2
\$\begingroup\$

First, a little backstory. My wife is planning to operate a booth at some craft fairs and we require power for a piece of equipment (Silhouette Cameo, a computer controlled vinyl cutting machine). We've done this once before with success at a show that had power provided. However, other shows do not provide power.

The Cameo’s AC to DC pack states that it provides 24v power at 1.25 amps.

My initial thinking is that you could run two 12v batteries in series to provide the power, then at the end of the day, let them recharge overnight and have them ready for the next day. I have a few questions on the practicality and feasibility of doing this.

First off, what kind of battery can I use? I’ve read a few things on the web stating that certain batteries are not good for this application (ie: car batteries). Is this an applicable battery?

https://www.amazon.com/ExpertPower-EXP1270-Rechargeable-Lead-Battery/dp/B003S1RQ2S/ref=pd_lpo_23_bs_tr_t_2?_encoding=UTF8&psc=1&refRID=NV9MYPMB3E5Q6JT65SZS

Secondly, as batteries don’t provide a straight up 12v as they describe, and the voltage differs based on the amount of charge they are holding, can this damage the appliance? If so, is this what a regulator can help with?

Pardon if anything I’m saying comes off as incredibly stupid as I only understand electronics at a very basic level. My hope is that connecting two 12v batteries directly to the Cameo’s DC input will work without damaging it, and if not, a little regulator chip ( like this: http://www.jameco.com/z/7824T-Major-Brands-Standard-Regulator-24-Volt-2-2-Amp-3-Pin-3-Tab-TO-220_51414.html ) would be enough to supply a steady current to the machine with no risk of damage.

Thanks in advance for any help.

\$\endgroup\$
12
  • 2
    \$\begingroup\$ It's viable. A few things to worry about are: how to not over-drain the battery to prolong its lifetime, how to check for battery level for it to not stop working suddenly, and how much your devices can tolerate voltage fluctuations. Not using a regulator could be fine in many cases or fry your device in others. If you want to regulate the battery as you state, I would use a ready made switching converter instead of buying a linear regulator IC. (due to ease of use and efficiency). \$\endgroup\$ – Wesley Lee Mar 24 '17 at 15:03
  • 1
    \$\begingroup\$ You could use a single 12V battery with a switching regulator (12 to 24V, 2A) - you will need a battery with about a 40AHr (or more) capacity rating to make sure it lasts the day. If you do use a car battery make certain it can't be tipped over (acid spill) or have its terminal short circuited. An in-line fuse (say 5A) is also a good precaution. \$\endgroup\$ – JIm Dearden Mar 24 '17 at 15:10
  • 1
    \$\begingroup\$ Thomas there are many better but more expensive choices. This looks like a good cheap solution. I would suggest you also get a cheap DMM to monitor voltage in each battery . The voltage level will drop quickly to 12.8V then 12.5 to 11.5 is the useful range but preferably should be recharged ASAP to avoid sulphation well below 12V the nominal 50% charge level (plate oxiding & aging). You may want 2 chargers, but possible with 1 if they are matched voltage before parallel charging. 2 is easier. The cutter can withstand this 10% voltage tolerance \$\endgroup\$ – Tony Stewart EE75 Mar 24 '17 at 15:26
  • \$\begingroup\$ 4.5 pounds each \$\endgroup\$ – Tony Stewart EE75 Mar 24 '17 at 15:30
  • \$\begingroup\$ Ya sorry for mixup @TonyStewart.EEsince'75, I didn't notice it picked the wrong name when I replied to the OPs comment. \$\endgroup\$ – Trevor_G Mar 24 '17 at 15:31
5
\$\begingroup\$

Just get two car batteries and put them in series. That should be close enough to 24 V to run the device. It is quite unlikely that ±1 V will cause any trouble, although there is no guarantee of that.

Since the power supply is rated for 1.25 A, you know the device won't draw more than that. Even assuming the full current draw for 8 hours, that comes out to 10 Ah. You don't want to run car batteries down very far, but even small car batteries can do well more than that. 30 Ah is a "small" car battery, and two of those should be just fine with 10 Ah drained from them. Overnight should also be plenty of time to charge them back to full.

\$\endgroup\$
3
  • \$\begingroup\$ I agree. But I'll add that you might also consider Gell cells, if you can find a pair (or combination) that will give you the amp hours needed. Based on their usage in UPS supplies and portable amplifiers, I've heard (but have not proven) that their chemistry is slightly more tolerant of deep discharge. But there is also the safety factor. Unlike with a car battery, you won't end up with a caustic acid spill if the setup is dropped. \$\endgroup\$ – Randy Mar 24 '17 at 16:18
  • \$\begingroup\$ I would agree except for the "get" part. If you're obtaining new batteries for the purpose, save money and weight by going deep-cycle types, and size them at least 3x the 5-10AH you'd expect to use. Or as an alternate, if you have cars with 3-6 year old batteries in them, replace those early, and keep the old batteries for this, this costs you a $10-15 core charge each. Another option is hit up golf cart repair shops and ask them for a set that won't make it 18 holes anymore; 4 x 6v or 3 x 8v will suffice. \$\endgroup\$ – Harper - Reinstate Monica Mar 24 '17 at 17:08
  • \$\begingroup\$ " It is quite unlikely that ±1 V will cause any trouble" - note that two car batteries in series will be +2.4V (an extra 10%). Still probably fine. \$\endgroup\$ – user253751 Jan 13 '20 at 16:30
0
\$\begingroup\$

First, instead of regular car batteries to get that ~24VDC, use deep discharge batteries. Second, 12V lead acid batteries are fully charged at between 14.4 and 14.8 VDC gassing point. Third, yes, two 12VDC batteries in series can properly supply this instance with plenty of power to get you through the day. You only have to supply the current requirements. Fourth, your goal would still be better realized with one 12VDC battery and a buck/boost supply or three 12VDC batteries regulated down to 24. Fifth option, use 18650 batteries stacked in series to make between 23 to around 26 VDC. Stack parallel banks to obtain the current you need for the day or use multiple sets. @user253751, how do you get +2.4 VDC from two 12 volt series-connected lead acid batteries? 2.24VDC each from 12 cells results in 26.8 VDC. A 12 volt lead acid battery produces barely any voltage above 12.2 after having been removed from a charger for 15 minutes.

\$\endgroup\$
5
  • 1
    \$\begingroup\$ "A 12 volt lead acid battery produces barely any voltage above 12.2 after having been removed from a charger for 15 minutes." That is simply not true. \$\endgroup\$ – StarCat Feb 3 '20 at 16:38
  • \$\begingroup\$ Okay, prove it. A lead acid battery will gas at 2.24 - 2.28 v/cell, but it will not remain at that voltage once charging power is removed. \$\endgroup\$ – Lee the Geek Feb 4 '20 at 17:25
  • 1
    \$\begingroup\$ 12.2 volts for a lead acid battery means less than 50% charged. A healthy, fully charged lead acid battery stays at about 12.6-12.7 volts for a very long time (weeks to months) once charging power is removed. No one claims a battery will stay above 13.5 V (2.25V/cell) once charging stops. Prove it? This is easy to check for yourself but I have some batteries here that have not been charged since november and are still well above 12.6V. \$\endgroup\$ – StarCat Feb 4 '20 at 18:23
  • \$\begingroup\$ I just checked a fully charged SLA battery and it's 12.3. Lead acid batteries attain 14.4-14.8 VDC while charging or else they aren't being charged to their full potential. batteryuniversity.com/learn/article/… \$\endgroup\$ – Lee the Geek Feb 4 '20 at 18:34
  • 1
    \$\begingroup\$ If your battery is 12.3 it's not fully charged. References: See energymatters.com.au/components/battery-voltage-discharge or forum.solar-electric.com/discussion/354532/… to name a few... \$\endgroup\$ – StarCat Feb 4 '20 at 18:49
0
\$\begingroup\$

It's a moot point anyway because the voltage going in on the power connector is regulated in all instances unless the regulator itself can't handle the added watt or two of dissipation or there are capacitors directly across the input rated for 25 volts, either scenario of which is highly unlikely. https://batteryuniversity.com/learn/article/charging_the_lead_acid_battery

\$\endgroup\$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.