Sylph Co. Posted December 21, 2006 Report Share Posted December 21, 2006 I'm building a guitar with a Fishman VMV bridge-mounted piezo pickups, the preamp for which (Powerchip), runs from a 9V battery. For various reasons, I am looking to power the guitar externally using a mains power adapter. Getting the supply out of the guitar will be done in a similar way to the "X-BATT", for reference. The problem I have is that when running from a battery, the circuit draws only as much current as it requires, which may fluctuate, but a power adapter supplies a set current. Is the exact level of current important, or can I measure/estimate and use a set current, or will I need additional circuitry to emulate the battery? Quote Link to comment Share on other sites More sharing options...
psw Posted December 21, 2006 Report Share Posted December 21, 2006 Welcome to PG Sylph Co. ...your first post I see... This is really a question for the electronics section, not the tutorial section...(you will get more responses there as this section is for answers)... Anyway...we have considered such things for the sustainer thread...there have been some circuits suggested. This will require a stereo lead for power, if your guitar is alraedy stereo (eg mag pickups and piezo signals) this will be a problem. If not, read on... So, you have a stereo lead with ground and hot signals and one which carries the power and the other the earth. This needs to be set up right as you don't want power going into the signal, but to the circuit...if you get that... I don't see why the X-bat or a DIY version could not be used with a wall wart like powersupply. It would need to be heavily regulated (as most digital pedals use) to avoid noise however (mains hum). You also want to be absolutelu sure nothing could short out in your equipment and pass mains power into your guitar from the powersupply...this could be deadly... Anyway...as I understand it, the circuit will draw as much, and only as much power as it needs. The current rating is the upper limit that the supply can reach. Normally circuits for inside the guitar are designed to draw as little current as possible as this is what will flatten a battery. I have used such powersupplies for testing my sustainers but they draw a bit of current. I have not tried the X-bat or a derivitive and although it sounds appealing for devices such as a sustainer, I am a little concerned and I'd want it fullproof! For various reasons, I am looking to power the guitar externally I'm guessing that finding room for a battery inside the guitar is probably the main reason not to consider onboard power for a device such as this, it should last a fair while, it is only a preamp afterall...not like the sustainer's poweramp. There are very neat battery boxes about that are ideal and often other places to hide a battery. I'd consider very carefully if this is really what you would want to do, and to do a lot of research to avoid a serious mistake (like blowing the circuit or pickups)...for instance, if somehow power were to get into a magnetic pickup, it would turnin into an electromagnet and draw as much current as it could, generate heat and eventually (and in not much time at all, ie before you realised it) burn out the windings...hmmm Anyway...that's my thoughts...I am still considering it for the sustainer project however as this device does draw a fair amount of power, so I am not ruling it out altogether... pete Quote Link to comment Share on other sites More sharing options...
TimS Posted December 22, 2006 Report Share Posted December 22, 2006 Current is what you get when you divide source voltage by a device's resistance. Since the resistance would stay constant, current would go up quite a bit when changing from a 9V battery to a 120V wall outlet. Naturally, a wall wart wouldn't transmit the full 120V, but if it lets through more than 9V, current will go up. Too much current will overload the circuit, so it's important to get it worked out. Quote Link to comment Share on other sites More sharing options...
jnewman Posted February 3, 2007 Report Share Posted February 3, 2007 (edited) Current is what you get when you divide source voltage by a device's resistance. Since the resistance would stay constant, current would go up quite a bit when changing from a 9V battery to a 120V wall outlet. Naturally, a wall wart wouldn't transmit the full 120V, but if it lets through more than 9V, current will go up. Too much current will overload the circuit, so it's important to get it worked out. That's why you buy a 9V DC wallwart . Purely from a power supply point of view, the wallwart type power supply supplies a fixed voltage at any current up to its limit. Based on this voltage, a device will draw a certain amount of current. If the current is below the power supply's rating, good. If it's above the power supply's rating, this causes a voltage drop to compensate, then the device will "brown out" or even shut down completely because of the decreased supply voltage. With a cheap power supply, you may also damage the power supply by overloading it. Edited February 3, 2007 by jnewman Quote Link to comment Share on other sites More sharing options...
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