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Electronics Forum / Basics / September 200612V DC to 220AC + calculations
Hi to all ! I need some help with calculation. I have 12 DC car battery and convertor for 220 AC with 150W output. Questions: 1.) If i have constant load on this converter and if i pull 150W all the time, how long can i use 12DC battery ( before new charge ) ? 2.) If i would have 24V battery at 15A, that would give me 360W. Again the same question... how long can i use this 360W output ? Please be so kind and give me some formula ( so i can make my own calculations), and some optimal results for such case. Thnx! Mark > Hi to all ! > > I need some help with calculation. > I have 12 DC car battery and convertor for 220 AC with 150W output. About 15 A assuming 80% efficiency. What is the amp-hour rating of the battery?  Signature_____ _ _ |_ _| | | | | | | __ _ _ __ ___ | |__| | ___ _ __ ___ ___ _ __ | | / _` | '_ ` _ \ | __ |/ _ \| '_ ` _ \ / _ \ '__| _| |_ | (_| | | | | | | | | | | (_) | | | | | | __/ | |_____| \__,_|_| |_| |_| |_| |_|\___/|_| |_| |_|\___|_| __ ____ / _| | _ \ ___ | |_ | |_) | ___ _ __ __ _ / _ \| _| | _ < / _ \| '__/ _` | | (_) | | | |_) | (_) | | | (_| |_ \___/|_| |____/ \___/|_| \__, (_) __/ | |___/ ... ... > Hi to all ! > [quoted text clipped - 15 lines] > > Mark Hi, Mark. A couple of things for your consideration. Car batteries are not made for deep discharge cycling. You might want to select a marine or deep discharge battery for this purpose The battery you choose has an amp-hour rating. Figure out your amps (15 is a good start as Mr. Simpson said, but I think he's being a little optomistic at 80% efficiency -- I'd guess 70%). The divide amp-hours by the number of amps you'll require. That will give you the number of hours. For instance, if you have a 225 a-h deep discharge battery you want to discharge to 20% of full charge, that would mean you have 180 amp-hours to burn. If your load is 15 amps, that means you have 12 hours at that load current before recharge is required. Take a look at this link -- it has most of the basic information you need: http://www.windsun.com/Batteries/Battery_FAQ.htm Good luck Chris > http://www.windsun.com/Batteries/Battery_FAQ.htm *** Thanx for this link ! I will start another topic on this thema but before i start it, just few explanations. Im trying to make my own battery so i was trying to get some main infos from this car battery ( output, power etc. ) so i can compare it with my battery. The second project is all about my own battery which can have output values from 5 - cca 50V ( the battery is really big + bunch of chem., but hey... :-)), we have to be optimistic :-)) ) Questions: 1.) Which is the "best" way to measure true DCA of some battery ( in my case my homemade 10 VDC ). 2.) Becuse im getting really small DCA ( sometimes i can't make normal measurment ), how to encrease this values ( charge pump ? ). 3.) Regarding your last post, im note sure what i should do in my case. I have really small output power ( if i have make the correct measurment) and i was thinking to make this exp. so i need your help //------------------------------------------------------------------------------------- Project: This is before i connect battery to some DC\AC converter ! . //------------------------------------------------------------------------------------ Idea: : AC lamp 500 W ( 220V, 50Hz ) Battery for this exp. : 38 VDC Battery DCA: : unknown or not verified Time to use : 10 hours before recharge -------------------------------------------------------------------------------------- As you can see i have missing power part ( I ) so i can not calculate how much el.power ( P ) do i need to output from the battery to converter. In end effect i want to get 500W for 10 hours, from the converter. How much power do i need on the battery size to make this work ? >> http://www.windsun.com/Batteries/Battery_FAQ.htm >*** Thanx for this link ! [quoted text clipped - 12 lines] >1.) Which is the "best" way to measure true DCA of some battery ( in my >case my homemade 10 VDC ). --- What do you mean by the "true DCA" of the battery? --- >2.) Becuse im getting really small DCA ( sometimes i can't make normal >measurment ), how to encrease this values ( charge pump ? ). --- If by 'DCA' you mean Direct Current Amperes, you can increase the value by changing the area of the plates or the chemistry of the battery. --- >3.) Regarding your last post, im note sure what i should do in my case. >I have really small output power ( if i have make the correct [quoted text clipped - 14 lines] >converter. How much power do i need on the battery size to make this >work ? --- At 220V, a 500W lamp will require: P 500W I = --- = ------ ~ 2.3 amperes E 220V from the output of the inverter. If your inverter is 70% efficient its input will need: Pout 500W Pin = ------ = ------ ~ 714 watts Eff 0.7 from the battery. If the battery supplies 38V, the current the inverter will need from the battery will be: P 714W I = --- = ------ ~ 19 amperes E 38V Since that current will be expected to be drawn from the battery for 10 hours before recharge is needed, the capacity of the battery will need to be: C = It = 19A *10hr = 190 ampere - hours. For that capacity, the discharge rate of the battery will need to be C/10. If it's lower (C/20 being common) the capacity of the battery will have to be increased somewhat. Google the various battery manufacturers' sites for data sheets for various chemistries. SignatureJohn Fields Professional Circuit Designer >>Questions: >>1.) Which is the "best" way to measure true DCA of some battery ( in my >>case my homemade 10 VDC ). > > --- > What do you mean by the "true DCA" of the battery? My question exactly, but I'm going to guess he means "DC amps" or how much current the battery produces. So, to the original poster: The answer to that is - it doesn't work that way. A battery (or even a single cell - we often toss that word "battery" around without realizing what it is really supposed to mean) has a maximum capacity in terms of how much current it can supply for a given period of time, but you have to realize that it doesn't deliver its energy at any one specific, fixed current. The best model for a battery (at least for the purpose of this discussion) is a voltage source in series with a fixed resistance. A "voltage source," in case you're not familiar with that term, means a theoretical source of electricity which will always maintain the stated output voltage (the potential across its terminals) NO MATTER HOW MUCH CURRENT IS DRAWN. Note that last bit, it's important. The resistance in series represents the unavoidable losses within the battery itself. Now consider what will happen when you place a load (another resistance) across the battery terminals (in series with the voltage source and the "internal" resistance). If that load resistance is very high, little current flows and almost all of the source's voltage appears across the load (since little current means little voltage drop across the internal resistance). A low load resistance draws a LOT of current, but drops proportionally more voltage across the battery's internal resistance (causing the voltage across the load to be reduced). If the load resistance is zero (a "short circuit"), the current is limited only by the internal resistance of the battery, and obviously there is zero voltage across the battery terminals (all the source voltage is being dropped across the internal resistance). That's the maximum current the battery can ever possibly supply, but clearly it's not doing very much useful in that case. So battery capacities are normally stated in terms of the "amp-hours" available at some specified load. You need to know the load current assumed for a given capacity rating, in other words. A battery might be specified as having a 60 A-h rating at a 5A load - which means that it can deliver those 5 amps for 12 hours (5A x 12 hr = 60 A-h). It's not quite as clean as that - obviously, the battery doesn't just deliver a solid 5 A for 12 hours and then immediately shut off - but the specifics are also generally available if you look in the battery specs. Real-world batteries also don't behave quite like the simple voltage-source-plus resistance model, either; for one thing, the internal resistance isn't quite constant, and the voltage tends to drop as the battery's capacity is used up (quite dramatically in some types). But this should give you a little better idea of what you're dealing with here. Bob M. > "DC amps" or how much current the battery produces. *** Yes, that's what i meen. I will try to create this battery till tomorrow afternoon and then i will post some results. Mark p.s. thank you guys for such response ! OK. These are some results from 1V battery. DC Volts : 1V DC Amps: 50 micro The power is between 70 and 50 microAmps , but the real value is cca 50 microAmps. I wil post some additional results in 2-3 hours, and then we can continue on this thema. Mark OK. After some testing and after some new design this is the situation: DC Volts : 1,02 V DC Amps: 500 micro ( 0.5mA) ------------------------------------------------ Becuse i have "old" 1KW and new 500W inverter i will try to make all calculations for 1KW inverter so i can see how much can i take out ( and how long ) from the 48 V battery. ------------------------------------------------ Suggestions, ideas etc, are very welcome :-) Mark >OK. After some testing and after some new design this is the situation: > [quoted text clipped - 8 lines] > >Suggestions, ideas etc, are very welcome :-) --- You're making no sense at all. Try separating what you're doing with the car battery and what you're doing with the battery you're building, so that there's no ambiguity. Better yet, run two threads. SignatureJohn Fields Professional Circuit Designer > Try separating what you're doing with the car battery and what > you're doing with the battery you're building, so that there's no > ambiguity. *** I totaly agree with you. I will make some additional test and then i will start new thread on 48V battery design and DC\AC convertors for such battery. Thank you for your time ! Mark |
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