Digital Camera DC power supply
entropy
Yah-Der-Hey (Wisconsin)
My camera has a "DC in 3.4v" plug. Now, I have a wall plug that fits in there, but it's output is 4.5v, .4a. Will my camera be smart enough to take as much as it needs, and not more so it doesn't blow itself out?
(PS - wasn't sure where to put this ... feel free to relocate)
(PS - wasn't sure where to put this ... feel free to relocate)
0
Comments
any takers?
Do you really need a DC wall adapter? Regular batteries not doing the job you want? I'd suggest that you take a look at building a battery pack (http://www.dansdata.com/cambattery.htm). Even if you don't want to build one, it's a good read. There's a part where Dan will write about DC wall adapters.
Let's take the base idea of a 4 pack of AA cells, which you are using, probably Lithium Manganese Di-Oxide cells(aka Lith cells or Lithium Di-Oxide cells). You have 2000mAH cells. They can each put out about 2000 mA for one hour. They put out, actually, about 1.2-1.3 V each, are run in two pairs and each pair puts out 2.4-2.6 V at 2000 mA for up to an hour. BUT, there are two pairs working at once. So, you have a 4000 mA hour combined capacity. Question then becomes, how long do they last??? Why do I ask that??? From the time the batteries are first put in service until camera dies will give us how many parts of an hour it takes to draw a cell pack worth 4000 mA about 94% dry due to the supply curve of Lith cells in cameras. My 2050 mA cells actually last about 25 +\- 3 minutes in use, that means they draw 3772 +\- 100 mA at a nominal rate of 2.5 V in 25/60ths of an hour, or actual draw of roughly 3772* 60/25 mA per hour. So, that means a Camedia 4000Z pulls about 9100 (to nearest 100 mAH rounded) mAH of DC power.
To get peak draw, take your cam and repeatedly format the storage media starting with freshly charged batteries and keep repeating the format . Start timing with time you turn camera on. Stop timing when cam display dies. Now you can plug in the time YOU got by substituting it for my 25 minutes, into same equation I used. You need a, about, 3.3 VDC by 9 AH power supply for your cam if yours lasts 25 minutes. Data transfer eats less power than clearing the storage media, which is what you are doing with the format repeat loop. You could alos take pics with flash OFF for a minimum draw and shoot for something in high third of range to get a better narrow base, but at guess you need about a 3.3V by 8-9AH transformer\compact linear power supply. Funny thing about this, is that if you have an otherwise unused PC power supply and can tap into one return and one nominal 3.3 V hot out of its motherboard connector and build a connector onto it that is polarity correct for your camera connection, you should be able to satisfy the cam quite well for what you want to do strictly at home. If you want to do this in the field, however, you will probably want an adapter.
Yes, I could go into converting back and forth from watts, but rarely bother when we are talking pure VDC. In this case you have base ratings, time to get draw demand, then buy that spec adapter or build or get a small linear power supply. OR, buy two more sets of the new Duracell 2050 mAH 4 packs or one 8-pack and be simply working your charger a LOT. My guess is, the cell sets will last at least as long as it takes to charge each set. I keep three sets avilable for my camera, simply change the things. The adapter for my Camera is available, but I paid a bit less for 8 2050 mAH cells.
Where??? Walmart.... About $18.00 plus a tib more for tax for an 8-pack here where I live ($17.83 pretax for an 8-pack, Duracell Ultras, new higher capacity). About $9.83 per 4-pack here. Depends on what you want and how much you want it....
I went into Radio Shack, 4 2050 mAH Radio Shack cells, Lithium, $18.99. Um, NOT.... The Salesman said "We don't HAVE to be competitive, and refused to coem even close to offering to mathc WalMart's price. BTW, Jameco has AA lith-Di-Oxide cells bulk also, depending on how many you want, might be cheaper to simply get a bulk pack from them than buy a linear power supply.
That's three ways to handle it.... Please feel free to take your choice.
I wouldn't. Pretty much every camera manual out there will tell you not to do that, and you do risk damaging the electronics. If your camera is cheap enough that you are willing to test the theory, feel free, but I would invest in a proper voltage adaptor to be safe.
Lower volts is usually ok, as the camera will still run. Higher volts can be bad. The biggest problem with using higher voltage, even if the amperage is lower, is something called "load-balancing." Load balancing basically means, the resistance in your camera versus the folw of electricity in should fall within a certain tolerance. Anything too low will not power the unit. Anything too high could overload the circuits drawing the electricity and providing the resistance.
Remember that volts x amps = watts. Would you stick a 20 watt bulb into a 100 watt socket? Maybe...but it will burn out faster than if it is in a 20 watt socket.
Now, remember what I said about resistance? As you may know, this is measured in Ohms. Find an Ohms Law calculator...here's one:
http://www.angelfire.com/pa/baconbacon/page2.html
Plug in your camera's 3.4 volt expectation, and just assume that it uses .4 amps. Push the Ohm's Law button. You get:
3.4v x .4 a at 8.5 Ohms resistance uses 1.36 watts of power.
Now do it again with the adaptor you want to use, 4.5 v:
4.5v x .4 a at 11.25 Ohms resistance uses 1.8 watts of power.
That's half a watt higher than it is expecting, and this is just assuming that:
a - the amperage the camera expects is .4
b - the camera can actually provide 11.25 Ohms resistance. If it cannot, something will overload.
Now, apply the resistance of the first equation against the voltage of the second. If you assume that the camera can provide 8.5 Ohms of resistance, then plug in 4.5 volts and 8.5 Ohms. You will see that the amperage you should then supply it at is .5 amps. Ok, safe enough, you figure, because you were going to give it .4. But look at the wattage. 2.38 watts, as opposed to the 1.36 we first calculated the camera might expect if we assume the amperage. That extra 1.02 watts of energy has to go somewhere. It does. It ends up heating up the circuitry...
Your call
Dexter...