Juggler 4 (CMOS Effects Order Switching) Clarification (4013 Flipflop Too)

[Andon]

Howdy! I'm looking to dip my toes into CMOS switching, specifically effects order switching via R.G. Keen's The Juggler 4, and I'm sort of stumped - it's sort of difficult for me to visualize exactly how it works, since with the 4053 you can use several pins for either inputs or outputs (on the same pin no less). I've tried breadboarding the Juggler 4053 design, but can't seem to get anything out of it, so I thought I might ask here to see if I'm thinking this through properly. I know this has been covered here before, but it's difficult to search through and find clear answers pertaining to pinout, etc. Here's the original image for reference (no direct posting, so they will be links):

http://www.geofex.com/Article_Folders/juggler/juggler4.gif

And here's how I interpreted it:



Pin 1: Main Output, tied to Pin 5
Pin 2: "N/C" (part of internal switch)
Pin 3: FX A Input
Pin 4: FX B Output, tied to Pin 9?
Pin 5: Output tied to Pin 1
Pin 6: Ground
Pin 7: Ground
Pin 8: Ground
Pin 9: Tied to LED, tied to Pin 4?
Pin 10: Tied to LED, tied to Pin 15?
Pin 11: Tied to LED, tied to Pin 14?
Pin 12: FX B Input
Pin 13: "N/C" (part of internal switch)
Pin 14: Main Input, tied to Pin 11?
Pin 15: FX A Output, tied to Pin 10?
Pin 16: 9V

The drawing seems to show pins 14/11, 15/10, and 4/9 connected, correct? Also, the original drawing shows pins 11, 10, and 9 connecting to the cathode of one of the LEDs, but not the other - is that intentional? Hoping to make sense of all of this!

Follow up sub question, when using the 4013 for controlling the 4053, pictured here:

http://www.geofex.com/Article_Folders/cd4053/cd4053-3.gif

Does the control signal just go into pin 14 of the 4053, assuming that it's also tied to pin 11 if correct? If using the LED layout from the Juggler 4, can I just forego the LED and 4.7K coming off of the righthand side 2N3904 collector (on the 4013) and just tie it to 9V? And with that, if I wanted to use the 4013 for momentary foot switching before the Juggler 4 is there a way to have the LED configuration from the Juggler 4 so that it "shows" which effect is first? Thank you for your time!

[FiveseveN]

The drawing seems to show pins 14/11, 15/10, and 4/9 connected, correct?

Not at all. This is where your confusion seems to stem from: what's inside the rectangle is a schematic approximation of what goes on inside the CD4053 (a bunch of switches and control signals). Pins 9, 10 and 11 are what tell the switches to flip to one position or the other, see the datasheet for more info.

Also, the original drawing shows pins 11, 10, and 9 connecting to the cathode of one of the LEDs, but not the other - is that intentional?

What happens if the control signal is connected to both LEDs? The Juggler would only work for the fraction of a second it takes for that SPDT to flip. So yes, very much intentional.

[anotherjim]

I'm not sure if there isn't a missing 10k pullup resistor to +9v from the common connection of control pins 9,10 & 11 in the Juggler example. The 10k would ensure the switch control pins get either 0v or +9v depending on the switch state. Without the pull-up, the LED voltage limits the control and it probably won't even react to it.

Does the control signal just go into pin 14 of the 4053, assuming that it's also tied to pin 11 if correct?

For the 4013, it will drive all 3 4053 control pins together and doesn't need a pull-up since it will output +9v or 0v. The 4013 is a logic "flip-flop" part and audio signals don't connect anywhere to it - it's replicating the Juggler footswitch but with a momentary SPST footswitch input instead of the Juggler SPDT latching footswitch.
If you want A or B indicator LEDs with the 4013 control, you would add another transistor identical to the one shown but with the input 10k connecting to pin 2 of the 4013 instead of pin 1. Pin 1 toggles between +9v or 0v every time you tap the footswitch. Whichever pin 1 is outputting, pin 2 does the opposite.

[PRR]

> I'm not sure if there isn't a missing 10k pullup resistor to +9v from the common connection of control pins 9,10 & 11

Isn't that the (upper) LED and the rectangle resistor? At least with leaky old red LEDs it would be sure to come above 7V, way above 2/3 of 9V supply.

[anotherjim]

When the lower LED is on, isn't it clamping the upper LED anode to its Vf?

[PRR]

You are right. It seems unreliable, subject to damp PCBs etc.

[anotherjim]

Maybe the lower LED was an add-on. Without it, yes, I'd expect the leakage would pull-up well enough. The 4013 driven version doesn't show a LED for the opposite switch.
It occurs to me, that if you use high-efficiency LEDs, the 4013 is perfectly capable of driving them without a transistor in between. The series resistor would be at least 6.8k.

[Andon]

Not at all. This is where your confusion seems to stem from: what's inside the rectangle is a schematic approximation of what goes on inside the CD4053 (a bunch of switches and control signals). Pins 9, 10 and 11 are what tell the switches to flip to one position or the other, see the datasheet for more info.

Thank you for this! I had been looking at the datasheet, but - again, given that some pins can be used as both inputs or outputs - I was still perplexed. So, knowing that, does the main output still connect to both pins 1 and 5 together? Also, are the FX A and B inputs/outputs correct on my visualization?

For the 4013, it will drive all 3 4053 control pins together and doesn't need a pull-up since it will output +9v or 0v. The 4013 is a logic "flip-flop" part and audio signals don't connect anywhere to it - it's replicating the Juggler footswitch but with a momentary SPST footswitch input instead of the Juggler SPDT latching footswitch.
If you want A or B indicator LEDs with the 4013 control, you would add another transistor identical to the one shown but with the input 10k connecting to pin 2 of the 4013 instead of pin 1. Pin 1 toggles between +9v or 0v every time you tap the footswitch. Whichever pin 1 is outputting, pin 2 does the opposite.

Thank you! So by control pins are we referring to the three inputs, pins 3, 12, and 14 - OR - pins 9, 10, 11?

Per your suggestion of using -Q on the 4013 to drive another LED, does this look correct?



Does -Q still need to connect to D (pin 5 on the 4013), and if not how should D/pin 5 be connected?

Follow up question: if one of the effects were a time based modulation such as a chorus that uses a rate LED to indicate the speed of the LFO, would it be possible to tie the anode of that LED (let's say D1 in the above example) to the LFO as per usual, or would another configuration be required?

Thank y'all again for helping me better understand this!

[anotherjim]

9, 10, 11 are the 4053 control pins. They are electronic switch toggle levers.

On the 4013, don't remove the wire pin2 to pin5. That is what makes it switch over every time the footswitch operates. 
When the power comes on, the Q output will be logical 0 (0v). The "bar Q" output (we say "Not Q) will be not Q which must be logical 1 (9v) and since it's connected to it, the "D" input (D for Data) is also logical 1.
When the footswitch is pressed, the "C" input (C for Clock) logic level changes from 0 to 1 and causes the Q output to whatever logic level is on the D input and it will stay "latched" at that level for now. This was a 1 remember? Now the Q output (yes, we do say Q) becomes 1 and the "Not Q" must now be 0 and D is also 0. When the footswitch is released - nothing happens, the 4013 ignores Clock going from 1 to 0 but will be ready for the next footswitch operation. I'll let you figure that out bearing in mind that the D input is now a 0.
If that doesn't explain, search for a tutorial about "D-type flip-flops".

You only need to add the extra transistor as you have drawn. Change nothing else.

Show us the modulation indicator schematic. It will be easier to add its LED cathode end to the relevant transistor for the 4013 indications.

[Andon]

Gotcha', makes sense on why the connection should remain! Here's what I had in mind:



Basically replacing the D4 LED (and the 4.7K R15 attached to it) on the LFO with the D1 or D2 LED on the 4013 switch. My idea was to use a bi-color LED (say, red/blue) with a common anode so that when the modulated effect (chorus, for example) were first it would be blue and flashing in time with the LFO, and when the other effect (fuzz, etc.) were first it would be red. If it happened to flash in time with the LFO regardless of which effect were first that would also be fine.

I've also included the CD4053 connections to make sure I'm picturing all of this correctly.

[anotherjim]

You can replace an A or B channel LED with the modulation LED. Doesn't matter what order the LED and its resistor are. You can have both channel and modulation LEDs turned on by the same transistor if you want, so long as they do have their own resistors.

Note that for the same LED and resistor values, the Modulation LED will not seem as bright since it doesn't get a constant voltage supply and that only momentarily reaches anywhere near 9v. I think it would be better to dim the channel indicators with larger resistors if you want any kind of match. Any of the Hi-Brite or High efficiency LEDs you can get will give plenty of light with larger resistor values giving more room to play with.

[Andon]

Had some more time to work on this over the last couple of days and here's where I'm at with added CMOS bypass switching (color coded and non-color coded for ease of reading):





Full size image heres: https://i.imgur.com/pGRM9IO.png / https://i.imgur.com/KTqHSrt.png

I think the bypass connections are right, but does this seem correct if I'm using a dual color LED with a common anode, or am I going about this the wrong way?

EDIT: Follow up question: could I use a single 4013 to control both switching schemes if I utilize Q2 on the other half of a single 4013, or is it a better practice to keep them separate?

[anotherjim]

Ok, so...
The Q and /Q outputs of the unused 4013 section should not be connected to anything. We mark them NC for "not connected" or ignore them. We do connect to ground the other unused pins because they are inputs and cannot be left disconnected.

I see nothing to stop you from using both sections of one 4013 for both jobs.

The dual LED's work better if they each have their own series resistor. So a common anode dual has the anode connected direct to the +supply and the cathodes go thru their own resistors on the way to the transistor.

[Andon]

Gotcha', so I'll just nix the second 4013 and use the first for both switches. Thank you!

Regarding the dual color LED, I'll make the necessary changes and add resistors to each cathode, but is it still okay to drive the anode from the output of the LFO, which is putting out/biased to 4.5V?

[anotherjim]

Gotcha', so I'll just nix the second 4013 and use the first for both switches. Thank you!

Regarding the dual color LED, I'll make the necessary changes and add resistors to each cathode, but is it still okay to drive the anode from the output of the LFO, which is putting out/biased to 4.5V?

Ah, I didn't see that. As long as you want either LED flashing from the LFO, it will be ok. But I thought you wanted either the fx with the LFO channel flashing a LED or the other channel without an LFO with static LED? I don't think that will be simple.

[Andon]

Ah, I didn't see that. As long as you want either LED flashing from the LFO, it will be ok. But I thought you wanted either the fx with the LFO channel flashing a LED or the other channel without an LFO with static LED? I don't think that will be simple.

That would be "ideal", but I also have no qualms with the LED flashing in whichever order so long as it does so. Also, I made the necessary changes and added resistors to both cathodes of the dual LED.

One final question on this setup: is it possible to also configure the same LED as a bypass/on indicator, or am I looking at needing a separate LED for that?

[anotherjim]

As some of the 4013 outputs are going to be at high voltage and lighting a LED, and we cannot predict which, a master bypass indication that kills those LEDs isn't so easy. We cannot force both Q and \Q outputs to low voltage output to switch all the LEDs off in master bypass, they always have to be different.

This is the "truth table" for the 4013. You can see there is no case where both Q and \Q output 0. There is a case that forces them both to 1, but that is no good to us here.

There is a sneaky way to do it. Every transistor base has the anode of a diode connected to it. The cathodes of all those diodes are connected together. When bypassed, the cathodes are switched to 0v. The idea is the diodes take away the current from the transistor bases and in turn, the effect LEDs will turn off. The diodes prevent them from interfering with each other. When the effect is selected on, the 0v is removed from the cathodes and the diodes have no effect on the transistors.
For this to work properly, the diodes must have a lower forward voltage drop than the transistor base-emitter junctions. Schottky diodes, such as BAT41 have a lower voltage and will be certain to turn the transistors off.

[Andon]

There is a sneaky way to do it. Every transistor base has the anode of a diode connected to it. The cathodes of all those diodes are connected together. When bypassed, the cathodes are switched to 0v. The idea is the diodes take away the current from the transistor bases and in turn, the effect LEDs will turn off. The diodes prevent them from interfering with each other. When the effect is selected on, the 0v is removed from the cathodes and the diodes have no effect on the transistors.
For this to work properly, the diodes must have a lower forward voltage drop than the transistor base-emitter junctions. Schottky diodes, such as BAT41 have a lower voltage and will be certain to turn the transistors off.

Hm, alrighty! So would you you use a BAT41 on the bases of all four transistor bases (so four BAT41 diodes), or just the bases of the transistors with the LEDS, or just the transistors going into the C inputs of each 4013 half (two BAT41 diodes in each instance)? I would think the former, since the half of the 4013 that's used for bypass switching doesn't have an LED currently?

Here's where we're at right now, no BAT41 diodes placed yet:



Full size available here: https://i.imgur.com/zIZhcdj.png

Thank you again for talking me through this! Related, I just received a copy of the CMOS Cookbook, and am slowly working my way through it.

[anotherjim]

Just for the LED driver transistors with the diodes. Only 2 I can see.
I'd forgotten you were planning a 4053 for the master bypass. What you need to switch the LED control diode cathodes to 0v is simply the Q output of the bypass 4013. That is pin 13 of U18 will also have the 2 BAT41 cathodes. In bypass, I think pin 13 will go to 0v and that should kill the LEDs.

[Andon]

So instead of connecting the BAT41 cathodes to 0V they should be connected to Q on the bypass - like this?



Full size available here: https://i.imgur.com/VP4EmpT.png