|
|
Post by cynical1 on May 11, 2011 19:12:47 GMT -5
Well, after a nearly 30 year self imposed ban on batteries in guitars or basses, I now have a project where the battery is not just critical, but essential. So, based on the knowledge that multiple batteries are tied in parallel for substation and telco UPS's to extend the time they can backup electronic devices, I assume the same thing holds true in a guitar with two 9v's set in parallel to extend battery life. But, is it worth the trouble to re-route the guitar to accommodate the larger battery box? For example:  The guitar is currently routed for the 3577 box, but I have room to route for the 3576 box. Any thoughts, experience or cautionary tales would be appreciated. Happy Trails Cynical One |
|
|
|
Post by ashcatlt on May 11, 2011 19:21:47 GMT -5
Can't remember which thread, but I do recall sumgai recommended against it recently. Something about one battery dying down and then sucking off the other and leading to premature (if not catastrophic) death of both.
|
|
|
|
Post by cynical1 on May 11, 2011 19:26:02 GMT -5
I agree. That's just the physics of batteries that once one fails you replace the entire series. Granted, no two batteries will ever have an equal charge unless they're on a metered charging system, but there is a tolerance...which escapes me at the moment...
So, to further qualify the question, if both batteries are new and put in service at the same time, is there a benefit to doubling up in parallel to extend battery life?
HTC1
|
|
|
|
Post by sumgai on May 11, 2011 23:20:23 GMT -5
...... but there is a tolerance...which escapes me at the moment... If there is such a tolerance, and I'm not saying there is, but if there was, it'd be on the order of millivolts of deliverable power. This is due strictly to the fact that no two batteries are ever so exactly alike that they appear to each other as the same resistance as their own internal resistance. IOW, the internal and external loads will never match, which means in turn that something's gotta give.  See above. The common use for two-battery boxes is for serial connections, delivering 18 vDC, or thereaboots to the circuitry. Said ciruitry does "meter" the voltage down to what it needs, which means that as the batteries are discharged, the available voltage will still exceed what the circuit needs in order to operate. Circuits may operate on 9, 12 or 15 vDC, all are common design voltages. Obviously a 9 volt circuit will not run out of power from a dual-battery setup nearly as quickly as a 15 volt circuit. But then again, sometimes a higher voltage is necessary to accomplish the mission. (That's often the case where you find a low-noise pre-amp with very high quality tone.) All of that said, if 9 volts is all that's needed for a circuit that's satisfactory in all respects, then a dual-battery box can still be used. (Considering of course that this 9 volt gizmo has no metering capability.) Enter the oft-misunderstood concept of steering diodes. In a nutshell, they prevent the two batteries from seeing each other - each battery sees only the circuit, and doesn't know the other battery even exists. However, these diodes do drop the voltage a bit. If the circuit won't operate very well below about 8 volts (a silicon diode (the venerable 1N4001) drops about 0.7 volts, a gernanium diode drops about half that), then you would probably do well to consider a switch to select which battery is currently active. Sort of like your old pickup's reserve gas tank selector knob.......  Depends on how often your circuit requires a battery change. In some cases, a simple rechargable setup might be the better solution.  HTH sumgai |
|
|
|
Post by cynical1 on May 12, 2011 0:27:30 GMT -5
Well, that saves dragging out the router...
The pictured battery box is just a molded plastic carrier...no intelligence or metering involved.
I'm not sure what the draw or power requirement is on the Fernandes Sustainer...haven't installed it yet... I was just in the process of ordering a new battery box and had a random thought...I really need to stop doing that...
Happy Trails
Cynical One
|
|
|
|
Post by sumgai on May 12, 2011 1:35:01 GMT -5
I'm not sure what the draw or power requirement is on the Fernandes Sustainer... If, during your conversations with Fernandez, you've never heard them say "18 volts is necessary" or "18 volts would be beneficial", then I'd say that what you've already got should be good enough. But then again, if you're Nutty enough to work here, you may not interpret "good enough" the same way as the nice folks at Fernandez..... HTH sumgai |
|
|
|
Post by cynical1 on May 12, 2011 8:11:37 GMT -5
The Sustainer only requires a single 9v battery, so going to 18v wouldn't give me any advantage in battery life...plus I'd have to dig a bigger hole in the body...so that's just filed in the "good intentions" file.
I have more evil ideas lurking for powering on-board devices for a guitar/bass...muhahahahahah...
HTC1
|
|
|
|
Post by ashcatlt on May 12, 2011 9:54:05 GMT -5
Can we revisit now the idea of a jack-borne short-protected regulated phantom power supply taken from a wallwart? I guess that depends on whether there's an easy way to temporarily short the connections in the battery box. Else you'd have to have the phantom box with you anytime you wanted to play the thing.
|
|
|
|
Post by sumgai on May 12, 2011 14:36:31 GMT -5
ash, Yes we can revisit that idea. Perhaps you should bring the rest of the readership up to date.... please?  sumgai |
|
|
|
Post by ashcatlt on May 12, 2011 19:02:23 GMT -5
Well I don't remember when was the last time we talked about it, and I don't think I've seen a schematic for a fully realized version, though I know there's at least one on the market. The idea is this: Most guitars with active electronics use a stereo (TRS) jack to disconnect the battery when it's not plugged in. It's exactly the same system that's used on the input jacks of most guitar pedals. The bottom of the battery is connected to the ring of the jack, while the sleeve is connected to the negative power supply input ("ground") of the active circuit. When a standard mono (TS) guitar cable is inserted, both the ring and sleeve of the jack contact the sleeve of the plug, completing the power supply circuit and allowing the thing to function. Now follow me here since the computer on which I draw thing is suffering from heat stroke.  If we was to remove the battery and install a device which shorts the + and - terminals of battery clip together, we could then take a TRS cable, apply +9V to the ring and 0V to the sleeve, et voilà! Circuit powered. Now we can have a box on the floor with a TRS input jack, a wallwart input, and a TS output. That part's easy. Problem is, there is at least potentially a point when we're plugging or unplugging where both the ring and sleeve of the plug will contact the sleeve of the jack. That's a dead short across the power supply, which could result in fireworks, smoke, and just general catastrophic failure. I = V/R and if R goes to 0 the I (the current demand) approaches infinity. No real world components can provide infinite current w/out failing. So, we need to add some form of protection to the power supply to limit the amount of current it will supply regardless of the downstream demand. I've seen relatively simple schemes for this using parts available at RadioShaft, but can't seem to find it right now. Luckily, we've got a couple EEs around, who could probably whip something up in their sleep. |
|
|
|
Post by sumgai on May 13, 2011 15:31:21 GMT -5
ash, The proper component to use here is a Polyswitch, a type of varistor that is sensitive to current instead of voltage. In brief, it's a resettable fuse. Too much current, and it opens up... when things drop back to the standard amount of current (its rated value), it looks like a straight wire to the rest of the circuit/power supply. It's wired just like a fuse, in series with one of the supply terminals. In this case, that would be the negative lead, the Ring terminal of the jack. I don't think RadioScholck has ever had them, but the usual gang of suspects (DigiKey, Mouser, et al) have 'em. One could use a 1Ω resistor in series with the negative lead (the Ring, not the ground), but that's not really the same thing. For short periods of time, it might limt the current enough to prevent damage to the supply, but I wouldn't depend on it over time, particularly if it was up for commercial sale. The bill is in the mail. sumgai |
|
|
|
Post by JohnH on May 13, 2011 15:46:59 GMT -5
Just to insert a comment on the original topic:
I'd expect that when two similar but slightly different batteries are paralleled, there would be a rapid flow of current as the one with slightly higher initial voltage attempts to pump charge into the lower one. Current will flow dependent on the internal resistance of the batteries, and therefore some power is lost. The better battery will drop in voltage and the weaker one will rise slightly (as it gets recharged, a bit, from the stronger one) Once the two batteries reach the same voltage, it all settles down probably at a voltage close to that of the initially weaker battery (a tad higher). Then no further waste current will flow, and the two will then share further duties to supply power to the equipment. So I think, its OK to do this, with two fresh batteries.
But I think a better way to get more millamp-hours would be to spring for a 9V lithium battery - same size no routing required, or, set up for 6x AAA or AA cells.
John
|
|
|
|
Post by ashcatlt on May 13, 2011 16:55:44 GMT -5
I was think also that for the short term short circuit situation we're talking about that a smallish resistor (small resistance, might want a bit higher wattage rating) would work fine.
The current limited supplies I saw had a regulator and some sort of transistor set up which didn't allow more than some maximum current to flow.
|
|
|
|
Post by sumgai on May 13, 2011 19:04:08 GMT -5
John,
You've described the action correctly and succinctly, but in point of fact, the internal resistance of a battery is directly related to the charge extant. Therefore, the two transducers will never exhibit an equilibrium in both charge and internal resistance.
But better yet, we can prove this in real-world time! Just plunk two batteries down on your work bench, hook 'em up in parallel, then come back tomorrow morning and take a voltage reading. Seems worth a shot, doncha think?
Though keeping in mind that the drain does take time, it's not quite so rapid as say, leaving the lights on when you lock up your car for the night. Nonetheless, the drain will be noticibly much more rapid that if the batteries had been just "left sitting on the shelf".
sumgai
|
|
|
|
Post by gumbo on May 14, 2011 8:41:46 GMT -5
Hmmm... sg, so you're saying that at all times after the batteries are connected in parallel, there will be a case where one battery will effectively 'drag the other down' because they won't (in 'real' time) always deteriorate equally anyway...  Just attempting to stay 'on topic' here.... ;D g-f-b |
|
|
|
Post by sumgai on May 14, 2011 13:06:12 GMT -5
gumbo, Yes, you have it correct, the action takes place without surcease, starting at the moment the two batteries are connected. This is why we have steering diodes. Because a diode conducts in only one direction, a pair of them connected back-to-back (or front-to-front, if you prefer) across one the terminals will ensure the neither battery "sees" the other one. Power for the circuit is then taken from the joint between the two diodes. Upon a bit of reflection, I find that I may have stated my case above a bit strongly. In the infinite universe, it is folly (according to Arthur C. Clarke) to say "absolutely". Better to say "most likely". Therefore, I will freely admit that it is possible, to some small degree, that a human being might find two batteries that do indeed have the exact same characteristic in all respects. Mating two such power sources in parallel might, in some small chance, not result in mutual discharge. But please don't ask me to compute the odds of that happening, I'd rather build that bridge with four lanes...... HTH sumgai |
|
|
|
Post by gumbo on May 15, 2011 6:44:39 GMT -5
Thank you...I think I've finally understood where the steering diode thingie goes in the overall scheme of things...I hadn't twigged that there were two of them... duh!
;-)
g-f-b
|
|
|
|
Post by Yew on May 19, 2011 15:54:20 GMT -5
John, You've described the action correctly and succinctly, but in point of fact, the internal resistance of a battery is directly related to the charge extant. Therefore, the two transducers will never exhibit an equilibrium in both charge and internal resistance. But better yet, we can prove this in real-world time! Just plunk two batteries down on your work bench, hook 'em up in parallel, then come back tomorrow morning and take a voltage reading. Seems worth a shot, doncha think? Though keeping in mind that the drain does take time, it's not quite so rapid as say, leaving the lights on when you lock up your car for the night. Nonetheless, the drain will be noticibly much more rapid that if the batteries had been just "left sitting on the shelf". sumgai I wouldnt want to work that out using calculus and sin waveforms |
|
