pheld

I wrote that pulling the plus _is_ wise. This is not a big issue, but I use the magnetic pickups 95% of the time and it's a bit annoying that the battery is draining quickly for no good reason. My solution is to feed power from an external adapter via a stereo cable. I may also use the leftover pickup-ring-switch from my coiltap setup as a circuit-breaker for the battery should i need to use it in the future.

I was kinda hoping that someone in or related to the design team may comment on design issues, but this may very well be case. I did some more measurements today and discovered that the digital interface is active even if nothing is connected. It may boil down to the fact that there are opportunities for optimisation in future hardware revisions. I'm sorry I didn't mention multiple batteries, but a battery seems ok to me when it delivers the prescribed number of mAh in each charge-discharge cycle. Pulling the cord when the guitar isn't used is a good thing to do, but I think that especially those who play mostly magnetic pickups would benefit a lot from an improved sleep-mode.

Wich of the facts presented in this thread make you suggest that this is a battery problem? Both batteries supplied with the guitar work the same, and both charge and discharge the amount of energy mentioned in their specification (calculated from voltage/current/time-plots). I.e the battery delivers the same number of mAh regardless of whether the variax-circuit is active or in sleep-mode, while the discharge-time only varies by a couple hours.

I'm not complaining that a sleeping variax is draining the battery, just trying to understand why the difference between active and sleep is so small. I have just completed tests sampling current and voltage over time while the battery discharges one time with the variax active and one time with it in sleep mode (turned off with vol on 0). The sampled data confirm that the energy delivered by the battery during the 2 test-runs is the same. The time from full-charge to voltage-cutoff occur was 11hrs13min with the variax on and 13hrs24min with the variax sleeping. I would expect power-saving to achieve quite a bit more than that. All it needs to do while sleeping is to maintain RAM-contents and state, something that most recent computing devices are able to handle with considerably less than 10% of its operating power-consumption. For reference: I have 2 guitars with active pickups. One has a built-in powerswitch on the volume-pot and thus doesn't pull any measurable current from the battery in 0-position. The other pulls about 2% of its maximum consumption when the volume is on 0.

The battery is as new as the guitar. It has about 10 charging cycles so far. I have no clue when it was made. I may replace it with an external powersupply feeding through a custom stereo-cable as part of some mods along with new magnetic pickups and coil-tap switches.

Any battery drains over time even if it is placed in a vacuum where electrons can't travel, because all known battery constructs have internal leakage. Lithium-ion in general is actually a lot better than many other technologies in this respect. The problem arise with rechargeable cells which shouldn't be discharged below a certain voltage, or they will loose their ability to take charge. To avoid that the equipment, and sometimes the batteries themselves, have a built-in circuit-breaker. A small amount of energy will leak through this circuit, but it is minuscule in most modern battery designs, and certainly insignificant compared to the load of an operational or even "sleeping" variax. A 1/4" jack used as a power-switch should be a stereo-jack, and wired so that neither the amp nor the magnetic pickups become part of the circuit. I.e. any leakage should happen within the internal circuitry. The variax supposedly has powersaving-features, unlike a plain guitar with active pickups so they don't compare at all. I've got other guitars with active circuitry that uses the jack as power-switch, but there's next to no drain on the battery with the jack plugged if the built-in circuit has powersaving features or other internal power-switching (hardware or software).

Which point I made in the original post. Any computer that requires some form of boot sequence will have to be kept alive to respond within reasonable time in a real-time environment. However, my experience with embedded computer design says that it should be possible to conserve 95% or more of the operating consumption by dropping the clock from GHz to kHz and still be able to wake the device up in less than 1ms when necessary. In this case the switch could generate a direct interrupt to the circuit and it should be able to react fully within a couple cycles. So I wonder why the battery drains nearly as quick when the variax circuit is off as when it is in full operation. It's not unusual for a device that operates for 10 hrs on a battery to stay alive in standby (active but running very slow) for 500hrs or more on the same charge. Measuring current I see about 15-20% reduction with the variax off and volume on 0, which is very far from the reduction seen on many other types of circuits in standby-mode. I'm happy with my JTV. Like with many new guitars it'll take a plek-job before I can get the action as low as I wish, but other than that I have no complaints. It's just that I'm reading about "power-saving features" and can't really find any.

Why does the battery drain so quickly with the 1/4" connected and variax switched off? Ideally there should be no need for power at all when the guitar is switched to magnetic pickups regardless of whether the instrument is plugged in or not. Also the "power-saving" feature that supposedly takes effect when the volume is rolled off doesn't seem to help much. My JTV59 drains completely over night if it is plugged in, is on magnetic pickups and with the volume on 0. Just like it would if I had been playing with variax-models all night. It's ok if I pull the plug, but that should imho not be strictly necessary. It may be that the variax-circuit requires a few seconds to boot so that it would be too slow switching back and forth between piezo and magnetics if the circuit had been completely powered off, but a power-saving circuit should still bring the consumption to near 0 when it is not used.