Wiring multiple quartz clocks together?
May 5, 2010 7:17 PM Subscribe
I'm an artist trying to wire 30 quartz clock mechanisms together so that they run off a single power strip. They each originally take 1 AA battery. I'm having trouble getting it to work. It's important that the clocks start simultaneously because the piece involves precisely planned movement.
I was planning on wiring ten together in series and using a 15v 1.5A dc plug- however, it seems the 1.5 was not enough and I would need a 2.5A.
I have also just tried wiring 6 together in series and using a 9v 2.5A, and was unsuccessful.
Any suggestions to help me wire this would be a huge help! Thank you!
I was planning on wiring ten together in series and using a 15v 1.5A dc plug- however, it seems the 1.5 was not enough and I would need a 2.5A.
I have also just tried wiring 6 together in series and using a 9v 2.5A, and was unsuccessful.
Any suggestions to help me wire this would be a huge help! Thank you!
Agree with Fishbike... it shouldn't be hard to find a 1.5V power supply, or use a set of 1.5V batteries in parallel, and power the clocks in parallel. Trying to put them all in series isn't likely to work out for something like this.
posted by Diplodocus at 7:31 PM on May 5, 2010
posted by Diplodocus at 7:31 PM on May 5, 2010
Response by poster: FishBike, how do I determine the amount of amps I'd need from the 1.5 volt power supply? Do I need to get a special power supply with a certain amperage? Thank you!
posted by davidriley at 7:40 PM on May 5, 2010
posted by davidriley at 7:40 PM on May 5, 2010
1.5A seems very high for a wall clock? (In series the amperage stays the same, the voltage changes. Hence your 15V.)
I second what the others are saying, run them in parallel. For bonus points, make all the wire lengths the same so resistance is equal across the circuits. (not critical) You may find you don't need much power at all, probably only a couple of batteries worth of mA. Page 4 here states current draw is 12mA / clock. So for 30 you need ~ 360mA. Double that with a power supply, but you could probably test the whole system with one NiMH battery (at 1.2V it may have some trouble) or a lithium battery. (Alkaline doesn't like high current draw, but give it a shot to see if it works)
posted by defcom1 at 7:41 PM on May 5, 2010
I second what the others are saying, run them in parallel. For bonus points, make all the wire lengths the same so resistance is equal across the circuits. (not critical) You may find you don't need much power at all, probably only a couple of batteries worth of mA. Page 4 here states current draw is 12mA / clock. So for 30 you need ~ 360mA. Double that with a power supply, but you could probably test the whole system with one NiMH battery (at 1.2V it may have some trouble) or a lithium battery. (Alkaline doesn't like high current draw, but give it a shot to see if it works)
posted by defcom1 at 7:41 PM on May 5, 2010
12 mA seems high. If this is the sort of clock that runs off of one AA for months at a time, I'd expect it to be 1 mA or less. So hardly anything.
posted by alexei at 7:50 PM on May 5, 2010
posted by alexei at 7:50 PM on May 5, 2010
An AA battery is 1.5 volts. Do not supply 15 volts. Supplying more amps than necessary is not a problem, but you do need to get the voltage correct. Amps in a well designed power supply are like reserve, the circuit will draw what it needs. The power supply however must supply the correct voltage as almost all circuits depend upon that and do not correct for higher voltages. Higher voltages can destroy circuits.
posted by caddis at 7:51 PM on May 5, 2010
posted by caddis at 7:51 PM on May 5, 2010
Hmmm. I see you tried to wire them in series. Not good. Go parallel, that way you can control the voltage. You have no idea home much current these draw and whether or not that varies. You can never really control the voltage properly in a series circuit.
posted by caddis at 7:54 PM on May 5, 2010
posted by caddis at 7:54 PM on May 5, 2010
Response by poster: Can I just buy this, cut the end off, and run all of the clocks in parallel off of one adapter?
This is a 1.5v 700ma AC>DC adapter:
http://www.amazon.com/RadioShack-1-5V-700mA-Power-Adapter/dp/B001AQUNGE
posted by davidriley at 7:58 PM on May 5, 2010
This is a 1.5v 700ma AC>DC adapter:
http://www.amazon.com/RadioShack-1-5V-700mA-Power-Adapter/dp/B001AQUNGE
posted by davidriley at 7:58 PM on May 5, 2010
Yes, but I would test the principle before buying it. You can test it with just one battery, alexei's comment of <1mA is more accurate than my 12mA estimate.
posted by defcom1 at 8:01 PM on May 5, 2010
posted by defcom1 at 8:01 PM on May 5, 2010
To really know, get a multimeter (they are cheap, say $15, and useful for lots of things in your life) put it into current measuring mode (on the amperage scale). this is a special mode of the meter and likely requires the leads to be placed into different holes on the meter (read the instructions). In this mode the meter completes the circuit, is placed into series with the battery. Read the amperage and peak amperage as the clock switches hours etc. Multiply by the number of clocks in your circuit and then by say 1.3 for safety and buy a 1.5 v power supply of that amperage.
posted by caddis at 8:13 PM on May 5, 2010
posted by caddis at 8:13 PM on May 5, 2010
First of all, everybody's right - parallel, not series.
Secondly, assuming an electromechanical clock mechanism, 15~30mA sounds about right, 12mA sounds a little low put possible, 1mA is ridiculously low. In the case of a mechanical mechanism, you're much more concerned with the peak current (& its effect on the average) that occurs when the mechanism operates - the electronics alone might draw <~1mA average depending on the design, but you've got a (relatively) high-current pulse occurring for a short period every second that ups the overall average.
Thirdly, have you checked your mechanisms still work? I wouldn't be surprised if you've destroyed one or more by testing them in series at 15v. In a series string, the voltage across any given mechanism will be a function of the apparent impedance at any given instance - if you've got 9 in the process of ticking over, and the 10th is idling about to tick, it'll end up 'seeing' the majority of your supply voltage, possibly destroying the electronics.
Personally, I'd grab the nearest available friend with a multimeter and a vague understanding of electricity, and start looking at the peak current drain of each clock. Or, you could just use a 1.5v supply with a decent peak current rating (educated guess - let's say 100mA peak current per mechanism x 10 mechanisms = 1 amp peak; sounds high, but remember there may be times all 10 clocks will tick over at the same instant due to drift from accumulated slight timing differences), and run with that. For testing purposes, 2 or 3 "D" cells in parallel should suffice.
posted by Pinback at 8:26 PM on May 5, 2010
Secondly, assuming an electromechanical clock mechanism, 15~30mA sounds about right, 12mA sounds a little low put possible, 1mA is ridiculously low. In the case of a mechanical mechanism, you're much more concerned with the peak current (& its effect on the average) that occurs when the mechanism operates - the electronics alone might draw <~1mA average depending on the design, but you've got a (relatively) high-current pulse occurring for a short period every second that ups the overall average.
Thirdly, have you checked your mechanisms still work? I wouldn't be surprised if you've destroyed one or more by testing them in series at 15v. In a series string, the voltage across any given mechanism will be a function of the apparent impedance at any given instance - if you've got 9 in the process of ticking over, and the 10th is idling about to tick, it'll end up 'seeing' the majority of your supply voltage, possibly destroying the electronics.
Personally, I'd grab the nearest available friend with a multimeter and a vague understanding of electricity, and start looking at the peak current drain of each clock. Or, you could just use a 1.5v supply with a decent peak current rating (educated guess - let's say 100mA peak current per mechanism x 10 mechanisms = 1 amp peak; sounds high, but remember there may be times all 10 clocks will tick over at the same instant due to drift from accumulated slight timing differences), and run with that. For testing purposes, 2 or 3 "D" cells in parallel should suffice.
posted by Pinback at 8:26 PM on May 5, 2010
I'm not sure you're going to get the effect you need with this set up.
"It's important that the clocks start simultaneously because the piece involves precisely planned movement."
It sounds like you're using the inexpensive clock modules from Clockkit or the like. These aren't really that accurate. There are atomic clock modules that would probably be more in sync, but they have rather sensitive receivers that sometimes have problems.
If the project needs synchronized rotation, you might want to go with multiple stepper motors controlled by a microprocessor. It would be much easier to keep the shafts in sync that way.
posted by Marky at 8:44 PM on May 5, 2010
"It's important that the clocks start simultaneously because the piece involves precisely planned movement."
It sounds like you're using the inexpensive clock modules from Clockkit or the like. These aren't really that accurate. There are atomic clock modules that would probably be more in sync, but they have rather sensitive receivers that sometimes have problems.
If the project needs synchronized rotation, you might want to go with multiple stepper motors controlled by a microprocessor. It would be much easier to keep the shafts in sync that way.
posted by Marky at 8:44 PM on May 5, 2010
agree with Marky. multiple clocks does not guarantee synchronization, even if you do start them at the same time.
posted by kenliu at 9:04 PM on May 5, 2010
posted by kenliu at 9:04 PM on May 5, 2010
A single AA battery will run a typical quartz clock for about a year. If you run 30 clocks in parallel off the same battery, they will draw (on average) 30 times the current, which should mean that they'll all run for about 12 days. So do you actually need to mains power this thing? Just wiring them all in parallel and running them off a battery might work fine. Certainly won't damage anything to try.
posted by flabdablet at 10:23 PM on May 5, 2010
posted by flabdablet at 10:23 PM on May 5, 2010
Yes, that's what I was going by-- if a ~2000 mAh alkaline AA battery runs one clock for a year, then that implies an average draw of only 0.22 mA. Now, peak current may be significantly higher, but it seems likely that even a single AA would easily handle it, even if peak is 100x average.
posted by alexei at 1:28 AM on May 7, 2010
posted by alexei at 1:28 AM on May 7, 2010
alexei: Now, peak current may be significantly higher, but it seems likely that even a single AA would easily handle it, even if peak is 100x average.
Yeah, but you've got to account for the ability of the power source to actually supply the peak current instantaneously - an AA alkaline may be 2000 mA/h (2 A/h) overall, but that's calculated at a lower overall rate. For example, a standard way of calculating these things is the C/10 rate (actually, I think most dry cells are calculated at a significantly lower rates e.g. C/20 or even C/50), which equates to being able to supply 200mA for 10 hours. Ask them to supply 2A for an hour, and they can't - the chemistry simply gets in the way.
Likewise, instantaneous current is similarly restricted - it takes a small (but finite) amount of time for the chemistry to 'get going'. I know for a fact that normal AA cells can't supply 1A peak in pulses < ~20msecs, and AA alkalines can just manage it (although not for long) - but C or D size can, due to the greater electrode surface area.
It's mostly down to the internal resistance of the cell, which is not linear under load - it's high to start with (as the cell chemistry isn't running yet), takes a finite time to drop (as the chemistry gets going), then gradually increases again. It's the first two factors which we're interested in here, and the greater electrode surface area of larger cells minimises them.
Or you could use an appropriately-rated power supply, which (largely) won't suffer from these effects (output impedance notwithstanding…)
posted by Pinback at 7:18 PM on May 7, 2010
Yeah, but you've got to account for the ability of the power source to actually supply the peak current instantaneously - an AA alkaline may be 2000 mA/h (2 A/h) overall, but that's calculated at a lower overall rate. For example, a standard way of calculating these things is the C/10 rate (actually, I think most dry cells are calculated at a significantly lower rates e.g. C/20 or even C/50), which equates to being able to supply 200mA for 10 hours. Ask them to supply 2A for an hour, and they can't - the chemistry simply gets in the way.
Likewise, instantaneous current is similarly restricted - it takes a small (but finite) amount of time for the chemistry to 'get going'. I know for a fact that normal AA cells can't supply 1A peak in pulses < ~20msecs, and AA alkalines can just manage it (although not for long) - but C or D size can, due to the greater electrode surface area.
It's mostly down to the internal resistance of the cell, which is not linear under load - it's high to start with (as the cell chemistry isn't running yet), takes a finite time to drop (as the chemistry gets going), then gradually increases again. It's the first two factors which we're interested in here, and the greater electrode surface area of larger cells minimises them.
Or you could use an appropriately-rated power supply, which (largely) won't suffer from these effects (output impedance notwithstanding…)
posted by Pinback at 7:18 PM on May 7, 2010
If there's any chance that simultaneous motor pulse currents from 30 clock mechanisms could overtax an AA cell (which my gut says they wouldn't, though obviously I've not tried this) then wiring a decent electrolytic capacitor (say 2200μF) across the cell should fix it.
posted by flabdablet at 5:43 PM on May 11, 2010
posted by flabdablet at 5:43 PM on May 11, 2010
Agreed. But given davidriley's apparent level of electrical knowledge, I thought it was easier to suggest using C or D cells for testing purposes rather than AA ;-)
posted by Pinback at 7:55 PM on May 11, 2010
posted by Pinback at 7:55 PM on May 11, 2010
This thread is closed to new comments.
posted by FishBike at 7:24 PM on May 5, 2010