Re: How big a battery?.... (long) Andy Fox (foxes@theriver.com) Tue, 02 Jun 1998 09:22:17 -0700 * Messages sorted by: [ date ][ thread ][ subject ][ author ] * Next message: Bruce Rattray : "'scope manual (fwd)" * Previous message: David Gauding : "Re: SLV FOR FIELD DAY" --------------------------------------------------------------------------- Hi Michael, There are a couple of ways you might want to approach this. The first is based upon your power requirements (recommended) and the second is based upon what you have for a photovoltaic (PV) module (next best). You generally want to think in terms of Energy, not Power. Power is an instantaneous phenomenon: that 60W light bulb uses 60 Watts at any given instant. Energy brings time into the equation (product of Power and Time). Your electric bill does not show how much Power you use (it's changing all the time); it shows the amount of Energy "consumed" by you during the billing period. You may have used 300K Watt Hours last month. At 10 cents per KWH, that would come out to $30. Maybe you used 10 KWH each day for 30 days, maybe you used 20 KWH each day for fifteen days. OK, on to the fun part. You may find it easier to use current instead of power, if all your voltages are the same. This is very common. Units are Ampere Hours (AH). Batteries are rated this way. I've made up some values for the following examples. They are for example only; they do not represent a real rig. First Method: How much power does your rig "take" during transmit? P(t) [4 W] How much power does your rig "take" during receive? P(r) [0.25 W] How much time will the rig be transmitting between charges? T(t) [10 H] How much time will the rig be receiving between charges? T(r) [20 H] Calculate the energy for transmit: P(t) * T(t) [40 WH] Calculate the energy for receive: P(r) * T(r) [5 WH] Add them up to find the total amount of energy: 45 WH Convert to Ampere Hours (divide by the voltage, I'll assume 12): [3.8 AH] Batteries are not 100% efficient; You will only get out of them about 85% of what you put in. You have just calculated what you need to get out of the battery. Now, divide by 0.85 to figure out how much how much to put in: [3.8 / 0.85 = 4.5 AH] Another thing about batteries. If you want them to last for a while, don't run them all the way down. For lead-acid batteries, 50% discharge is a good figure. Divide by 0.50 (or multiply by 2) to find the size of the battery: [4.5 / 0.50 = 9 AH] So, for the virtual rig above, a 9 AH battery would be swell. How large a PV module do you need? It depends. How much time do you have, where do you live and what season is it? I can count on six "peak sun" hours per day (yearly average) in sunny Tucson. Sure, the sun shines for more hours than this, but it's going through more atmosphere in the early and late part of the day, so it is attenuated somewhat. There are charts that show "peak sun hours" for the planet. Fire up that search engine! If you want to charge the battery in the example above up from halfway, that's 4.5 AH. --- intermission --- I'll stop here for now. I need to get some work done. Besides, this is one looooong e-mail already! -- 72/73 de Andy, KK7HV - QRP-L #1286 - Tucson, AZ