Calculating overdrive's output impedance

[AM]

I am doing some service on an old favorite of mine, my Red Llama clone. While I am at it I'm thinking of changing the tantalum output cap to a film one.
I can only find a 1uF value in film instead of the 10 uF stock value.
I would like to calculate the output impedance of the circuit so I can decide if I should leave there the stock value (10 uF) or change to a lower value 1 uF.
I tried to work it out using the Ohm law but I can't really figure it out. The 10k volume logarithmic knob stays around half way most of the time and I have installed the 1M resistor between lug 2 and ground.
Can somebody help me calculating the output impedance of the circuit please? Thanks in advance.
schem: http://gaussmarkov.net/layouts/redllama/redllama-schem.png

[anchovie]

That 1meg resistor isn't really doing a lot in terms of the pot resistance. The highest value it will ever be in parallel with is 10K with the pot at max - 1/(1/10000 + 1/1000000) = 9.9K. I expect it's just there as an anti-switching-pop measure, it's too high a value to have much of an influence on the pot taper.

I'd say that changing the output cap to 1uF would be fine, whatever's next in your chain would have to have an incredibly low input impedance to create a noticeable high-pass filter.

[mac]

Before desoldering, try a small film or ceramic cap in parallel with the tantalum, >0.047uf. I always do this to help crappy electros caps in the upper range.
Unless you need to replace it, of course.

The highest film value I got is 6.8uf x 100v, but they are too big for many applications, 2x2x1cm.

mac

[AM]

That 1meg resistor isn't really doing a lot in terms of the pot resistance..... I expect it's just there as an anti-switching-pop measure, it's too high a value to have much of an influence on the pot taper.

You are absolutely right. That's what it is. Just anti-switching-pop measure.

I'd say that changing the output cap to 1uF would be fine, whatever's next in your chain would have to have an incredibly low input impedance to create a noticeable high-pass filter.

Can you calculate the output impedance of the circuit with:
1. The 10 uF output cap
2. The 1uF output cap
I know that the volume pot acts as a variable resistor but there should be some way to calculate min, half way and max impedance values.

The highest film value I got is 6.8uf x 100v, but they are too big for many applications, 2x2x1cm.
mac

Yeah, I saw those too. They are way too big to fit nicely on the board. 1uF is probably more than enough to preserve frequency response anyway.

[AM]

I just wanted to add that I have found a way to calculate the output impedance empirically. I haven't tried it yet though.
What I really want is to be able to calculate it (even approximately) by just looking at the schematic.
Does anybody know how?
Thanks for reading my post.

[PRR]

The OUTPUT impedance is the CMOS chip with NFB, and not easily known, but surely under 1K.

But that's NOT what you really want to know.

The CAP must be sized for that <1K, plus 10K pot, plus some load from 1Meg plus whatever it plugs into next. The "next" could be a 1Meg Fender or a 68K fuzzer.

The worst-case load is 8.6K, and total resistance including 4049 is probably close to 9K. Therefore the cap should be sized for 9K at the lowest frequency of interest. (Assuming pot is never full-up, 9.9K is probably more accurate, but not much different.)

What is the the lowest frequency of interest? Well, look at R1 C1. NFB around IC1B forces pin 5 impedance low, so we guess 100K+0.068u as 23Hz. Ample enough. There isn't much point in making C5 and 9K much lower. In fact without even doing the pi and RC math, you could compare 100K to 9K, about 11, then multiply by 0.068u to get 0.755uFd. C5 could be "much" larger than 0.75u and it would not matter compared to C1 R1 low-cut.

The original designer noted that 10uFd electrolytic (or tant) is not much more expensive than 1uFd, and dropped a 10uFd in there. You find that Film over 1uFd is very expensive (and big).

Agree with James-- I think 1uFd will be utterly fine for guitar. For bass, the 1dB added loss on the bottom note _might_ be detected (though it is far less than speaker and room variation), 2uFd might be a wee-teeny bit better.

[AM]

Paul thanks for the detailed info.
Just to clarify and make sure I understood you I would like to double check a couple of things with you.

The OUTPUT impedance is the CMOS chip with NFB, and not easily known, but surely under 1K.

So should I think of the pedal's output impedance value as 1K?

The CAP must be sized for that <1K, plus 10K pot, plus some load from 1Meg plus whatever it plugs into next. The "next" could be a 1Meg Fender or a 68K fuzzer.

What follows next is:
1. Sometimes a long cable run to the 1Meg Fender
2. Sometimes a short cable to the input of a mixing desk for some direct recording. The signal is then fed to cab sims etc.

Basically I want to make sure that the output impedance is good enough to drive the long cable to the amp or (in more rare occasions) the desk  input.

What is the the lowest frequency of interest? Well, look at R1 C1. NFB around IC1B forces pin 5 impedance low, so we guess 100K+0.068u as 23Hz. Ample enough. There isn't much point in making C5 and 9K much lower. In fact without even doing the pi and RC math, you could compare 100K to 9K, about 11, then multiply by 0.068u to get 0.755uFd. C5 could be "much" larger than 0.75u and it would not matter compared to C1 R1 low-cut.

I've actually substituted that C1=0.068uF with a 0.1uF I had. Sorry, i forgot to mention this.

[PRR]

> So should I think of the pedal's output impedance value as 1K?

This is hard to know, and not very important for this question.

If these were real opamps with NFB, the raw output impedance is a few ohms.

The 4049 CMOS is two small MOSFETs. IIRC in their linear range they act like a few hundred ohms. There may be effective NFB around the last one, which would make output impedance lower.

You asked about "changing the ...output cap". For this question, you want the TOTAL resistance, source and load. In many common cases, source is low and load is high, so we can simply use the load impedance.

The 10K is the major resistor. Another one or two 1Meg loads hardly matters. A 22K desk-input matters some. If the pot is only half-up, it matters little. Model the "half-up" 10K pot as two 5K resistors, you'll see the load is around 9K.

Now add the source. If the effect were op-amp, say 1 ohm, or a naked CMOS, say 300 ohms, or even something 1K, it hardly matters against the 9K.

But.... aren't we over-thinking? What is the worst can happen? Put the cap in. Play. Listen.

[AM]

But.... aren't we over-thinking? What is the worst can happen? Put the cap in. Play. Listen.

Did that already. I spent sometime playing and I think everything works fine. I haven't tried it at stage volume level yet but at studio volume level it works fine.

I guess my phrasing wasn't very clear in my initial question. The reason why I kept on asking about the output impedance of the pedal and not just the IC is because while I was fixing it I was reading this at the same time:
http://www.muzique.com/news/boosters-are-not-buffers/
There is an explanation in this article about why volume pots at the end of a circuit are not so good for driving cables etc.
So, I was trying to figure it all out at the same time. New cap value, the interaction with volume pot, etc.
Since my pedal is unbuffered I was trying to figure out what's going on regarding this issue and possibly come with a value of 1k , 2k , 10k etc according to which position the volume pot is set.
I know that when I use it together with my Boss delay with a buffered output of 1k I can pretty much drive any reasonable cable length.

[liquids]

This is an interesting discussion.  Since I am less versed on the less techie side of things, a few things to input (seems there is some communication breakdown here)  and questions at the same time...please correct me where I am wrong, Paul.  

1)  That 1M resistor to ground does nothing useful...if it's a 'pulldown resistor' on the output, the pot is itself functioning that way.  However, I realize that certain caps (say, electrolytics) leak DC and that would have an affect on popping when switching...not sure about tantalums. But the 1M is probably not necessary at all....or it's a 'afterthought' kind of fix or addition...or lack of understanding that the volume pot should be doing all the 'pulldown' needed at the output.

2) The impedance out and frequency response issues are always an interesting discussion when the volume pedal is last.  This is a simple calculator, FYI:

http://www.muzique.com/schem/filter.htm

The circuit, where pin 2 meets C5, at that point, the output impedance is negligably low, as paul says surely under 1K.  

However, you can think of the pedals (worst case scenario) output impedance as the value of the volume pot in this circuit - ~11k.

But for frequency -- you don't need such a large 10uF cap to retain full frequency in an ideal situation, such as if it were simply going into another similar stage.  But it isnt.  It's followed by a volume pot - which is a series resistor + a resistor to ground.  The series resistance affects output impedance.  The resistor to ground affects frequency response.

The 10k pot is probably why the cap is 10uF, because of the use of the 10k pot...And/or the 10k pot is used because of the use of the 10uF cap - they work together in determining freq response and impedance...

Is the volume pot audio taper?  I would think so, but I realize it could be 10k linear in this case.  I'll assume audio for the following.

With the pot 'half way up' on an audio taper pot, this means that there would be ~9k resistance in series with C5 (the output cap), and a 1K resistor going to ground: 9k+1k=10k.    This would mean the output impedance would be around the pot's series resistance -- 9k (lets say 10k worst case scenario).  

But the frequency response would hence be represented by 10uF/1k to ground.   Which, with that calculator, has a roll off point at 160hz?  If so, that is affecting the low frequencies....but you probably are used to that sound if so...a smaller cap would affect the bass even more, in that instance.  Again, this is assuming a log/audio taper volume pot at halfway.  Would need adjustment if it's a linear pot.

Anyhow, if that is true, you can't get much' better' frequency response here unless you use a 22uF electrolytic cap instead, which should have negligible affect on the lows...but back to DC popping on output switching.   Or, you can instead use a larger value volume pot.  But that will add more series resistance, making the output impedance poor.  It's a balancing act.

Anyhow, AM - I think in short, up to the cap, the output impedance of the 'circuit' out of C5 is negligible (say 1k at worst).  That is basically regardless of whatever cap you use.  In this case, the cap affects frequency response in conjunction with the volume pot.  IF you wanted to use a 1uF cap, you could. But you might want to use a larger value volume pot  if you do - A100k, A250k, so that the lows aren't as audibly affected along the volume pot range.

However, the 10k pot as it is going to have the least 'loading' affect on the next thing in the signal path, in terms of output impedance..  The maximum output impedance the circuit (because of the 10k pot) is going to have is ~11k, worst case scenerio.  That's 'acceptable' output impedance for almost anything you would use, though not ideal all around...but it's worked thus far, I assume.

However, if you go for a larger value volume pot, you will be putting the whole pedal's output impedance in a range that will 'load' the 22k desk input etc.

I think your best solutions are:
1)  leave it as is - use a 10uF tantalum, and if you like the sound/response, the output impedance is fine for what you are doing. Don't mess with it simply going to a 1uF will not affect output impedance, but it will affect frequency response.  But you probably want a buffer feeding the desk input all the time anyway!
2) Change to a film 1uF (or smaller) output cap in conjunction with a larger value volume pot, and follow it with a buffer, built in or otherwise.  This will keep things more consistent and ensure best frequency response and no loading on whatever input you follow it with.   In this case, you really need a buffer after the volume pot that can handle the strong signal (think op amp running off of a charge-pump derived power supply for maximum cleaness) and put it between the pedals out and whatever you feed it (desk in).  This will be a consistent low output impedance load feeding the desk input (or otherwise), without concern that the impedance being affected by where the volume pot is set.

Edit: that article you posted is a good one...op amps, and these CMOS chips however, by themselves, are an instance where straight out of the device, they have low output impedance, even when boosting.

[AM]

Is the volume pot audio taper?

First of all, thanks for your input. It is highly appreciated.
The pot is logarithmic. It's an Alpha pot. I hear that cheaper pots are not really logarithmic etc but that's another topic altogether. So, yes, it was bought as a logarithmic pot.

But the frequency response would hence be represented by 10uF/1k to ground.   Which, with that calculator, has a roll off point at 160hz?

Are you sure you haven't entered the 1uF value instead of the 10uF? Using the calculator I get 159.2 Hz for the 1uF/1kHz combo and 15.9 Hz for the 10uF/1kOhm combo.

The circuit produces a bass-rich sound anyway, so cutting a wee bit off at the end is not too much of a problem really. I've seen people placing way lower value caps there. I personally like the fuzzy character of it so I wouldn't like to cut more bass.

2) Change to a film 1uF (or smaller) output cap in conjunction with a larger value volume pot, and follow it with a buffer, built in or otherwise.

If I keep the pedal without an output buffer but make sure it's always followed by a Boss style buffered pedal wouldn't that achieve the same effect?
If yes, I'd rather do that since space is a bit limited inside the enclosure.

If my calculation of 159.2 Hz for the 1uF/1kHz combo is right I could just keep the 10k pot there. Maybe this could be the best compromise, no? Have a bit of bass scooped out but using all parts I have in stock already without taking the impedance too high.

[liquids]

Yeah, I don't know what I incorrectly entered,, but yours is correct.  Mine had a zero missing somewhere.   

Log pot is good, better than linear for volume of course......cheap log pot is close enough, we all use them.      So assume halfway on the pot is 9k series resistance, 1k to ground.

If you follow it with a boss pedal all the time, than there you go...should be adequate, as the boss has a buffer on the input and output, so it takes the 'pressure' off the impedance of the pedal...which again, is probably ~10k.  Functions the same (actually, easier) than a stand alone buffer or a mod that ads a buffer...

So yeah it comes down to, if you can rig it up and compare the 10uF to the 1uF with the 10k pot...and hear the difference.  Try it, maybe you'll like it.

If not, since you are always following it with said buffer, go ahead, leave it 1uF and make the volume pot a A100k, since the buffer means you need not worry about output impedance, and you're back to the same sound but using a 1uF film cap, if that is preferable.

[AM]

Matthew, Thanks man!
You really helped me with your posts. By the way, your signature message is way cool....and right!
Also, Paul, Marcelo and James. Thanks a lot! You guys made things much clearer for me.

[PRR]

> please correct me

Not fully accounting the amp/CMOS output impedance effects.

> good for driving cables etc.
> 1k , 2k , 10k etc according to which position the volume pot is set.

Simplifying assumption: the pot is higher impedance than the thing driving it, the load is bigger than the pot.

Check: the CMOS is a few hundred ohms, most loads are at least 10K.

"Look into the output", what do you see?

Try three extremes: zero, (electrical) mid-way, and full-up.

• At zero the pot is zero ohms to ground, zero output impedance. (Also zero signal, so this is sorta moot.)


• At full-up, the pot is zero to the thing driving it, which is often "small" impedance, so the output impedance is "small".


• At mid-way.... the 10K pot has 5K to ground, 5K to the thing driving it. With assumptions, this is two 5K paths in parallel. The output impedance is nearly 2.5K. This is the worst possible case.

A better calculation adds the thing-impedance to the pot impedance. Say the CMOS is 500 ohms. 500+10K is 10,500 ohms. Worst case is halfway up the total 10,500 ohms. This gives 5,250 ohms two ways, or 2,625 ohms output.

> any reasonable cable length

Define "reasonable". 30 feet? 300 feet? 3,000 feet? (Past a mile gets into other problems.)

30 feet of typical 30pFd/ft cable is 1,000pFd. 1,000pFd against 2,500 ohms is 63KHz. This is so far in excess of the guitar-zone that we won't worry if "2,500" may really be 2,625 (60KHz).

300 feet is 10,000pFd and now our low-pass is near 6KHz. Starting to nibble the guitar band. The soundperson may want to boost 6KHz a skosh. However an unbalanced 2.5K cable through a paying venue will cross enough power lines to pick up buzz before the high-droop is noticable.

Low-end: pot at zero, the 10uFd cap must drive 10K of pot, which it will do to 1.6Hz. Full-up, it drives the pot in parallel with the external load. Assume 10K. Total 5K. Low-droop has moved up to 3.2Hz. This may be ample for most audio. In fact, knowing that 10uFd often works for 1K loading, we expect problem with pot-taper or CMOS distress before bass-loss.

With 1uFd pot the bass corner shifts to 17Hz-32Hz. Again generally ample for all but the most precise audio.

If the cap is 1uFd, the pot is full-up, the output is dead-short (or say someone stuck a 100 ohm mike across your line), a full analysis must account for the amp/CMOS output impedance. Say 500 ohms: the bass extends to 1uFd+500r or 318Hz. That sucks, but the low-low-low voltage usually sucks worse.