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 Forum index » DIY Hardware and Software » Lunettas - circuits inspired by Stanley Lunetta
Ring Modulators
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-minus-



Joined: Oct 26, 2008
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PostPosted: Tue Jul 06, 2010 3:01 am    Post subject: Reply with quote  Mark this post and the followings unread

Thanks. I have gone with the pulldown resistors just to be sure.

Just tried the same signal into the 4011. I'm noticing no difference with this. It's the same as if only one input is hooked up.

I had no luck with the 4 diode version. I'm using transformers which are 3K Ohm/3K Ohm. I tried 1N4148 diodes and some other unknown diodes I have, but not much luck. The carrier signal seems to have no effect on the sound really. Which is why I have gone with the NAND ring mod. It appears to be doing the job...
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-minus-



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PostPosted: Tue Jul 06, 2010 3:53 am    Post subject: Reply with quote  Mark this post and the followings unread

Further... I have noticed a difference in sound or tone between three different brands of 4011 IC too.

I was using a Texas Instruments one at first. I just breadboarded a second identical ring mod with a Phillips 4011 and the sound is HEAPS better! Just tried a THIRD circuit using a NXP 4011. This performs up there with the Phillips one. Switching IC's between the breadboards makes no difference.

....so the moral of this Lunetta bedtime story is, not all IC's are alike! Wink

unless it's something to do with the batch of course.
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RF



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PostPosted: Tue Jul 06, 2010 7:32 am    Post subject: Reply with quote  Mark this post and the followings unread

minus - I wonder on those 4011's... are some buffered and some un-buffered?
It would be interesting to know which you are using - Check the suffix on the part#...
4011A or 4011UB = unbuffered
4011B = buffered

bruce

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-minus-



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PostPosted: Tue Jul 06, 2010 7:58 am    Post subject: Reply with quote  Mark this post and the followings unread

Ah! Good point! I'll just check....

Texas Instruments: CD4011BE

Phillips: HEF4011BP

NXP: HEF4011BP

But what does this all mean? Confused Are these the ones I should be after? Or have I been using the wrong type?

Thanks Bruce! Fancy meeting you here... Laughing
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RF



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PostPosted: Tue Jul 06, 2010 8:39 am    Post subject: Reply with quote  Mark this post and the followings unread

'morning -minus!
I think the secong letter might be to identify the manufacturer or package type...

Here's a very basic primer on logic naming...
http://www.key2study.com/elogic.pdf

Now I'm interested in what results you would get from an un-buffered 4011.... Wink

bruce

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Top Top



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PostPosted: Tue Jul 06, 2010 9:10 am    Post subject: Reply with quote  Mark this post and the followings unread

-minus- wrote:
Thanks. I have gone with the pulldown resistors just to be sure.

Just tried the same signal into the 4011. I'm noticing no difference with this. It's the same as if only one input is hooked up.

I had no luck with the 4 diode version. I'm using transformers which are 3K Ohm/3K Ohm. I tried 1N4148 diodes and some other unknown diodes I have, but not much luck. The carrier signal seems to have no effect on the sound really. Which is why I have gone with the NAND ring mod. It appears to be doing the job...


If you are getting sound out with only one input connected on the passive ring modulator, then something is wrong. You shouldn't get any sound unless you have audio going to both inputs.
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-minus-



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PostPosted: Tue Jul 06, 2010 9:58 am    Post subject: Reply with quote  Mark this post and the followings unread

Top Top wrote:
-minus- wrote:
Thanks. I have gone with the pulldown resistors just to be sure.

Just tried the same signal into the 4011. I'm noticing no difference with this. It's the same as if only one input is hooked up.

I had no luck with the 4 diode version. I'm using transformers which are 3K Ohm/3K Ohm. I tried 1N4148 diodes and some other unknown diodes I have, but not much luck. The carrier signal seems to have no effect on the sound really. Which is why I have gone with the NAND ring mod. It appears to be doing the job...


If you are getting sound out with only one input connected on the passive ring modulator, then something is wrong. You shouldn't get any sound unless you have audio going to both inputs.


Yes, that is what I thought too...

Blue Hell's truth table suggests that there would be sound with only one input connected. (His diagram is on the previous page of this thread):

A | B | X | Y | Z | out
--+---+---+---+---+----
0 | 0 | 1 | 1 | 1 | 0
0 | 1 | 1 | 1 | 0 | 1
1 | 0 | 1 | 0 | 1 | 1
1 | 1 | 0 | 1 | 1 | 0


NO input on A, with an input on B, = an OUTPUT

An input on A, with NO input on B = an OUTPUT


So the question is, what was mosc on about when he said this:

Building one will be self-illuminating. But simply put, if there is nothing on one input there will be nothing on the output.

X * Y = Z
Question
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Top Top



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PostPosted: Tue Jul 06, 2010 10:42 pm    Post subject: Reply with quote  Mark this post and the followings unread

-minus- wrote:
Top Top wrote:
-minus- wrote:
Thanks. I have gone with the pulldown resistors just to be sure.

Just tried the same signal into the 4011. I'm noticing no difference with this. It's the same as if only one input is hooked up.

I had no luck with the 4 diode version. I'm using transformers which are 3K Ohm/3K Ohm. I tried 1N4148 diodes and some other unknown diodes I have, but not much luck. The carrier signal seems to have no effect on the sound really. Which is why I have gone with the NAND ring mod. It appears to be doing the job...


If you are getting sound out with only one input connected on the passive ring modulator, then something is wrong. You shouldn't get any sound unless you have audio going to both inputs.


Yes, that is what I thought too...

Blue Hell's truth table suggests that there would be sound with only one input connected. (His diagram is on the previous page of this thread):

A | B | X | Y | Z | out
--+---+---+---+---+----
0 | 0 | 1 | 1 | 1 | 0
0 | 1 | 1 | 1 | 0 | 1
1 | 0 | 1 | 0 | 1 | 1
1 | 1 | 0 | 1 | 1 | 0


NO input on A, with an input on B, = an OUTPUT

An input on A, with NO input on B = an OUTPUT


So the question is, what was mosc on about when he said this:

Building one will be self-illuminating. But simply put, if there is nothing on one input there will be nothing on the output.

X * Y = Z
Question


The diodes and transformers passive ring modulator is a true ring modulator (the name comes from the "diode ring" in the design) and will only output a sound when it has a signal on both inputs. It gives you the sum and difference of the two frequencies at the inputs, without any of the original signal (when it is working correctly). For example, a 500Hz and 400Hz signal on the two inputs will output a 100Hz and 900Hz output. The mathematical and non-tonal relationship is what gives ring modulation it's "random" seeming and bell like sound.

The "pseudo" ring modulator made from gates may act differently. I am thinking that it does not actually give sum and difference because I don't think it is capable of producing a frequency higher than either of the two inputs, but I could be wrong about that. Putting truth tables into practice in terms of the resulting frequencies is still a bit esoteric to me.
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-minus-



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PostPosted: Tue Jul 06, 2010 11:28 pm    Post subject: Reply with quote  Mark this post and the followings unread

Ah, I see.... I thought your question was pertaining to the 4011 NAND ("pseudo" ring modulator), not the diodes and transformers passive ring modulator (true ring modulator).

Esotericism aside, as mosc pointed out, it's nice to have several ways to do the same thing. I'm quite happy with the sounds I'm getting from the pseudo version. Nonetheless I would like to investigate the true ring modulator further. I might try it again on breadboard incase I was overlooking something.

I'll post a sound clip later. I'd be interested to hear what sound you are getting from yours as a comparison.

Thanks.
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mosc
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PostPosted: Sat Jul 10, 2010 12:15 pm    Post subject: Reply with quote  Mark this post and the followings unread

BTW, a minor point, but I will make it nonetheless.

You do NOT need a pull up or pull down resistor in CMOS. Just tie the pins directly to ground or Vdd. CMOS draws no current through the inputs, except on transitions. There is no need for the resistors. Why waste them? TTL and some other old logic families is a different story, but CMOS is virtually ideal WRT input currents.

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Top Top



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PostPosted: Sat Jul 10, 2010 12:30 pm    Post subject: Reply with quote  Mark this post and the followings unread

mosc wrote:
BTW, a minor point, but I will make it nonetheless.

You do NOT need a pull up or pull down resistor in CMOS. Just tie the pins directly to ground or Vdd. CMOS draws no current through the inputs, except on transitions. There is no need for the resistors. Why waste them? TTL and some other old logic families is a different story, but CMOS is virtually ideal WRT input currents.


I have found the need for them in situations where you are mixing several outputs through diodes to one input.

One example is when using a 4017 and a 4093 on a project I just did - the output pins of the 4017 were protected by diodes so that the current only goes one way, then summed and sent to the gate inputs on 4093 oscillators.

When I didn't have pulldown resistors (on the 4093 inputs), the oscillators would just run wild and make noise even when they weren't activated by a step. The pulldown resistors kept the gates acting as they should and only opening up when a pulse from the 4017 was present.

Also in other cases where you will be using a switch to decide whether to allow a signal to go through to a gate. If you don't have a pulldown resistor, then when the switch is off (and the connection is left floating neither high nor low), the gates tend to act up and just start spitting out stuff from that output.

If I just had the inputs grounded, then when I sent a gate signal to the inputs, they wouldn't respond. The resistor allows it to be pulled down when not in use, but still allows the signal to the pin when needed.

I have not found them necessary in any situation when making direct hardwired connections though.
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electri-fire



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PostPosted: Sat Jul 10, 2010 2:23 pm    Post subject: Reply with quote  Mark this post and the followings unread

mosc wrote:
You do NOT need a pull up or pull down resistor in CMOS. Just tie the pins directly to ground or Vdd. CMOS draws no current through the inputs, except on transitions. There is no need for the resistors. Why waste them?


Just to avoid confusion:
We use these in a patchable situation where unused inputs do not stay unused permanently.
Indeed, permanently unused inputs can be be directly grounded or attached to +V.

Oops, Top Top answered that, I overlooked it. I'll just leave it though, as we can't stress this point too much.
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mosc
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PostPosted: Sun Jul 11, 2010 1:28 pm    Post subject: Reply with quote  Mark this post and the followings unread

Ahh, glad you made that point. Of course you need them if you are going to over ride the inputs with a patch.
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Lysol



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PostPosted: Fri Mar 16, 2012 8:06 am    Post subject: Reply with quote  Mark this post and the followings unread

I'm really glad I read this thread!

Last night i was messing around with a passive ring mod circuit with 4 germ diodes and 2 transformers but it wasn't working very well (massive volume drop and some carrier bleed through).

I'm going to give it another shot tonight, maybe see if I was doing something wrong. But if not I'm just gonna try that psuedo-ring mod with a NAND IC (i'm pretty sure I have a couple extra 4093's - i'm assuming one of these would suffice?). I don't need anything very sophisticated and I'm sure it would suit my purposes just fine. I'll report back!
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JingleJoe



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PostPosted: Sat Mar 17, 2012 6:55 am    Post subject: Reply with quote  Mark this post and the followings unread

A 4093 would work but I think the signal is just getting XOR'ed so a 4070 gives you the equivalent of four 4093 ringmods in one chip.
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Lysol



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PostPosted: Sat Mar 17, 2012 8:34 am    Post subject: Reply with quote  Mark this post and the followings unread

don't 4093s and 4070s both have 4 NAND gates each with two in's and one out? What would be the difference in using one (or even a 4011) over the other for an XOR psuedo ring mod?
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PostPosted: Sat Mar 17, 2012 1:05 pm    Post subject: Reply with quote  Mark this post and the followings unread

4070 is Exclusive OR gates, XOR, and 4093 is NAND gates.
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Lysol



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PostPosted: Sat Mar 17, 2012 6:23 pm    Post subject: Reply with quote  Mark this post and the followings unread

ah thanks. damn, i probably could've realized that fairly easily. good to know for the future!

i have an extra 4093 though, so i'll probably just go ahead and see how that'll work.
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Lysol



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PostPosted: Thu Mar 22, 2012 7:25 am    Post subject: Reply with quote  Mark this post and the followings unread

wired up a 4093 to make an XOR, it works but i'm trying to wire in a "depth" pot for the psuedo ring mod effect to no avail.

I tried having the output of the synth* go to both the left term of a pot (for "dry") and one of the in's of the XOR. Middle term goes to the output jack. right side is the output ("wet") of the XOR. The signal is always going through the ring mod circuit, the idea is you're either mixing what's going to the output jack with either the "dry" or "wet." This is how i wired the depth pot for my filter and it works. Did not work for this.

*the output of the synth i'm referring to is coming off the filter depth pot (left side "dry," middle "out," right side "wet") so you have the middle term of the filter depth pot going to both the left side of the ring mod depth and the in of the XOR.

Since this didn't work i brainstormed some and thought about how the "carrier" (a 40106 oscillator) is really loud, too loud really. So i thought maybe i could wire up a "volume" control for the "carrier," a pot essentially just controlling how much gets sent to the second input of the XOR. The output of the synth is always just going into the other XOR in.

That ALMOST works. The filter depth has to be fully wet or the synth cuts out and you only hear the carrier oscillator. The filter effect is fairly dampened as well. And the carrier "volume" doesn't really offer any control - it's either grounded or practically full signal as soon as you start turning the pot. Regardless of the filter effect issue, don't know why I can't get control over the carrier depth/volume.

Damn. Any ideas? Sorry if my descriptions are vague or confusing, I can provide some shoddy schem sketches is it would make things easier.
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JingleJoe



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PostPosted: Thu Mar 22, 2012 2:48 pm    Post subject: Reply with quote  Mark this post and the followings unread

You seem to have misunderstood or overlooked how these chips work; they are digital/logic, only giving out and responding properly to inputs of 0 and 1, high or low, ground or +V. So you can't get any inbetween or "depth" and feeding them lower voltages or voltages inbetween 1 and 0 will cause problems for the chips.
A simple volume control could be added to the output or even filters, for a volume control simply wire a pot as a voltage divider with the "top" (one side of the resistive trace, not the wiper) going to the output of the chip, and the "bottom" to ground. Add a resistor in series with it between the output and the pot if it's range is too loud for most of the turn.

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Lysol



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PostPosted: Fri Mar 23, 2012 6:11 am    Post subject: Reply with quote  Mark this post and the followings unread

Quote:
You seem to have misunderstood or overlooked how these chips work


hmmm, well yes, perhaps! I am still fairly new at diy synthesis! Confused

i was just confused because since the filter "depth" pot (really just a "mix" pot i suppose) i wired in worked fine i figured one would work for the psuedo ring mod as well. But obviously not i guess, different ICs, different circuits, different results.

i've wired in volume pots before, i have a master volume on this synth. perhaps i will see how a volume pot for the output of the ring mod will work for my purposes.

anyways, it's the weekend and i will hopefully be able to get enough experimenting done to get it sorted out. Thanks for your replies.
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nathanxl



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PostPosted: Thu May 03, 2012 6:59 am    Post subject: Reply with quote  Mark this post and the followings unread

jnuaury wrote:
yes, you can out other signals through these chips
some quirky things can happen

you can always use a 4049 to "square up" your input signal too
this works great if you want to run an external source into a divider or use your source to modulate the lunetta style non-vco oscillators


This is something Im interested in. Sorry to be off topic.
I havent been able to find any info on using the 4049 to assist in an external source modulating an oscillator. What would be involved in achieving this?
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knutolai



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PostPosted: Sun May 06, 2012 2:39 pm    Post subject: Reply with quote  Mark this post and the followings unread

Hi
I'm pretty new to ring modulators and pseudo ring modulators. The XOR modulator really caught my interest, but a couple of things are still unclear to me.
Is the XOR ring modulator applicable to a line-level (read synth) or instrument-level (read guitar) device? Would the second input be for a carrier oscillator?

Im currently designing a guitar FX pedal for noise and I wish to add multiple ring modulators to the circuit. A XOR design would be favorable on single IC contains 4 or more gates.


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Psyingo



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PostPosted: Sun May 06, 2012 3:55 pm    Post subject: Reply with quote  Mark this post and the followings unread

All signals need to be logic level, if you want to run a guitar into it you will need to square the guitar signal up which will heavily distort it. You will not get analog voltages out.
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