Crazy "Amp in a box" idea: need help...

[The Scum]

That's almost the circuit I described above...other than R2.

To my eye, it looks like R2 is in there as a "slug resistor," used when someone doesn't have an audio taper pot handy, so they use a linear taper, and hang that resistor to make it act more like audio.

Does the source of that pic have any info about R2? I believe that if you use an audio taper pot, it's unnecessary.

As for tolerance and power rating:
-you don't have to match anything critically, so tolerance isn't a huge issue (it might be more important if you were making a stereo device, and needed the left and right to match each other).
-with a 100K pot, it's a high resistance, so it doesn't draw much current (especially if it's in parallel with a speaker), therefore doesn't dissipate much power...quick back of the envelope calc: to dissipate .25 W, it would take ~150 V to push 1.5 mA into a 100K load. You can measure to be sure, but looks safe on first blush.

You're going to be hanging this thing on the speaker output of a little tube amp? We may need to look at how it gets grounded - some speaker outputs are floating. Have you picked a specific amp yet?

[greatmagnet]

Okay I understood some of that....

The place I found the original schematic (the one I posted yesterday was re-drawn by myself just to make it very clear where things are going in/out of on the potentiometer) did not make any mention of whether it as audio taper or not so you're probably right...it may not need the other resistor at all.

quick back of the envelope calc: to dissipate .25 W, it would take ~150 V to push 1.5 mA into a 100K load. You can measure to be sure, but looks safe on first blush.

I think I know what this means. You're saying my 100k pot won't explode and catch fire if I put 1/4 watt or less through it?

You're going to be hanging this thing on the speaker output of a little tube amp? We may need to look at how it gets grounded - some speaker outputs are floating.

No actually the signal will first come out of a (very low wattage) tube amp...pushed hard, then into a WeberMASS attenuator, and THEN out of the Weber's line out and into the potentiometer circuit you suggested. That final circuit is merely there to bring line level down to instrument level so I can put the whole thing before my stomp boxes.

Have you picked a specific amp yet?

Because I'm trying to get this into a compact form factor...and because they actually sound pretty surprisingly great (for overdrive tones anyhow), I'll be using a Blackheart Killer Ant which actually uses a 12AX7 as a power tube and puts out a bit less than one watt fully driven. From there it'll be into a Weber MicroMASS (for use on 15 watt or less amps) to capture the entirety of that pushed amp tone, and then finally your pot circuit.

I'll be re-housing all this in a single metal chassis of course.

[The Scum]

I just looked at the MicroMASS, and it raises one question.

It says that it works as a dummy load with the attenuator controls all the way down, which looks like it would be right for your application.

But does that also attenuate the line output?

If the line out is active regardless, then the above circuit should work.

To address your other q's:

I think I know what this means. You're saying my 100k pot won't explode and catch fire if I put 1/4 watt or less through it?

Sorta...the Mouser/Alpha pots I use a lot of are rated at 1/4 W. No guarantees about where they smoke or flame, but that's what they say is safe. So to make a 100K load dissipate 1/4 W, it takes the "pressure" of a ~150V signal, which will push 1.5 mA into the pot.

By means of comparison, with number that are more likely for your amp: 1W into 8 ohms means an 8 V signal, and .35 A.

8V is far less than 150V, giving you a wide margin of safety.

If you'd actually want to study the math, I can post how I did it. If It'll make your eyes glave over, there's a phantom power flame war I need to attend to

[greatmagnet]

Ha! I won't keep you from your flame war...we'll discuss math some other time after I've had a Red Bull. Regarding your question...

I just looked at the MicroMASS, and it raises one question: It says that it works as a dummy load with the attenuator controls all the way down, which looks like it would be right for your application. But does that also attenuate the line output?

...it's funny you should ask because I had that same question myself...plus I was curious if I could safely run the line outs with NOTHING plugged into the speaker outs.

so I called them to ask: the answer to your question is NO. You can turn the rheostats all the way down and it will handle the entire dummy load without speakers plugged in AND the volume of the line out is totally unaffected.

[The Scum]

Looks to me like a green light, then.

[greatmagnet]

Nice! So just:

? Input jack "tip" to pot input.
? Pot wiper to output jack "tip".
? Input/output jack "sleeves" tied together and tied to pot ground as well.
? No second resistor at all.

Correct?

[The Scum]

Yup.

You might also tie the case of the pot to the ground points...like is done inside guitars...or put the whole thing inside a metal chassis, and make sure the chassis is grounded.

[greatmagnet]

I'm going to stud-mount a Floyd Rose tremolo to the circuit's outer housing and ground to that.

[natwalk]

Just a quick noob question of my own as a sideline to this discussion, is the circuit described on http://www.epanorama.net/circuits/line_to_mic.html basically a very simple, low feature re-amp device? If so what changes do you need to make to have it work as a re-amp, if any?

Sorry for the partial hijack, and probably dopey question,

Nate

[greatmagnet]

...is the circuit described on http://www.epanorama.net/circuits/line_to_mic.html basically a very simple, low feature re-amp device?

That's actually a REALLY good question and it made me realize that essentially what I've been trying to make this whole time IS a passive re-amp device...although I wasn't thinking of it in those terms. But that's exactly what a re-amp device does, right? It takes a guitar signal that's line-level and then brings it down to instrument level again.

Hope someone here responds to your question...seems like it SHOULD work, but it ALSO begs the question of what ELSE is being put into those ACTIVE re-amp devices (the original ReAmp and the ripped-off Radial version) that makes 'em cost $150-$200???

[greatmagnet]

Also, not to be a thread hog, but...

For the record, I put mine together last night (the variable version with the 100k pot) and it actually works really perfectly...although the pot is not working as it was "supposed" to in theory: allegedly with the pot all the way counterclockwise it was supposed to be ZERO volume, and at 100% clockwise it's line level.

As it turns out, the latter is true, but at 100% counterclockwise it's NOT all the way off...it still puts out some volume.

That said, the 100% counterclockwise position seems to be just what my stomp boxes want to "hear" and works perfectly for my needs. I cranked my vintage Supro head all the way up, put it through my WeberMASS with the rheostat all the way down, went out of the Weber's line out, then into the new variable pad box I built. From there I put it through a digital reverb stomp box (if anything's gonna clip in a shitty way with the wrong input level it would be a digital stomper) and finally right into the front end of another amp (a solid-state combo) set up for "clean".

My cranked all-amp Supro tone was totally intact and sounded fucking awesome!

here's the final box from underneath...

...and FYI the input is where my index finger is.

[The Scum]

The issue with the counterclockwise position not being fully off probably has to do with the pot. If you turn it down, then measure the resistance from the wiper to ground, you'll probably find some tiny amount. Ideally it should be zero, but not all pots will get you there.

As for reamps, they do a handful of things:
-lower line level to guitar level (which a pad can do)
-bump up the source impedance - guitars are often in the 100K Ohm source impedance sorta range - line outputs are often 50 or 100 Ohms. A pad can do this too, but it takes some thinking.
-Balanced to unbalanced conversion - done very gracefully by a transformer.
-act as an inductive source. Guitar pickups are inductive transducers - they're a mile of wire wrapped around some magnets. The transformer in a reamp box just happens to be an inductive source itself.

As for cost - a silkscreened custom metal box probably runs $25-ish. The knob and connectors probably run $15. The transformer probably runs $30. Making $75 in parts, with a high margin of error in my guesses here. Then add labor, profit, freight, operating expenses. Common calculations for small-volume electronics are parts times 4 or 5 (and the sad truth of the matter is that when you get into high volumes those numbers look more like 6 or 8, even though the parts costs are greatly decreased when you buy 10,000 pieces of something). That turns $75 into $300+. So a $250 reamp might actually be a relative bargain.

[greatmagnet]

That's a great explanation. I had kinda guessed to myself that active reamps have a transformer which was probably a big part of the cost...good ones ain't cheap.

[greatmagnet]

The issue with the counterclockwise position not being fully off probably has to do with the pot. If you turn it down, then measure the resistance from the wiper to ground, you'll probably find some tiny amount. Ideally it should be zero, but not all pots will get you there.

You know, I did this last night, although I didn't have access to the interwbz at my house so I was doing what you told me from memory and accidentally measured resistance from in-to-out, instead of out-to-ground.

That said, I think it basically gave me the same answers: with the pot turned all the way clockwise, I got a resistance measure of zero. With the pot turned all the way COUNTER-clockwise (off), I got a resistance of 95.2.

That basically means that it's a 100k pot and that ideally at the counter-clockwise position I should be reading 100k resistance but instead...as per your statement about pot reliability...4.8k is "leaking" through. Right?

This kinda leads me to two more questions:

1.) Although it's an accident based on a crappy pot, that 4.8k that is leaking through actually seems to be exactly the amount of attenuation I need! SO if I were to make a FIXED attenuator based on that finding, does that mean I need about 95k of resistance?

2.) If I DID want to make another varaible resistance box that is more accurate (goes all the way down to a full 100k of resistance), where do I find a more reliable pot? Are pots sometimes sold with a stated degree of tolerance? Much in the same way that a metal-film resistor is stated to be accurate within 1% whereas most other are 5% or more?

[ashcat_lt]

Measure it again. 95.2 is only 4.8% off from 100K, and is therefore within tolerance for even a 5% component. To put it another way, without knowing the total resistance between the the CW and CCW lug (it won't be exactly 100K) you really have no idea how much resistance is left when you turn it all the way down.

The answer to your question #1 is that it is the ratio of the resistances to either side of the wiper that determines the amount of attenuation. If you were to replace the pot with a 95.2K resistor in series with output, you'd also need a 4.8K resistor to ground to make it a voltage divider. Otherwise you won't get much real attenuation. You'll just be increasing your output Z, which could affect hi frequency response.