reset when the button isn't pressed. Because there are actually 2 inverters in series (U1a & U1d)
you could also connect the reset input of the 4040 directly to the button (before U1a). But U1a helps with debouncing and if you're not using the
spare NAND gates for anything else you might aswell add it. There is also a small chance the 4040 would react a bit weird because the reset input
doesn't go high immediately as a result of the resistor and capacitor.

[Musica_En_Fuego]

here is the complete schematic. Even though every looks about right. I'm getting some weird problems. it used to work just fine, but now for some reason it doesn't function properly. It may be the jumper wires/alligator clips.


Thank you for the explanation of the circuit. I didnt know that the RC circuit could affect the reset pin of the 4040.

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[Steveg]

Hi Musica En Fuego,

I have some concerns about that circuit.
The switch on the far left shorts V+ directly to earth. That won't be good for your power supply. Possibly you need a high value resistor between the top of the switch and V+.
The DPDT switch in the centre shorts the outputs of the 4040 directly to earth which will probably destroy the 4040. Just leave that terminal of the switch un-connected.

That link from Pin 2 of the 386 to earth doesn't look good. Without checking the pin-out of the 386 I can't say what effect it will have but it could block the signal.

I'm not sure what the circle of dots at the bottom represents. As far as I can tell the trimpot should connect directly to the potentiometer.

Cheers!

[PHOBoS]

yes, there are several errors in that schematic which Steveg already mentioned. Maybe you've actually connected things differently, but I can only go by how it is drawn.
1. The button/switch creates a dead short between V+ and GND.
2. The capacitor and resistor connected to the other input of the NAND gate have no use connected like that. (have another look at the schematic I posted)
3. The DPDT switch does indeed short the outputs of the 4040 to GND.
You actually don't need a DPDT switch there, you could just switch the input of the LM386 amp between the output of the NAND oscillator or the 4040.
If you do use a DPDT switch I would add a pulldown resistor to pin 10 of the 4040, although that's not really needed in this case. Also you'd normally add
a pulldown resistor to the point where all the diodes are connected together but in this case the level pot for the LM386 already takes care of that.
The LM386 is actually connected ok, although it looks a bit weird how it is drawn. The circle of dots seems to be a rotary switch but there's nothing connected
to the other contacts so I am not sure what its function is, but maybe you have some plans for that.