|
|
Post by davidofchatham on Aug 30, 2021 11:24:39 GMT -5
I'm starting to learn lead guitar and I'm practicing with a Fender Mustang Micro headphone amp, which (not surprisingly) has no boost switch. So I'd like to use controls on my guitar. I'm trying to do it with the volume control right now, but I think it would be nice to put a volume cut switch on latching-push pot such as the Bourns PDB185-GTR (because it seems like it would be easier to use than a push-pull pot), so that I can just hit it rather than trying to rotate precisely back and forth. I've read about the blower/bypass-type options, but I'd like it even better if I could do a fixed cut in addition to whatever the current volume setting is. That's largely because there's also no gain control on the FMM, so getting a sound I like with one model may require a different volume pot setting than with another. I realize, by the way, that depending on the gain staging, I may only get more saturation, rather than more volume. I'm assuming that a fixed cut would have roughly the same effect on the overall volume at different pot settings, which may not be correct. Is this idea a non-starter? |
|
|
|
Post by newey on Aug 30, 2021 20:44:18 GMT -5
I'd like it even better if I could do a fixed cut in addition to whatever the current volume setting I don't see why this wouldn't work. The push/push pot (and you're not alone, many people prefer those to the push/pull ones) would be wired off the volume control, so that pushing the pot to the "latch" position would engage a fixed resistor. If, say, with the pot at "10", the resistor would drop it to about "5" on a linear pot, then turning the pot down to about an "8" would give you a "3"-ish setting, and so forth. Recognize that this is never going to be an exact cut to a predetermined level, as even linear pots aren't perfect, but you can probably come pretty close. Once you've settled on a pot value, I would recommend some testing, external to the guitar using a resistor substitution set-up, such that you can try different values and "zero in" on appropriate resistances. |
|
|
|
Post by reTrEaD on Aug 31, 2021 9:18:26 GMT -5
Hi davidofchathamI think the simplest method would be to place a resistor between the wire from the pickup or selector switch that normally feeds the CW lug of the volume pot and the lug itself. Disregarding the load presented by the amplifier, if the fixed resistor is roughly equal to the resistance of the pot, this would result in a -6dB cut. That would be the 'normal' position of the push-push switch. The 'boost' position of the push-push would directly connect the wire from the pickup or selector switch to the CW lug of the volume pot, bypassing the fixed resistor. To refine that a bit, you could experiment with a small capacitor in parallel with the fixed resistor, if you find the 'normal' position to be lacking in highs compared with the 'boost' position.
If you need less than -6dB cut, just decrease the size of the fixed resistor. If you need more than -6dB cut, it would probably make sense to use a full divider before the volume pot and a more complex switching arrangement. But since the push-push is a DPDT switch, you'll have enough poles to work with. |
|
|
|
Post by davidofchatham on Sept 1, 2021 10:47:28 GMT -5
I don't see why this wouldn't work. The push/push pot (and you're not alone, many people prefer those to the push/pull ones) would be wired off the volume control, so that pushing the pot to the "latch" position would engage a fixed resistor. If, say, with the pot at "10", the resistor would drop it to about "5" on a linear pot, then turning the pot down to about an "8" would give you a "3"-ish setting, and so forth. Recognize that this is never going to be an exact cut to a predetermined level, as even linear pots aren't perfect, but you can probably come pretty close. Once you've settled on a pot value, I would recommend some testing, external to the guitar using a resistor substitution set-up, such that you can try different values and "zero in" on appropriate resistances. Sweet! Have I missed an obvious reason why this is less popular than the bypass/blower switch? Are there other push/push pots I should be looking at? I read about Bourns pots somewhere hereabouts, saw that one looking at their catalog, and thought it would be much easier to use for this than push/pull. I didn't know they were widely available. I don't know yet how to do a resistor substitution setup. (This whole thing is actually quite daunting because I barely know how to solder. I've been stymied in a few projects by my lack of skill there.) For a starting point, I'm thinking I should get a working multimeter, then try to measure the difference in resistance with the actual volume settings I tend to choose when switching between rhythm and lead. That should get me close, right? Hi davidofchathamI think the simplest method would be to place a resistor between the wire from the pickup or selector switch that normally feeds the CW lug of the volume pot and the lug itself. Disregarding the load presented by the amplifier, if the fixed resistor is roughly equal to the resistance of the pot, this would result in a -6dB cut. That would be the 'normal' position of the push-push switch. The 'boost' position of the push-push would directly connect the wire from the pickup or selector switch to the CW lug of the volume pot, bypassing the fixed resistor. To refine that a bit, you could experiment with a small capacitor in parallel with the fixed resistor, if you find the 'normal' position to be lacking in highs compared with the 'boost' position. Ooh, it didn't occur to me that I would need treble bleed for this as well! Thanks for that tip. I will have to get back to you guys about the wiring specifics once I get farther along. |
|
|
|
Post by reTrEaD on Sept 1, 2021 11:01:23 GMT -5
Ooh, it didn't occur to me that I would need treble bleed for this as well! Thanks for that tip. I'm not entirely certain you'll need to compensate for the series resistor. The added resistance combined with the cable capacitance will create somewhat of a high-cut filter. But it will also reduce the loading on the pickup(s), so we could expect a brighter sound due to that factor. They are opposite effects. Not sure if they're equal and opposite, though. |
|
|
|
Post by Yogi B on Sept 1, 2021 13:47:33 GMT -5
Ooh, it didn't occur to me that I would need treble bleed for this as well! Thanks for that tip. I'm not entirely certain you'll need to compensate for the series resistor. The added resistance combined with the cable capacitance will create somewhat of a high-cut filter. But it will also reduce the loading on the pickup(s), so we could expect a brighter sound due to that factor. They are opposite effects. Not sure if they're equal and opposite, though. It varies, see the graph from a previous post of mine. With the addition of a 500k series resistor we're effectively switching from a 500k pot with full range to a 1Meg pot with reduced range. Also of note is that when the volume is turned to 10: for the regular control this is one of the points where any external loading would have the least effect; whereas the -6dB setting with the series resistor switched into the circuit at same setting (volume pot on 10), it is the point of the sweep where external loading has the maximum effect. However when used with davidofchatham's headphone amp, there's another reason why we mightn't need to compensate for cable capacitance — there is no cable: (However Fender could've just included a cap on the input in order to simulate cable capacitance.)
It's probably very impractical, but this has got have me thinking instead of the addition of one fixed resistor, what would be the required value & taper of two variable resistors such that the loading would remain consistent with a 500k pot in both normal & -6dB modes? This wouldn't solve the aforementioned quirk at full volume, but would mean that the -6dB settings would be exactly equivalent to the regular settings but with the volume control turned down a little more.Edit: what did I say in the first half of this message? With the addition of a 500k series resistor we're effectively switching from a 500k pot with full range to a 1Meg pot with reduced range. So, that but halved! A 500k pot in the normal mode and 250k series resistor + 250k pot (or equivalent) in -6dB mode. A dual gang 500k pot could do this — annoyingly although bourns make dual gang push- pull pots, they do not make dual gang push- push pots. |
|
|
|
Post by newey on Sept 1, 2021 14:57:17 GMT -5
Have I missed an obvious reason why this is less popular than the bypass/blower switch? d of c- Originally, you asked for: So, we have been talking about a fixed volume reduction, via a resistor which is switched in/out of the circuit. With the resistor switched "out" of the circuit, you'd still have your volume control, operating normally, in the circuit. This would be a bit of a "boost" of sorts when going from the resistor setting to the "no resistor" setting, but it would be different from a "blower switch" or bypass, which bypasses the Volume pot (or, more commonly, bypasses all the pots). Such a switch not only adds a bit of extra output, because the resistance of the pots is no longer in the circuit, but also gives a treble boost at the same time. The extra higher frequencies are perhaps the bigger reason to add a blower switch than the volume increase - more treble for lead work is the idea. In your scheme here, the resistance of the pot is still in the circuit with either switch setting, so you won't get that extra bit of high end that comes from bypassing one or more of the pots. |
|
|
|
Post by davidofchatham on Sept 2, 2021 9:57:29 GMT -5
I'm not entirely certain you'll need to compensate for the series resistor. The added resistance combined with the cable capacitance will create somewhat of a high-cut filter. But it will also reduce the loading on the pickup(s), so we could expect a brighter sound due to that factor. They are opposite effects. Not sure if they're equal and opposite, though. It varies, see the graph from a previous post of mine. With the addition of a 500k series resistor we're effectively switching from a 500k pot with full range to a 1Meg pot with reduced range. Also of note is that when the volume is turned to 10: for the regular control this is one of the points where any external loading would have the least effect; whereas the -6dB setting with the series resistor switched into the circuit at same setting (volume pot on 10), it is the point of the sweep where external loading has the maximum effect. However when used with davidofchatham's headphone amp, there's another reason why we mightn't need to compensate for cable capacitance — there is no cable: (However Fender could've just included a cap on the input in order to simulate cable capacitance.) I think Fender did do this or something like it. I haven't noticed a big difference in the sound of the guitar through the Mustang Micro, nor have I seen others comment about it. But I might be wrong. As for the rest of the discussion, I'm afraid I don't grasp it just yet. Are you saying that the taper will be very odd with the volume cut engaged? Or is the volume cut going to sound different depending on where the volume pot is set? Or both?... Even though I'm not quite following the discussion, you've answered something I'd wondered about, whether there is a dual-gang push-push pot that would allow me to add another tone control, for independent pickup control, without switching to independent volume controls and without giving up the volume cut switch on the volume. (If I understand what would be required to keep the volume control after the tone control in this scenario.) Maybe by the time I want to completely rewire my guitar, there will be one! Originally, you asked for: So, we have been talking about a fixed volume reduction, via a resistor which is switched in/out of the circuit. With the resistor switched "out" of the circuit, you'd still have your volume control, operating normally, in the circuit. This would be a bit of a "boost" of sorts when going from the resistor setting to the "no resistor" setting, but it would be different from a "blower switch" or bypass, which bypasses the Volume pot (or, more commonly, bypasses all the pots). Such a switch not only adds a bit of extra output, because the resistance of the pots is no longer in the circuit, but also gives a treble boost at the same time. The extra higher frequencies are perhaps the bigger reason to add a blower switch than the volume increase - more treble for lead work is the idea. In your scheme here, the resistance of the pot is still in the circuit with either switch setting, so you won't get that extra bit of high end that comes from bypassing one or more of the pots. Gotcha. I was just second-guessing myself, wondering if I was missing a reason why I shouldn't do this, since the blower switch is relatively common as compared to this idea (in that I haven't found anyone else talking about it). I can see why that treble boost aspect would be attractive. |
|
