Batboard

obviously bigger is better, but how do i know how long the system will run on the UPS? for the sake of discussion, i'm talking about a UHF repeater that consists of a pair of cdm750's, an MTR2000, and a Zetron 38A. if i was looking for at least an hour run-tim with heavy comm traffic, approximately what size UPS should i be looking for?

You have to compute the total power draw of the system while in transmit mode, then figure out how big the UPS must be to last for a minimum of an hour at that rating. To get a longer run time doesn't mean you need a bigger UPS but that you need a bigger battery in a certain-sized UPS to run it longer. I once purchased a Best Power 500ME which was a 500-watt unit, but for a few bucks more they increased the size of the battery so it became an 800-watt unit. The thing weighed about 125 pounds but it was bulletproof, and I eventually sold it to my company when my need for it passed. It's still in use today about twelve years later and I believe it's now on its fourth set of batteries.

There are a lot of UPS companies out there, and the good stuff costs a lot more than the junk. You get what you pay for.

Actually, the better approach would be to have the equipmnent running off of a battery bank that is hooked to a charger. This would be the same as an online UPS, but much cheaper. You can get float chargers just for this purpose for not a whole lot of money, and batteries aren't expensive. I managed to snag a few Power Conversion Products 12VDC charger units, for free that, we were taking out of service at a large oil company that I used to work for. So look around and see what you can find. I am assuming that all your equipment will run off of 12VDC, if not then it can get a little more complicated.

Let me know if you have any questions.

Jeff

Wholeheartedly agree with John Wayne (presuming all equipment is 12VDC).

Size the battery (or bank of batteries) so that the nominal continuous draw is equal to or less than 5% of the nominal battery rating in AH.

Figure transceiver draw as follows:

Xmit Current Draw x .1
Rx Current Draw x .1
Squelched Current Draw x .8

Add all three.

Batteries should be G27 or G31 gel cells, wired in parallel. Charger should be StatPower 10A or 20A.

Works this way:

While there is utility power, the StatPower will charge the batteries (if needed) and will carry all DC loads.

When utility power dies, batteries pick up load so that equipment never sees loss of power. Batteries will be 50% discharged at 10 hours, which is the max you should discharge them. When utility power back, StatPower will properly recharge the batteries. StatPower can be left connected to batteries continuously without overcharging. Gel cells, because they do not gas, do not need (and will not accept) any maintenance and can be located in occupied spaces.

Commercially available APCs, in my judgment, are worthless, as they undersize the batteries, deeply discharge batteries not designed for deep discharge, and then improperly recharge the batteries, ultimately cooking them to death. Also, they are grossly overpriced.

System described above gives 10 hours of standby, should function for 5-10 years before replacing the batteries, and will cost on the order of $500.

I have found most computer UPSs charge the battery VERY slowly, so if you add a bigger battery that could be a factor.
And the APC ones charge so slowly, one cannot even run a 1/2 amp scanner off the battery.
Also be aware if running things off the battery of a UPS. One brand floated the battery high,,,, ie you could measure 70V AC from the pos or neg terminal to any of the 3 AC legs...BLEW up a RIB that way!!!

I'm working on a ghetto-hacked UPS project. I have a 60A 48V power supply that puts out about 56V (it takes in 40A @ 120V), four lawn batteries (175 CCA, probably 15 - 20 aH), a telco fuse panel with built in diodes, and a 1400VA UPS that used an external battery box, no longer present.

The way this sucker is going to be rigged, the 48V power supply can power the UPS in inverter mode and/or charge the batteries. The UPS will not be able to charge the batteries (muhahaha) and will divert the power feed to standard AC if present. It's a standby UPS with the ability to run in double conversion mode... which is great, because it means you can rent the cheap generators.

One of these days, I need to see if I can power part of it on a inverter... but that's next to useless...

johnwayne described exactly what i had in mind. perhaps calling that a UPS is using the wrong name. a few concerns though. i don't think my repeater station is capable of operating on ~12VDC. it's a UHF MTR2000, 435-470Mhz 100W. i think the 30-40W units operate at 14V, and the 100W units take 28V. is that right? so, is there a DC transformer i can use, should i use an inverter to get it back to 110VAC and power it that way or would i have to get creative with my battery wiring to give some components 14 and others 28? i really like to K.I.S.S. so the less complicated the better.

also, the power requirements for my equipment are as follows:

MTR2000:
stby = 1A @ 28VDC or 0.6A @ 117VAC
tx = 13A @ 28VDC or 5.4A @ 117VAC

CDM750: (2 of these, 1 transmits as a VHF cross-band patch)
stby = 0.3A @ 14VDC
rx = 1.5A @ 14VDC
tx = 8A @ 14VDC (25w)
tx = 13.5A @ 14VDC (40w)

Zetron 38A:
350mA @ 13.8VDC

RKG, can you give me some help with the math on calculating the AH required in the batteries? 100AH sound right or am i way off? and is a 20A power-supply enough or should i be looking for a slightly larger one?

thanks

I don't much about the MTR2000, but if you can't power it directly from the battery bank then you can use an inverter. Keep in mind the inverters in the price category for this project are at best about 85% efficient. OK, let's get out our calculators:

I will wildy assume ther MTR2000 is transmitting 33% of the time, so that is 20 tx minutes per hour and 40 rx minutes per hour. So that is an AC load of 2.04AH, we'll round it to 2.1A over one hour. Now, we have a inverter at 12VDC nominal, running at say 70% efficiency:
(2.1)(10)(1.3) = 27.3A draw from your battery string over a one-hour period

Now for the CDMs. The calculation is easier since we don't have to deal with inefficient conversions. I don't know how exactly your cross-band is set up, but I'll assume tx 33% of the time just like the repeater. So we have a draw of of 14.37A over one hour; round to 14.5A;

For the Zetron it's simple: it will draw 350mA over one hour.

So we have:
27.3A from the MTR2000
14.5A from the CDMs
0.35A from the Zetron
------
42.15A over one hour.

These calculations use a lot of assumptions about your usage. It is worth noting that I used 12VDC nominal instead of 13.8VDC. The batteries will float about 14VDC or so, but will drop when they are used in a power outage situation. I am guess that this equipment will work down to 10.5VDC or so.

Your max current draw at any one instant is going to be the sum of all the possible simultaneous current draws. 13A for the MTR + 13.5A for one TX CDM + 1.5A for one RX CDM + .35A for the Zetron = 28.35A.

A 100AH battery string would keep this going for about 2 hours.

If you want this all in a spreadsheet then I can cook one up later, but I am running late for supper right now!

Hope this helps.

Jeff

jeff, that helps ALOT. thank you. at 2hrs i would have discharged the batteries about 85% though, is that right? to keep the batteries above the 50% mark as RKG would require ~170AH, correct?

and is the math as simple as 170AH for 2hrs and 340AH for 4 hrs, etc?

anybody know a good place to get the proper gel-cell batteries?

Depending on the battery, it could be rated in a few different rates. The most common are the 20-hour and the 10-hour rate. These refer to the number of amps a battery can supply for 20 or 10 hours at a specified temperature, usually 80F, before the ouput voltage drops to 10.5V, multiplied by 20 or 10 hours. Confused yet? Well, in addition to that, when you increase the current draw, then you decrease the number of recoverable amp-hours before the battery will reach 10.5V.

Here is an example: A battery rated for 50Ah at the 20 hour rate. This battery will supply 2.5A for 20 hours no problem before dropping to 10.5V. However, if you draw 10 amps per hour from this battery, it will last less than 5 hours before dropping to 10.5V. Usually you can get a chart from the battery maker that has the discharge times. It is a non-linear function.

So, you have your 100Ah battery string. You are drawing ~43A over one hour. That is 86A over 2 hours. 86 is less than 100, but you need to derate the battery because you are drawing more than the textbook 2.5 amps per hour from the example above. That is why I ballparked 2 hours, even though it is going to a lot be less depending on the style of battery, which I will now get in to.

You can get batteries that are desgined for continuous high current draw, and they are less affected by this "penalty" for drawing high current. A good example of these is the large batteries used in electric forklifts, called traction batteries, that are supplying thousands of amps per hour. You don't need anything this fancy for your two-way shack.

Just found a good example battery. Check out http://www.power-sonic.com/ps-121000.pdf. You can see the different rates from 20 hours down to 15 minutes. Take your current draw of 43 amps per hour. That is pretty close to 55.2A for the 1 hour rating on this particular battery, even though it is down to 9V. So let's fudge the numbers and say that because you are only drawing 43A and the not the spec sheet's 55A, that it will get down to 10.5V and not the spec sheet's 9.0V. So now we know that we will 2 of these batteries to supply a 43A load for 2 hours.

So now with these calculations and information, we can see that FOR THIS PARTICULAR BATTERY, a string of 2 will keep you up for 2 hours.

Sorry I told you one 100Ah battery would work for 2 hours on my last post, that was first-year EE basic math theory, and you can't understand this post unless you understood my previous one. This is more of the later-on real-world application stuff they don't teach you in college.

Hope that helps. Let me know if you have any questions or need more explanation.

Jeff

i understood both of your posts 100% the discharge characteristic charts on the power-sonic page were very helpful too. pretty straight forward. ok look at the last battery on the bottom of this page, http://www.batterystuff.com/battery/batteries.htm the 8A8D. i'm thinking that two of those would be maybe 5Hrs worth of run-time. now take a look at the last float charger on the bottom of this page, http://www.batterystuff.com/battery/12volt.htm the Samlex 12V45A 2 Bank. something like that is what i'm looking for, right? then all that's left is a DC/AC inverter for the MTR2000. i figure i need at least a 650W continuous inverter (5.4A @ 117V = 631W) i'm still looking for one of those. how am i doing so far?

Let's say that 8A8D has about same characteristics as the one I used in the example. Its real capacity at your draw is ~120Ah, times 2 is 240Ah. divided by 43 = 5.5 hours. Voltages, and thus run time, will be slightly lower since you a paralleling them and one with a lower voltage will discharge the other until equilibrium is reached. But I would say that 5 hours is a good estimate for two of those.

That Samlex SEC1245 unit looks ok. I read the instruction manual, but can't tell what exactly the difference would be in hooking the two batteries to the two + terminals, or just paralleling the batteries and hooking the string to the "consumer" + terminal. For what it's worth, I'll sell you that unit for $349.

For the inverter, I would go ahead and get something like a 700W continous. I don't know how picky an MTR2000 is about having a pretty sine waveform on its AC feed. True sinewave inverters are not cheap, but they are good, and you get what you pay for. I've seen stuff not work when hooked up to cheap inverters that use a modified sinewave. Figure about $1000-$1500 for a true sine wave inverter. Maybe someone will chime in and say that a modified sinewave inverter works on an MTR2000, and in that case you can figure about $300-$500 for a good inverter.

You mentioned that the MTR2000 runs on 28VDC (or is that 24VDC?). If all it cares about is having 28VDC, and there is no smarts in the power supply or anything fancy, then you can use a DC-DC step-up converter. This would eliminate the need for the inverter, and you wouldn't have as much inefficiency. I'd have to do some digging to find one that does what we need, but it's a possible option.

Another thing that we haven't talked about is a low voltage disconnect, or LVD. This will disconnect the batteries from the load when the voltage gets too low, about 10.5V. This will prevent damage to your batteries by making sure that they are not taken down too low. It will also help your equipment that doesn't like to run at say 8VDC. They are good to have since they are relatively cheap and protect your investment.

Feel free to give me a call at the office tomorrow if you want to discuss this on the phone. I know internet communication can be frustrating and ambiguous at times.

I would also be happy to give you a quote on all of the components that you'll need. No pressure; I just might be cheaper than other places.

Thanks,
Jeff

jeff, check your waltel mailbox.