0
\$\begingroup\$

I have a small house measuring ~ 8m x 4.5m. It is in Indonesia and power outages are expected daily. Probably the daily power outage is on average less than half-an-hour, but sometimes the outage can be the whole day (9am till 5pm say).

I have various appliances that run on 12V:

  • a wifi router/modem (1.5A)
  • a burglar alarm system (1A)

I also want to install a CCTV system. There are two options for this, either a 48V NVR with POE sending out 48V to cameras, or a 12V NVR (I believe 15W, plus the hard drive) and the cameras (around 6W each) being individually supplied with 12V.

Anyway, I want to have these things as 'always-on'.

So I am looking at:

  • deep cycle conventional 12V lead acid battery
  • 12V charger

  • maybe 220V inverter

Note that I am not backing up the whole house, and I am remodelling the house including replacing all wiring, so whether I have an 'always-on' circuit on 12V or 220V is not particularly an issue.

Anyway what I can see is that if I power 12V directly from 12V than obviously that's more efficient than converting 12V to 220V then back to 12V, but I am not sure how much of an issue this is in practice as the battery obviously needs to be somewhat oversized in any case.

On the other hand presumably I will have transmission losses with such a low voltage, and I will need to fit special 12V sockets and provide a whole-house 220V -> 12V transformer for when the system is running on mains, so the infrastructure costs aren't obviously lower than keeping everything on 220V.

The main advantage seems to be a slightly smaller battery. How much smaller?

\$\endgroup\$
  • \$\begingroup\$ I'd lean towards 12V operation and locating as much as possible of battery & 12V equipment in close proximity. The 50% or so extra capacity you get from your battery will give you either less-deep cycle (helps battery life) and/or longer holdup in bad weather or very long outages. | For best battery life you probably don't want over 20% DOD most days with lead acid. With LiIon or LiFePO4 it can be deeper but still 100% DOD is not advised. About 50%-60% total capacity with lower Vmax and higher Vmin is ecommeded. \$\endgroup\$ – Russell McMahon May 26 at 8:22
  • \$\begingroup\$ @RussellMcMahon You can get batteries designed for deep discharge. Most lead-acid batteries aren't, because they're made for use in vehicles where the alternator keeps them charged most of the time, but if you go out of your way to get deep-discharge batteries you can get ones that can handle it better. \$\endgroup\$ – Hearth May 26 at 11:08
  • \$\begingroup\$ What about the rest of the appliances that are not 12V? If you count your refridgerator, do all the 12 V devices even reach a percentage of power usage that you'd remotely care about? \$\endgroup\$ – Marcus Müller May 26 at 11:30
  • \$\begingroup\$ WIth 86% for each stage best case in 3kVA sizes that's 73% overall best case but due to load fluctuations and idle drain you are lucky to get 50%. Bigger systems will be more optimized for efficiency. \$\endgroup\$ – Sunnyskyguy EE75 May 26 at 14:06
  • \$\begingroup\$ @Hearth Deep discharge batteries are essential in this task BUT if you want economically acceptable lifetimes they need to be operated at the depth of discharge levels that I indicated. You'll note that he mentions "deep cycle" batteries - which are the same products. .The sad reality is that lead acid technology in any acceptable cost form does not handle true high DOD levels well, even when designed for deep cycle use. Manufacturers data sheets demonstrate this well. \$\endgroup\$ – Russell McMahon May 26 at 20:55
0
\$\begingroup\$

You can estimate this fairly easily using $$ P_\text{IN} = \frac {P_\text{OUT}}{\text{Eff.}} $$ for each device. So, let's say a sine inverter has an efficiency of 75% (I didn't look this up) and the mains to DC converter has an efficiency of 85% then the power required for a 12 V device powered on mains rather than directly will be $$ P_\text{IN} = \frac {P_\text{OUT}}{\text{Eff.}_1 \times \text{Eff.}_2} = \frac {P_\text{OUT}}{0.75 \times 0.85} = 1.57 P_\text{IN}$$

I've never had to do this but I would be inclined to use DC where possible. That way you have no losses for the 12 V devices and one conversion loss for each of the others.

\$\endgroup\$
  • 1
    \$\begingroup\$ Note that you're not accounting for the distribution losses on the 12V side, which may be significant. Typical 120V wires in the wall will be 12AWG, according to my research. Assuming you keep the same wire gauge, you are looking at a resistance of 0.32 ohms per 100 feet roundtrip. That doesn't sound like much, but at 1.5 amps, that's a drop of half a volt per hundred feet, or about 4% losses at 12V. So, as long as the stuff on the 12V side stays very low-draw, it might be a good idea. But if you go up to 5 amps (which is only 60 watts), now you are dropping 1.5V or 13%. \$\endgroup\$ – Glenn Willen May 26 at 8:05
  • \$\begingroup\$ One added complexity is that a 12V lead battery will not be exactly 12V, the charger could output in the order or 14V. Some failure may let you see even higher voltages, say disconnecting the battery from the charger. A bit or overvoltage protection is a good thing. \$\endgroup\$ – ghellquist May 26 at 8:16

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.