Trigger CR Question Nr 2

[zipzap]

Hi Forum
The shematic down there shows the logic for the trigger rows of my sequencer. I´ve been trying around the A-Version (for about a year, this was my first real project), it was never running really good.
Now i´m going for the B-Option, wich is really unshakable. Independent from clockspeed, duty cycle, delays.
I found out that i don´t need the Buffers, which makes things even easier.
Question is: Do i need the resistors going to ground between the cab and the diode? It´s working well without, but i don´t know if i maybe risc something like creating negative voltages that cannot be shortened or something else that i can´t even imagine.
Anybody knows about this?

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[blue hell]

Leaving out the resistors in principle makes the input voltage for the gates undefined, which could cause occasional trouble. Also the impedance on the port inputs is very high, and it is lowered by putting the resistors in, this means it will happen less often that your refrigerator is making the sync pulses

The wrong voltages you mention will happen with or without the resistors, but when the capacitors have small values (lets say some tens of nF's) the built in protection diodes on the gates' inputs will deal with that. For large capacitors you'd want to have external protection diodes and/or current limiting resistors.

The thing is that when you blow the internal protection diodes you'll not notice much, the part will become slower and it will draw more current on its input but it will probably just continue working (as the gates are not exactly stressed to their speed limits in this application).

[zipzap]

Hello to Hell
Of course with gates i need pulldown resistors. I think there´s some misunderstanding. It´s the Gates i want to get rid of. So the thing in the shematic with the mark 40?? should be left away.
The cab is going direcly through the diode to the Gatebus. That goes to an pulldown resistor and a 4538 monostable multivib.
i just don´t know if i need a resistor between the cab and the diode for some reason.
The cab i juse right now is 22nf. [/quote]

[blue hell]

Ok, yes I misunderstood then.

Without the pulldowns on the anode side of the diode it's not quite clear how current should flow on the negative edge of the input pulse, it would have to be the diode leakage which is small and not very accurate. On the next upgoing edge the capacitor will not be full discharged and so some extra current will flow into the bus line then and (so) the voltage will be higher. Doesn't seem to be a problem to me except that the pulses will have a less (probably mariginal) wel defined width.

I was thinking though, when you leave the gates in but make them open collector gates or TTL compatible and let them be inverting ones and then use an inverting amp after the bus you'd be able to leave the diodes out - that would save some work and some cost. The bus will be pulled low then instead of driven high, so you'll need a pull up on the bus instead of a pull down.

[zipzap]

Thanks. So i can save a few resistors. The diodes are already soldered to the swiches so changing anything at that point would really lead into work!
I´m happy that this whole idea came to me. I think it´s a lot better than the and-ing of gate and clock that you see in many schematics.
If you keep the gates it would even be possible to have a gatelength pot for every step (i´m not gonna do that, besides there are other ways as well for vc-gatelength)

[zipzap]

Funny how these things go. yesterday i had this on the exp.board and it was running perfect. Today i do some soldering and nothing works.
The thing that was different yesterday was that i had the cab still at the input of the gate. i just sticked the wire going back to the diode to the same row (wire coming from the cab+input of 4584).
Without the gateinput i do seem to need the resistor we talked about!
Glad i solved this, was kind of annoying. But i learned once more how easily one can get tricked by those little electrons.
Ok, resistor needs to be there. On the scan you can see a complete step of the sequencer. All the steps are summed behind the diodes for cv and gate 1,2.
2 more questions i have, maybe you know:
1. It seems to work if i connect all the steps at the place marked 1. Then send a resistor from there to ground. If my logic is the same as cmos-logic that means that i could omit the diodes going to the gatebus, wire all the swiches together, send them to the 4538 (Gateout) with a pulldown r and be done.
i don´t know, but wouldn´t the cabs have some impact on the cv pots of the other steps then?
2. The Place marked (2) There should be a resistor there going to ground (at least in the shematic i got this thing from). Has it got to do with noise? the cv is kind of shaky sometimes (if i move my hand near the sequencer).
I guess a small cab to ground can cure that. But what´s the resistor for?
The problem is that i didn´t use just one potsize. I thought they are voltagedividers so it doesn´t matter and i use what i have. so they go from 10k to 500ka. Now if i put some 100k at (2) they all start to act different.
learning, learning

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[zipzap]

Ok, the first question is nonsense. Lets forget about it quickly! One swich would turn on the trigger for every step... of course.
I was only experimenting with two steps. This can happen if you try to go to fast and don´t look at the whole thing.
Remains only question 2.
cheers