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Does Varitone cut or boost midrange?

#21
Rob22315 said:
Where does the energy for the boost come from?
Click to expand...
We don't care about energy too much since voltage transfer replaced power tranfer in audio more than half a century ago. While energy conservation laws most certanly apply, there is peak above unity around p'ups resonant peak. That is natural response of p'up.

bongomania said:
Resonance is a feedback-like condition where a specific frequency will amplify itself.
Click to expand...
Pretty much like that. In frequency selective electronic circuits, like EQs, it is actuall feedback (Q control in parametric eq changes feedback in filter core of EQ). In pickup, reactive part of impedance acts as energy storage. If we had ideal L-C circuit (that is, zero resistance) and "ping" it with very short voltage pulse, it would oscilate for eternity at frequency f=1/(2pi*sqrt(LC)), beacuse there is lossless transfer of energy between inductor and capacitor
 
#22
Rob22315 said:
Where does the energy for the boost come from? TINSTAFL In a passive system, there's only one source for energy and that's the pickups themselves.
Click to expand...

The pickups. It's a resonant LCR circuit. Keep in mind that it's boosting the lows, while the highs were removed.

Also check out the Villex passive booster. It does work, and it's totally passive.
 
#23
SGD Lutherie said:
The pickups. It's a resonant LCR circuit. Keep in mind that it's boosting the lows, while the highs were removed.

Also check out the Villex passive booster. It does work, and it's totally passive.
Click to expand...

That still doesn't explain how it's done and I'm not saying it can't be done, only that the explanations so far are lacking. LCR circuits do not create higher voltages than the input unless they are underdamped and ringing in an underdamped system is a short lived effect. As I said before, it's possible to have a circuit at the resonant frequency lose energy slower than is being pumped in which case the energy will build for a time. This is basically a pay-it-forward scheme which should be limited to a very narrow frequency band and be a short lived effect because the energy from the string will eventually fall below the level necessary to overcome the losses. Do you have a schematic? I could probably dust off the braincells holding my memories of circuit analysis to figure this out.

I know about transformers and DC to DC converters and I once saw a clever arrangement of capacitors that acted as a passive voltage doubler but nothing so far has identified the underlying mechanism for this passive booster. Here's another example - fluorescent paints and dyes look brighter than ambient because they are, the pigment converts the available but unseen ultraviolet into visible light.
 
#24
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Crude model of pickup: DCR=8k, L=4F, C=20pf, source (blue line) is 1KHz square wave, output is green. When pickup gets loaded with 200K, ringing is much reduced, put there are still overshoots of around 50mV-100mV on edges
edit: slight asymetry in response is beacuse SIMetrix defaults to 0V-1V square wave, and I've done this in a hurry, so there is heaviside response on top of response to square wave
 
#27
Sine sweep, from 500Hz to 50kHz in 100ms:
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Detail of sweep around resonance:
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This one would be much more usefull. It's volatges and currents with no load and 200KOhm load. With out load most of energy induced by oscilating strings is spent at resonant frequency, with load most of energy is spent below resosnant frequency, and peak in response is quite damped:
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#28
mongo2 said:
The way my ee prof. explained passive filters 30 years ago is that nothing is really boosted per se, the resonant frequencies are simply not attenuated as much as the others.
Click to expand...

He didn't explain that in the context of a passive variable reluctance transducer (magnetic pickup). The controls and pickup are interactive on a passive bass or guitar. The pickup is a complex impedance and inductive device. So you can set it up to resonate a bit at certain frequency ranges.

Turn down a passive tone control on a bass and you will indeed hear a slight boost when its on zero. You can see this boost on a VU meter. This is because the resonant peak was moved down into the bass range, and you get a larger peak.

It is a boost that is more than what the pickup was doing in full range operation, but it's still very small compared to active systems.
 
#29
recnsci said:
We don't care about energy too much since voltage transfer replaced power tranfer in audio more than half a century ago. While energy conservation laws most certanly apply, there is peak above unity around p'ups resonant peak. That is natural response of p'up.


Pretty much like that. In frequency selective electronic circuits, like EQs, it is actuall feedback (Q control in parametric eq changes feedback in filter core of EQ). In pickup, reactive part of impedance acts as energy storage. If we had ideal L-C circuit (that is, zero resistance) and "ping" it with very short voltage pulse, it would oscilate for eternity at frequency f=1/(2pi*sqrt(LC)), beacuse there is lossless transfer of energy between inductor and capacitor
Click to expand...
Lossless I can agree with but I better see a transformer in there somewhere for a voltage boost.
 
#37
mongo2 said:
The way my ee prof. explained passive filters 30 years ago is that nothing is really boosted per se, the resonant frequencies are simply not attenuated as much as the others.
Click to expand...

It's possible you may have misunderstood him. That's a description of a non-resonant filter.

A metaphor that might be useful is an echo. The guy up on the mountain yells "hello!" and the mountains echo back "ello ello ..." Where did the energy come from for the mountains to make that sound?
 
#39
bongomania said:
It's possible you may have misunderstood him. That's a description of a non-resonant filter.

A metaphor that might be useful is an echo. The guy up on the mountain yells "hello!" and the mountains echo back "ello ello ..." Where did the energy come from for the mountains to make that sound?
Click to expand...

The echo is much quieter than the original so no boost is occurring. That's not a good analogy.

I'd like to see graph of the response vs the input if some can post a reading from one of these. That or a circuit diagram.
 
#40
recnsci said:
Crude model of pickup: DCR=8k, L=4F, C=20pf, source (blue line) is 1KHz square wave, output is green. When pickup gets loaded with 200K, ringing is much reduced, put there are still overshoots of around 50mV-100mV on edges
edit: slight asymetry in response is beacuse SIMetrix defaults to 0V-1V square wave, and I've done this in a hurry, so there is heaviside response on top of response to square wave
Click to expand...

Just found this, thanks for posting. Reading the time scale in sec, this is a very short lived effect and, I suspect, limited to a narrow band. I still don't see this as practical for a broad range of notes on a guitar.
 

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