Gate Sequencer

Started by JonasL, June 13, 2012, 10:46:38 AM

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JonasL

Hi,

I'm planning on building a gate sequencer from this schematic, but I don't know if it will work if I just follow the schematic. Don't I need to put resistors in front of the leds? And shouldn't I power and ground the IC? I can just use 9v as power source, right? Any help on this is much appreciated!


R.G.

What is the provenance of the schematic?
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

JonasL


Jdansti

It looks like R1 is the current limiter for the LEDs. Similar to R1 in this schematic: http://www.electronics-circuits.com/tech/2004/10/running-lights-with-cd4017/.
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R.G. Keene: EXPECT there to be errors, and defeat them...

JonasL


esnabez

Don't I need to put resistors in front of the leds?

Normally, yes, you should use current limiting resistors with LED's. In this circuit though, if you add series resistance, you'll also be limiting the gate pulse's current. The resistor in the diagram is not a current limiter for the LED's, it's part of the diode 'AND gate' circuit (see here for an example: http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/diodgate.html). The AND gate basically separates the 4017's adjoining pulses.

There's a great 4017 step sequencer and info at: http://casperelectronics.com/finished-pieces/sequencers/step-sequencer/

And shouldn't I power and ground the IC?

Definitely. Pin 16 is +VDC and Pin 8 is ground.

I can just use 9v as power source, right?

Yes, I've run 4017 sequencers from 9V batteries and they work fine.

JonasL

Thanx man, that's all I needed to know! Could you tell me what resistor value I should use when using 8 leds? (for 8 steps)

esnabez

A 1K resistor will work all the way up to 10 LED's.

Processaurus

I don't see how it could work as drawn.  There is no path for current coming from a high pin on the 4017 through an LED, and switch, to make it to the output, because there are reversed biased 1n914 diodes in the way, as well as a resistor to V+.

For it to work as desired, remove all of the 1N914 diodes, connect all of the switches directly to the circuit output, and take R1 and tie it to ground, rather than V+.  That will let current flow through the LED's to switches to R1 to ground, and the voltage of the gate sequence is present at the output.

JonasL

And how about the clock connenction to the output? And the "AND gate" esnabez mentioned?

esnabez

I realize that it looks like the gate out is blocked by the reverse-biased diode. If you take a look at how AND gates work, you'll see that it's using the output state (hi or low) of the 4017 and the output state of the clock to determine the final output state. If you remove the AND gate, you will not have separate pulses when two consecutive steps (switches) are turned on.

Having said all this, I should mention that diode AND gates are crude and very finicky, and although I initially had some success using them, I now use 4081 AND gate IC's because they're predictable and reliable.

Processaurus

No, it really won't work.  Mickey Mouse Logic still needs a path for current, and two diodes connected (even if one is an LED) cathode to cathode as shown will not pass current.  It is an open circuit.

The thing I described indeed doesn't cut the sequencer outputs to the gate pulse...  To get that one would need a real AND gate.  You could use a quad NAND gate and use two of the gates for an AND gate and two for the clock.

esnabez

Ahhh... I think I see where we're at odds... As originally shown, there will be no current flowing through the LEDs, so they won't light up. I was more concerned with the actual output pulse (which would be there) more than the LED indicators lighting up. My apologies. I've modified the schematic so that the 4017 output goes separately to the LEDs and the AND gate. I agree that the diode AND should be replaced with a real one...

Processaurus

That's looking more like the right idea, now the 4017 can pull an input of the diode AND gate low, however, the clock cannot do the same.  If the 1N914 diode between the clock input is replaced with a wire, then the clock can pull the diode AND gate low, and the gate output would be split up into pulses corresponding to when the clock is on.

JonasL

I've built the sequencer a while ago, but it didn't work... All it does is act like some strange volume knob. The more switches you turn on, the louder the signal gets. Are the leds crucial to make the sequencer run properly, beacause I left them out.

Processaurus

#15
Quote from: JonasL on August 22, 2012, 12:04:05 PM
Are the leds crucial to make the sequencer run properly
Yes.  You cannot (or rather shouldn't) connect logic outputs together directly, because they are low impedance in both hi and low states.  It's like hooking two amplifier heads up to one cabinet.  A low state from one output pin will fight a high pin in a match to the death.  The low pin will often win, depending on the chip, as CMOS chips often can sink more current into a low pin than they can source out of a high pin.

EDIT:
The LED's do two things: they indicate the step (optional), and they are diodes , which provide a one way street for current, which is the correct way to mix logic outputs together so they don't try to drive each other (not optional).

crane

Quote from: Processaurus on August 22, 2012, 02:17:47 PM
Quote from: JonasL on August 22, 2012, 12:04:05 PM
Are the leds crucial to make the sequencer run properly
Yes.  You cannot (or rather shouldn't) connect logic outputs together directly, because they are low impedance in both hi and low states.  It's like hooking two amplifier heads up to one cabinet.  A low state from one output pin will fight a high pin in a match to the death.  The low pin will often win, depending on the chip, as CMOS chips often can sink more current into a low pin than they can source out of a high pin.

EDIT:
The LED's do two things: they indicate the step (optional), and they are diodes , which provide a one way street for current, which is the correct way to mix logic outputs together so they don't try to drive each other (not optional).
+1 - withuot LEDs IC's sinking power will fight with IC's sourcing power. You can substitute LEDs with diodes but LEDs add more sexapil to such a device.