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logical pain - 2x4=6
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zipzap



Joined: Nov 22, 2005
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PostPosted: Wed Jan 25, 2006 7:00 am    Post subject: logical pain - 2x4=6
Subject description: ever tried to cascade 2 4017s to one 8034?
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Hi There
last night i was experimenting with two 4017s, trying to get a sequence longer than 10 steps. Later i want to use 4053 swithes to make my 2 sequencers cascadeable.
The way i did this was to use any step of the first 4017 to enable, to stop that 4017 and send a short pulse to reset the other one....... Look at the shematic...
(Sice i use a resetswitch for every step that disconnects the step from the cv-outs i don´t get any voltage from the enabled seq - just in case you were wondering).
Anyway, it works, but its acting strange. In the first cycle they run at the same time- logic. the one first hitting the step that is selected for reset halts and the other one is reset to step two_____ Shocked
In other words, if i use step 9 for reset at both 4017s, i was expecting to get a 16 step seq.
Insted i get 14 steps, the first ones of each ic being scipped.
I tried 100p to 100n caps with 10 to 100k for rcs. just wont work the way i want it to.
If there are any logical people around, please help me help myself,
cheers


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Scott Stites
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PostPosted: Thu Jan 26, 2006 11:03 am    Post subject: Reply with quote  Mark this post and the followings unread

Dunno if this is the case, never actually tried it with a CD4017 (step counts higher than 10 are sooo much easier when a different device is used) but...

If you look at the block diagram for the CD4017, the clock input is NANDed with the clock enable bit. When the clock enable line is high, the output of the NAND is prevented from clocking the innards of the CD4017. When the clock enable is low, the output of the NAND is allowed to clock the CD4017.

If the clock should be high at the time that the enable drops to low, it looks to me like the NAND is made true, and that would present a clock pulse to the inside of the 4017, letting it clock forward.

On your schematic, this is what the timing is - the clock goes high, this sends a signal from the opposite 4017 causing the CD4017 in question to reset to the first bit. Very, very shortly after, the reset makes the bit tied to the clock enable drop low - if the clock is still high, then internally the transition of the enable bit from high to low allows the 4017 to clock forward from the first bit to the second. This happens so fast, the first bit is never observed by our paultry human eyes/ears.

4017 afficiondos - does this sound plausible or am I all wet?

Cheers,
Scott
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zipzap



Joined: Nov 22, 2005
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PostPosted: Thu Jan 26, 2006 2:50 pm    Post subject: Reply with quote  Mark this post and the followings unread

That sounds kinda- logic. I hat to read through what you´re saying three times, but it seems logic. So what is the solution?
Could it be that i just have to make the resetpulse longer than my clockpulse? (or shorten my clock?)
How long does any cmos trigger have to be anyhow?
Well, having written this i just went to my setup and tried different rc combinations. With higher cabs and resistors one 4017 started at step one, but the were no longer cascaded but running independant (somehow).
Most important: It works (at the moment) After trying for an hour i got them running now with the old values. It seems like one of the 4017s didn´t like to drive both the rc-network and its enable. I put a buffer between the ic and the cab and now its running...
just strange that doing this with one ic-rc combination solved things for the whole circuit. And still its strange why the first steps were left out when they arent any more. As i say, i´m using the same rcs.
Mystery...
Sometimes i wish i could strech the time and whatch the electrons hitting oneanother, just to see whats going on.
About the other ways of making longer sequences: In this case i have these two indepaendant 2-10 step sequencers in one box. Good for polyrhythms and stuff. As An option i want this cascade mode.
I´ve read about using a cd40193+4067 (eg. music from outer space)
What i like about their project are the different playmodes like pingpong.
I want to check it out when i´m finished with this project.
I already got the perfect housing for it: An old Graphic Eq with 24 Faders (20k lin) with a LED in each faderknob. Pretty.
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Uncle Krunkus
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PostPosted: Thu Jan 26, 2006 3:44 pm    Post subject: Reply with quote  Mark this post and the followings unread

I recently re-discovered this circuit which works perfectly.
I'm doing a stripboard layout which includes the 4066s and buffer transistors. Let me know if you want a copy posted.


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zipzap



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PostPosted: Fri Jan 27, 2006 1:58 am    Post subject: Reply with quote  Mark this post and the followings unread

Cool! I would like a copy to see how you solved it. This looks like the cascade circuit in the 4017 datasheet. The problem i see is that ic2 is reset by step 1 of ic1. Thats why it has do be not connected, otherwise you would get a high step 10 in the sequence during steps 1-9.
As I am using those two ics for independant sequencers also i want to be able to use ic2-step1. So in cascade mode aditional swiching has to be done. Anyway, my circuit works now (don´t know why), giving me 2x 10 steps or 1x18 steps by the swith of 2/3 of a 5053.
I will use a buffer transistor for each step also as it has to drive a pot, 2 rows of trigger and the resetcircuit. I guess i can then ommit the buffer i put in yesterday. One 4066 is soposed to join cv and trigger 1/2 of the two sequencers befor the outputcircuit where they can be processed with glide, offset and gatelenghth.
About CMos swiching: A closed 4066 is said to present an resistance of 90 ohms. Do you think that can be ignored? Say seq1 goes to outputboard 1, seq2 is joined by the 4066. I would have to use an summing opamp i guess. Should i then just put a 90 ohm resistor to the output of seq1 so they both give the same signal level ?
Have a nice day!
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Uncle Krunkus
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PostPosted: Fri Jan 27, 2006 5:08 am    Post subject: Reply with quote  Mark this post and the followings unread

I'm not exactly sure, but I think if the input impedance of the circuits you're running is high enough it won't matter very much. I only just did my self taught course on loading circuits, and it still hasn't completely sunk in yet!! Laughing
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zipzap



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PostPosted: Fri Jan 27, 2006 11:20 am    Post subject: Reply with quote  Mark this post and the followings unread

Input impedance in this case is the resistor i use for summing, right? So if the two cvs are joined at the inverting input via say 100k resistors than 90ohm don´t make a big difference. is that what you´re saying?
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Uncle Krunkus
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PostPosted: Fri Jan 27, 2006 2:48 pm    Post subject: Reply with quote  Mark this post and the followings unread

Yep.
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