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 Forum index » DIY Hardware and Software » Lunettas - circuits inspired by Stanley Lunetta
Simple Volume Control Question
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Captain_Clams



Joined: Jul 02, 2012
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PostPosted: Mon Jul 02, 2012 7:21 am    Post subject: Simple Volume Control Question Reply with quote  Mark this post and the followings unread

I'm new to this fascinating subject of electronic noisemakers. I spent the past week developing a firm grasp on a 555 timer in "astable mode". Now I am looking into attenuating the volume of the result with a potentiometer. Before I will purchase a potentiometer, I have a few questions that never seem to be directly answered in the articles I've read on the topic. I would be very obliged if someone could shed some light on it for me here:

1.) When attenuating a signal with a potentiometer, does the change in load resistance pose any threat (of distortion or some kind of thermal failure maybe) to the signal source?

2.) From an article found online:
"The source impedance should normally be no greater than 1/10th (0.1) of the pot's stated resistance."

If I should base my potentiometer value off of the signal source impedance, what is a typical signal source impedance for a silicon chip? How do I measure that with a multimeter?

3.) And finally, perhaps the most difficult question to phrase: If a potentiometer is at maximum resistance in the full-counterclockwise position (stop me if I'm wrong), then couldn't the signal being attenuated become inaudible before reaching the full-counterclockwise position? If so, what can be done about that? It is typically expected that a signal will become inaudible only upon "touching" the counterclockwise position.

Feel free to ask me to refine any questions, I have probably written them with a faulty understanding of all of the variables involved. Thanks in advance.
Captain Clams
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JingleJoe



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PostPosted: Mon Jul 02, 2012 10:29 am    Post subject: Reply with quote  Mark this post and the followings unread

It would be helpful to have a circuit diagram of the potentiometer arrangements you speak of. You see, for a volume control I'd wire up a potentiometer as a voltage divider, then the voltage of the signal is cut down the further you turn the wiper along the track. From the wiper connection I add a capacitor of about 1uF or more and a 100k resistor to ground ... tell you what, circuit diagram time! (see attachments). I just use this for everything, potentiometer values ranging from 100k to 2k have been used, the standard audio amplifier may need more attenuation as in my experience most are expecting signals between 50mV and 1Volt. This can be acheived by adding another resistor to the input, in series with the pot, as stated on the attachment.
Here is a really good web page about potentiometers which goes into more detail about making volume controls:
http://sound.westhost.com/pots.htm

quick answer to number 3: most things you will input anything to should have a high input impedance, you can allways add a 1k resistor to any output if you are worried about overloading it.


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Captain_Clams



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PostPosted: Mon Jul 02, 2012 11:21 am    Post subject: Reply with quote  Mark this post and the followings unread

It takes an old fella long enough to use the "quote" marks on this forum. I don't know if I can get any schematics to you. But if you don't mind, I'd like to pick your brain cause you seem to know what you're talking about.

Quote:
for a volume control I'd wire up a potentiometer as a voltage divider


I am under the impression that current is what directly translates to "volume" in an audio system. Although of course voltage and current are directly related, and in most cases you won't have current without a voltage difference somewhere. (Correct me if I'm wrong on that point)

Well, when you put it that way it makes good sense. As resistance increases, current approaches 0. If you have a high enough value potentiometer, at full setting the current is "essentially" zero, as in I = V/R.

Quote:
capacitor of about 1uF or more and a 100k resistor to ground


Ok, I'm generally familiar with the concept of a pull-down resistor, but why the capacitor? I've seen this formation several times but I do not understand its function.

Quote:
most things you will input anything to should have a high input impedance, you can allways add a 1k resistor to any output if you are worried about overloading it.


I think that's a good idea. It still presents a variable load to the signal source, but it ensures that not too much current is drawn.

Now how do I choose a good potentiometer value? Is it necessary to measure the voltage and current of the output signal? What if I get a potentiometer where below the 10 o'clock position it's inaudible and above the 12 o'clock position it's at full volume? Maybe this is something that's never an issue. Like I always say, correct me if I'm way off base.

Thanks, friend
Captain Clams
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JingleJoe



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PostPosted: Mon Jul 02, 2012 2:21 pm    Post subject: Reply with quote  Mark this post and the followings unread

Using a potentiometer to limit the current will work, but not quite as well as the voltage divider method in my experience.
You are absolutely right about the relationship between V, I and R. However the volume of a wave as far as I am concerned is the amplitude of the voltage, that is the peak to peak voltage of the wave.
Current will be determined by the imedance of the load and/or output impedance of the circuit in question. I don't remember the exact details but it has something to do with the maximum power transfer theorem; if you don't match impedancs the current and voltage will be lower or out of phase, however this doesn't matter so much in the application here, you just need to make sure the input impedance of whatever you connect this volume control to, is nice and big (like 100k, thats a good base figure, most op amps are 1mega-ohm or more so no worries if you use a buffer)
I can see this in my mind's eye, writing it is so hard Laughing
If the impedances of input and output are inverted, so the output is high and input is low then the voltage divider effect of the impedances causes less voltage than you would ordinarily see at output, to be formed accross the load.
I hope I am making sense, I'll leave it there for tonight, I am too tired to think much more Neutral

P.S. the capacitor blocks DC and lets through the AC, a DC offset can discombobulate some amplifiers or other circuits. However in some unusual cases you may want to maintain your DC offset. The 100k resistor is a dummy load and sets the output inpedance to under 100k (I think) the main thing I notied it does is to stop your jack plugs from making the amplifier go "KA-TCHK" when you plug them in Smile

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Stream Operator


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PostPosted: Tue Jul 03, 2012 1:02 am    Post subject: Reply with quote  Mark this post and the followings unread

Good advice, Joe - I can always learn something from your posts. And welcome Captain!

I'll just chime in about that maximum power condition Joe mentioned which I remember from undergrad school. The maximum power condition states that the most power will be delivered to a load when the load impedance is the complex conjugate of the source impedance.

The whut? The complex conjugate, meaning that our source impedance is R+jX and our load impedance is R-jX, where the jX is the reactive component: inductive or capacitive. So for a purely resistive source and load impedance we have maximum power transferred at Rs=Rl.

Now to unconfuse the issue, I'll repeat what Joe said in hopefully more clear words being this: Inside of the system, at the sound generation and manipulation and all until we hit the output drivers, we deal with signals as voltages typically. That's because most of our devices work on sensing and driving voltages (opamps, logic gates, comparators, microprocessors, etc). It's only when we get to the output jack that we are normally concerned with power transfer and such.

So this goes to reinforce the statement you referenced, Captain, that the source impedance should be less than 10% of the load impedance. That's because we are working with voltages and our potentiometer will be wired as Joe said as a voltage divider and ideally the voltage divider should be driven by a source with as low an impedance as possible.

By the way, the magic value of potentiometer in audio systems is 100k. If in doubt use 100k, buy 100k, stock 100k, and well, 10k is used a lot in consumer products and such for user controls like thumbwheels on game controllers and such, but in audio it's 100k. Most popular value you will see on schematics is 100k.

Does that help?

Les

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Captain_Clams



Joined: Jul 02, 2012
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PostPosted: Tue Jul 03, 2012 4:20 am    Post subject: Reply with quote  Mark this post and the followings unread

As an aside, JingleJoe, I went to your website and watched your videos. That's some inspiring stuff. I feel like I've found a pocket of people on the Internet that have struck an immaculate balance of science and enjoyment that I have never quite experienced before. Thanks for helping me get up to speed on the science so I can bask in the enjoyment.

Quote:
Using a potentiometer to limit the current will work, but not quite as well as the voltage divider method in my experience.


What exactly is the difference in wiring? How can one reduce the voltage but not the current? Are we actually reducing both, but only focusing on the voltage change? Do we simply not care about the amperage at this stage in the design because we will later amplify it?

Quote:
"KA-TCHK"

Haha, I was having that problem with my power switch. With the oscillation at a reasonable volume through my "home system", when I switched the power off and on the woofer cones visibly whacked back and forth, a bit unsettling. That's probably just the quality of switch that I'm using, though.

Thanks for chiming in, Inventor!

I'm glad you said that about 100k potentiometers, because I'm about to make a purchase from Mouser to fill up my hardware drawers, and that's exactly the kind of stuff I need to know. A guy can sit around and read all the specs and datasheets, but none of the datasheets tell you if the part is useful in a general sense to your category of device. That comes from experience, of which yours I am trying to draft, if you don't mind. Wink
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elektrouwe



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PostPosted: Tue Jul 03, 2012 5:27 am    Post subject: Reply with quote  Mark this post and the followings unread

placing the capacitor on the output side of a volume control is not a good idea. A rule of thumb is:
keep audio attenuators always DC-free
That means a decoupling C must be in front of the potentiometer.
Why ?: often ( = always with single supply CMOS circuits) your audio signal is composed of a DC-signal ( 1/2 supply voltage) + the pure audio AC-signal. Both DC and AC parts of the signal produce noise when you turn the Poteniometer (proof: connect a DC-power supply to the circuit shown above, turn the pot. up and down and listen to the noise, then repeat the experiment with C @ input side) So with only the AC-part of the signal attenuated , there is only 1 wiper-noise source instead of 2!
2 more arguments: 1) electrolyte caps are more stable with proper DC-bias voltage, which is the case when + is connected to a CMOS output and - goes to the potentiometer. If the electrolyte cap is output sided and the wiper is turned to GND, there is no more + bias voltage. Things get worse when you connect an amplifier to the output which itself has a + bias voltage. Then your cap gets reverse polarized which should be avoided in any case. 2) Wiper in middle position means Rpot./2 in series with output C. Wiper up means Rpot.=0 in series with C - if your load impedance is not very high you get a free (but probably unwanted) tone control.
But hey, we are talking about Lunetta noise circuits not about HIFI, so let's be happy without this C Wink
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Captain_Clams



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PostPosted: Tue Jul 03, 2012 5:58 am    Post subject: Reply with quote  Mark this post and the followings unread

These are my misunderstandings/fears with what you're saying:

If a capacitor is linked from the squarewave signal path to ground, it would only pass the sharp changes to the ground (you could consider that a breif AC), while leaving the steady "DC" sections (high state or low state) in the signal path to go to the potentiometer, slightly rounded off by the removal of the sharp state change. Any DC offset would stay in the path, and the sharp transitions would be smoothed over.

Then only way to kill a DC offset of a squarewave would be to put the capacitor in series before the potentiometer. That would remove the offset, but it would also remove the steady DC sections of the squarewave (the sustained high and low states). A squarewave through a capacitor would just be a series of spikes, albeit with no offset.

So, in light of that ramble, I don't see any good way to remove a DC bias from a squarewave signal.
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Draal



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PostPosted: Tue Jul 03, 2012 6:22 am    Post subject: Reply with quote  Mark this post and the followings unread

Welcome Capt. Very Happy

I am the farthest from scientific on this forum. I abide by some universal basics: which pins go to ground, which go to v+. Where are the Ins and Outs?
Did I remember to add pull down resistors on the inputs? Did I follow that groovy mixer schematic right?

I'm an artist by trade and as a result quite horrible at math, excluding geometry for some reason. My point: even an ignorant cmos abuser as myself can build these beasties and enjoy hours on sonic mischief without blowing things up.

Sorry I don't have much helpful advise except have fun and show us what you build Smile . I use 100k A pots on my mixer/output section usually.

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elektrouwe



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PostPosted: Tue Jul 03, 2012 6:32 am    Post subject: Reply with quote  Mark this post and the followings unread

Captain_Clams wrote:

Then only way to kill a DC offset of a squarewave would be to put the capacitor in series before the potentiometer..

That's what I'm talking about

Captain_Clams wrote:
That would remove the offset, but it would also remove the steady DC sections of the squarewave (the sustained high and low states).

No. A short ( let's say 20ms) horizontal line in a signal ( eg square) is NOT DC
if T = Rpot. x C >> Tsquare_period the shape of the square is (almost) not affected.
A squarewave frequency of 50 Hz = 20ms passing a C with 100uF followed by a potentiometer of 100k (RxC =10sec) will remain
a nice squarewave without DC
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Captain_Clams



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PostPosted: Tue Jul 03, 2012 6:36 am    Post subject: Reply with quote  Mark this post and the followings unread

Quote:
That's what I'm talking about

Sorry I got mixed up because JingleJoe was recommending a capacitor and resistor in parallel to ground. And then when you said it should go before the potentiometer, I thought we were still talking about going to ground. Embarassed

Now your post is making sense to me. Thanks.


And Draal, I've seen some of the fantastic things that you've built. That's another striking thing about this art. Electrical engineering is not a "prerequisite," but an option.

I failed my first year of the Calculus in college because I wanted to understand why the equations worked so badly that I couldn't perform the operations without first spending several hours to get aquainted with them. It was almost a matter of envy to me that some bastard out there was able to come up with these equations, and I can't even understand them when they're written out in front of me. Other kids did fine because they just learned the inputs and the outputs of the equations and were able to pass the tests. I became bogged down, failed, but I've learned my lesson now. Some things you're better off accepting, and using because they simply work. Understand them later if you want to. But that's optional. That's how I defeated the next several years of it.

Now I'm in a similar spot. The past often provides the best wisdom, so perhaps I should ask less technical questions and start building some diagrams.

I'll let you know what happens. Very Happy But first I got to put in my order with Mouser.

Edited to respond to new post
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Captain_Clams



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PostPosted: Tue Jul 03, 2012 7:01 am    Post subject: Reply with quote  Mark this post and the followings unread

Sorry to double post, but I was just thinking about something else.

Quote:
A short ( let's say 20ms) horizontal line in a signal ( eg square) is NOT DC


This sounds like Fourier Transform business. In that viewpoint though, nothing is DC. "True" DC is just properly aligned AC. But for practical reasons, yes, I see what you mean that a short cycle of a high state would not behave like DC. The issue then becomes at what point does your squarewave become so long that the capacitor does impact perceived high and low states? What if I was using a 1Hz square to trigger some other events? Would a capacitor affect its defined high and low states?
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elektrouwe



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PostPosted: Tue Jul 03, 2012 7:44 am    Post subject: Reply with quote  Mark this post and the followings unread

Captain_Clams wrote:
This sounds like Fourier Transform business

it is just as easy as using a cap in front of the pot. for audio signals
and the pot. without cap. for modulation&control signals.
For slow modulation signals (<1Hz) the timing constant would be so high,
that you had to wait minutes after switching on your gadget until the signal reaches steady state. In this case an op-amp adder would be better to cancel DC offsets
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Captain_Clams



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PostPosted: Tue Jul 03, 2012 7:50 am    Post subject: Reply with quote  Mark this post and the followings unread

Thanks. I'm with you on that count.
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JingleJoe



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PostPosted: Tue Jul 03, 2012 2:39 pm    Post subject: Reply with quote  Mark this post and the followings unread

elektrouwe, thanks for the info! I never thought of a few of those things, but the inadvertant tone control was something I noticed using the cap-before-pot arrangement and decided I didn't want Laughing Anyways, I'll work around that somehow and connect the cap in series with the pot in the future.
I suppose the 100k resistor would also be ommited in that arrangement wouldn't it? Ultimately saving parts, jolly good! Smile

P.S. Captain clams, I have the exact same "problem" with understanding things! I feel almost guilty when I use a circuit or equation without atleast some deeper knowledge about it's origin or function. However I take solace in what feynman says, knowing he felt something like this too http://www.youtube.com/watch?v=Uxa1gLt5YKI The difference is, we have the banana but don't know how to get it!

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elektrouwe



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PostPosted: Tue Jul 03, 2012 2:53 pm    Post subject: Reply with quote  Mark this post and the followings unread

JingleJoe wrote:
e
I suppose the 100k resistor would also be ommited in that arrangement wouldn't it?

yes if you have a log.pot. else a 15..22k load resistor behind a linear pot bends it almost to log. behaviour (me and my friends made a great ebay deal with 1000s of 100k lin.pots so we use this trick very often Smile )
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JingleJoe



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PostPosted: Wed Jul 04, 2012 3:44 am    Post subject: Reply with quote  Mark this post and the followings unread

Just to clarify regarding the 15 - 22k resistor, you mean it is connected from wiper to ground? Or accross the entire resistive element of the pot?
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Captain_Clams



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PostPosted: Wed Jul 04, 2012 3:48 am    Post subject: Reply with quote  Mark this post and the followings unread

http://sound.westhost.com/pots.htm

At the very bottom of that article, there is a graph showing what electrowe is talking about next to a schematic of how it's achieved.
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elektrouwe



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PostPosted: Wed Jul 04, 2012 4:12 am    Post subject: Reply with quote  Mark this post and the followings unread

JingleJoe wrote:
Just to clarify regarding the 15 - 22k resistor, you mean it is connected from wiper to ground? Or accross the entire resistive element of the pot?

yes, a pull down Rload. from wiper to ground .
imagine 3 wiper positions in an example with Rpot=100k, Rload=15k
1) wiper to GND=> R has obviously no effect; output = 0
2) wiper in middle pos. => voltage divider 50k / (50k parallel 15k)
3) wiper in upper pos. =>Vout=Vin, Rpot. and Rload do not affect signal,
which answers your 2nd question Wink
by carefully selecting Rload you can bend the attenuation curve quite close to
log. behaviour.
See Capt.Clams nice link for more pot-bends
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JingleJoe



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PostPosted: Wed Jul 04, 2012 6:31 am    Post subject: Reply with quote  Mark this post and the followings unread

That's the link I posted before Rolling Eyes I forgot about that part!
Thankyou for the nice explanation, that makes it much clearer now Smile

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