Millenium bypass - maybe this application can't work?

Started by blackieNYC, November 11, 2017, 10:51:00 AM

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blackieNYC

Leave it to me to find a way to turn two clever and proven ideas into a failure. I'm trying to combine Mark Hammer's idea of the passive attenuator pedal (as a "boost" when it is off kind of thing) with RG's Millennium 1 bypass.  I think this is going to be flawed, if it works at all.
This is like a looper where you are inserting a voltage divider:

The millennium 1 wants a fairly low resistance to ground at the output of the effect -  The article says the Rat bypass needed 10k or less, and the Millennium 1 can handle more.  How much more?   I was going to use a 1Meg pot as the attenuator - the 100K in the diagram seems like it could introduce an impedance loss.  Shouldn't I use 1 Meg?
Also, I would like the thing to light up when attenuated. That my mess things up too.
Apologies for the lack of research - my breadboard is stuffed so I just went ahead and starting building this thing, in a 1590A with a 9volt battery. (Never do this.) I only have room for a 2p2t stomp.  The other alternative shown at the Beavis archive is a variable shunt to ground, but this also looks like an impedance compromising circuit.  I've been so diligent in using 1M anti-pop and/or biasing resistors, I don't understand how I could get away with so much lower a value as a passive means of attenuation.
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R.G.

Quote from: blackieNYC on November 11, 2017, 10:51:00 AM
The millennium 1 wants a fairly low resistance to ground at the output of the effect -  The article says the Rat bypass needed 10k or less, and the Millennium 1 can handle more.  How much more?   I was going to use a 1Meg pot as the attenuator - the 100K in the diagram seems like it could introduce an impedance loss.  Shouldn't I use 1 Meg?
The standard I used in developing the Millenium Bypass was ensure that 1M was a >low enough< resistance to force a light-off change. Anything less than 1M will turn the light off. The trick is using those high leakage diodes. Signal diodes were made quote - fast - unquote by introducing gold into the silicon, which reduced the charge carrier recombination time, but also made the diodes leak more than non-gold-doped junctions by a couple order of magnitude, well up into the nano-amperes.  :icon_exclaim: :icon_question: :icon_exclaim:
(Sorry, can't help it.)
What that means is the the reverse-biased high leakage diodes leak a few nanoamperes into the gate of the switch FET, and turn the light on, except when the addition of the resistance to ground pulls the gate of the FET into off voltage territory. The 1M resistance I demanded as a minimum threshold will happily do this. Somewhere between maybe 5M and 100M, things get dicey, and this depends on exactly how much the high leakage diodes leak.

The Millenium Bypass is pretty simple - open circuit to ground, light is on. 1M or less to ground, the light is off.

What's not simple is the attenuation. Guitar pickups can be modeled as a resistor (the resistance of the actual winding wire) in series with an inductor of 1 to several henries, and a capacitor across the resistor/inductor. This is actually too simple a model, but it tells you how to think about it. Not knowing what pickup is attached to your attenuator, and therefore what the values of resistance, inductance, and capacitance are means you can't easily tell what value of attenuator loading will do what to the treble loss. You can only bound it by making more or less reasonable guesses about what the values in the pickups are, and what the guitar tone controls may have done to that. A single coil pickup has a DC impedance of the wire resistance (duuuh, of course) and not much bigger in bass frequencies. This rises to about 100K (for average values of single-coil  :icon_lol: ) at 10K. So to avoid treble loss being noticeable, the industry has informally settled on 10X that, or 1M, as the negligibly loading value for pickups. Humbuckers and the other high signal variants have dramatically higher inductance and even more noticeably treble loss, even at 1M, but the pickups themselves are so treble-averse that we don't generally notice anyway.

So the game is how to avoid making treble loss more noticeable than it already is. If you have a resistive load - like a simple pot - it's easy. Don't let your loading go under 1M at any position of the pot, and you're there. That brings in the question of what is connected to the output of the pot. You don't in general know this. Could be the input of an amp, generally 1M for tube amps, but it also might be a fuzz face input at 10K or a bit more a wah at about 68K to 100K, a poorly designed other pedal. This loading appears at the output of the attenuator, and for a simple pot makes the impedance seen by the guitar driving the attenuator vary, thereby making the treble loss of the guitar vary with pot setting. ACK!!

The way to be sure about this is to buffer the attenuator input so the guitar (if it's really a guitar, not a pedal or some other buffer) sees a fixed, and high, load impedance, while the buffer drives the attenuator with a much lower impedance and as a practical matter doesn't care about the attenuator's loading changes.

Then there's the effect of the variable OUTPUT impedance of the buffer and its effect on whatever is being driven by the attenuator, but that's the subject of another post.
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.

anotherjim

So, 470k/500k attenuator pot? Fenders & most s/c pickups won't mind that, although of course you can't rely on the following load being 1M. but if set to some attenuation, some of the pot resistance is between the source and the load.

Or... fit 1M pot. If the MP is sensing the wiper (output side), then when set to attenuate, it must be <1M.

blackieNYC

Thanks gentlemen.
I should say, this will most certainly be late or last in the chain.
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PRR

Seems to me a 3P2T switch is the answer for the light?

The pot: there's no right answer. 100K, 250K... too low one way, too high another way, and you may not mind. I do suggest you mock it up and play a lot before you cast it in marble.

But since you need LED power, why not put a buffer in there? Or a switched gain stage, unity and more.
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blackieNYC

I do not have room for a 3p2t.
And the passive idea lead me down the "small as possible, why not" rabbit hole.
Looks like you can fit a 3p2t next to a 9v in a 1590a (the kind of sentence that makes my wife roll her eyes) if you grind away the lid lip, but I committed myself to the 2p2t.  I'll have a look at that though. Big washers make everything pretty. Could have gone with two AAAs. Hindsight - the view one gets with your head in your hind.
Buffer - yeeaah..., No.  Then it's a real pedal, and a boring one at that  :icon_wink:  this is end-of-chain, and I always have a pedal on (otherwise, why make 'em).  Which means the passive will probably work!  I'll try this, do some buffered frequency sweeps and report back. Thanks!
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blackieNYC

the LED stays on a bit too much.  I  have a MPF102 in there.  a 4148 diode, 4.7K and a 1K.
On, my circuit draws 6mA. when Dim it draws just under 1mA.  1 month per battery.  Eh, I can live with that I suppose, but it is not dim enough.  Is this a FET swapping thing?
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thermionix

I'm late to the party, but I'll throw this out there anyway:

http://www.smallbear-electronics.mybigcommerce.com/low-profile-3pdt-switch/

Description says "shorter body" but it is considerably smaller than common 3PDTs in all dimensions.  I use these switches regularly with good results, have yet to experience a failure.  Holes in solder lugs are pretty small but 24AWG fits through.  Bigger wire can be tacked on sideways.

anotherjim

Increase the LED series resistor. How bright do you need it? 6mA is more than I ever allow an active LED to run. Series resistors I pick are 6k8 or more, depending on efficiency of the LED.

blackieNYC

I love that 3p2t, but the 2p I'm using is also low profile.

I did try a 10k pot on the incoming 9v. It appeared to reduce the brightness of ON more than the dim OFF, making the two modes closer in illumination.  Non-linear behavior perhaps.

Thanks again, all.  Maybe all I need to buy is a new 1590a!   And put this one in the Drawer of Shame. (Yes, I once described it as the Bag of Shame.)
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anotherjim

If you go back to it, try the mosfet version. Much lower on resistance than a Jfet in their channels. Save the jfet for places where only a Jfet will do.
QuoteI did try a 10k pot on the incoming 9v
That would mess everything up, The 1k is the thing to increase. The 4k7 across the LED needs increasing proportionally since it's shunting current away from the LED and dropping voltage at the LED anode and will cause excessive loss of brightness (personally, I don't see a need for it at all).

On the breadboard, the MP gate is very high impedance & will pick up noise. That is a common cause of dim lit when should be off. Should be ok boxed up.

blackieNYC

So I now have 5.7K in series with the LED.  I did not change the 4.7K across the LED.  The dim looks really off and the ON looks good.

Passive attenuator frequency sweep - what factors are in play here? because the results are underwhelming.
I have a 1M pot as a voltage divider.  wiper is the output.  The attenuator is last. There is a buffer before it.  I tried two. One BMP output stage, the other a commercial pedal without TB. Cable on the input is 1 foot of canare mic cable, as is the 15 feet of output cable. The output goes to a fender amp.
If I attenuate 100 Hz by 2.5 dB, the output compared to input curve is not good.
dB reduction of the passive attentuator @ freq:
-2.5dB @100Hz, -4 dB @1000Hz, -8dB at 2500Hz, -10dB @10K.   And yes, it sounds like it. A little darker. Not a ton.
A rolloff at 1K with say 200K on the top of the divider means .8 nanofarads of capacitance? That's a lot, no?
If I rolloff zero dB, with the pot all the way up, its a 1 meg resistor to ground and the response is flat. Which means if I were to dial this thing back by 6-10dB it is quite dark.
Does a passive attenuator really suck this badly?
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anotherjim

Maybe not surprising. A 1M at 50% R puts 500k in series with the cable capacitance.

1M pot feeding something also like 1M is getting "logarized" somewhat. A Lin pot could have a better feel.

If a lowish output impedance is feeding it, 1M pot looks unnecessarily high. Sadly, say a BMP output would lose bass, because the output cap is already feeding its 100k vol pot and you don't want to make that 100k "look" any lower. Buffered bypass pedals could, but don't necessarily have to, drive into 10k. I think all the Boss ones I've seen, like DS-1, can drive 10k easily.

A treble bleed cap between pot feed and wiper might help fight the dulling you have. Try 1nF and up or down from that.

blackieNYC

100 k pot maybe?  This nice cable is rated at 150 PF per meter. Seems like a lot, but with 50 k the filter rolls off at 10kHz.  Basics, duh. I thought a 1m lot would be more neutral than 100k
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blackieNYC

Having switched from 1 Meg to 100K, it works fine.  The passive attenuator rolls off 3dB at around 5K measured, when cutting 6dB. This would make a fine end-of-chain "booster" when turned off.   And if I were to buy Lake unbalanced instrument cable from Redco Audio, This would be quite flat.   It has a capacitance that is less than half what my expensive canare cable is.  I am not paid for this endorsement in any way.
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PRR

That Lake-wire is 28pFd/ft, which is what I expect for basic audio coax.

If your Canaire is twice as much, it is low-Z stuff, maybe for TV band.
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