First, you should know the units in which power usage is defined:

• Volts (V)

• Amps (A), milliAmps (mA) -- Units of current

• Watts (W) = Volts * Amps. This is the only unit that is officially called "power". Volts are Volts and Amps are Amps.

• Watt-Hour (WHr) = a constant one watt over the period of one hour

With these units, you can calculate everything; how much power your device takes, how much power you need to run for 5 hours, and how much power density there is in that battery.

How much power does your device take?

Using your standard multimeter, you can measure both the voltage going into your device, and the amps. I recommend you start by setting your meter to its highest mode (typically 10 amps). Most devices have a high startup current, and if you use a lower mode, you will probably blow a fuse if you're lucky, or worse, permanently damage your meter.

Advanced method for measuring current

• To avoid ever blowing up a meter again, use this trick instead. Put a 0.1 ohm meter in line with your device, and measure the voltage across it. Make sure to use a large enough wattage resistor, and don't touch it! It could easily get hot enough to burn you, if you're working with high currents.

• How do you calculate current from the measured voltage? Use the formula Volts = Amps * Ohms (V=IR) and rearrange it to Amps = Volts / Ohms. Since Ohms in this case is 0.1, you simply multiple your Volts by 10 and you have Amps.

• How large of a resistor should you use? Use the formula Watts = Volts * Amps (P=IR) and substitute Volts from the formula above and you get Watts = Amps * Amps * Ohms. For example a 0.1 Ohm resistor rated for 5 watts would be able to handle... 5 Watts / 0.1 Ohm = Amps ^2. sqrt(50) = Amps. Amps = 7.07.

• How do you know if you've exceeded your wattage rating? Volts = Amps * Ohms. Ohms is fixed at 0.1 and Amps should not exceed 7.07. Therefore, your measured voltage across the resistor should not exceed 7.07 Amps * 0.1 Ohms = 0.707 Volts.

So now you know how many Volts and Amps are going into your device. How is this useful? Simply multiply the two, and you get a Watt rating. Most PC/104 devices require a regulated 5 Volts input, so simply muliply: 5 Volts * 2 Amps = 10 Watts.

How much power can a battery provide?

Batteries come in various voltages, and are rated for specific Amp-Hour or Watt-Hour ratings. Typically batteries come in Amp-Hour ratings, which tells you absolutely nothing useful until you combine it with Voltage. For example: Which battery is better? Battery number one with a 1 Amp-Hour rating, or battery number two with 10 Amp-Hours? The answer is battery number one, because it has a higher Wattage rating than number two.

Do the math: 100 Volts * 1 Amp-Hour = 100 Watt-Hours (think 100 watt lightbulb running for one hour). 5 Volts * 10 Amp-Hours = 50 Watt-Hours (think 50Watt lightbulb running for one hour).

So take a typical NiCd C size battery from http://www.digikey.com/ (Choose Catalog -> Batteries -> Nickel-Cadmium). Their catalog lists several different capacities. Naturally, we want to play with the largest capacity batteries, because they'll give us the longest runtime. This leads me to choose a Panasonic 2300mAHr battery. Nominal voltage for a NiCd battery is 1.25 Volts, so 1.25 Volts * 2.3 Amp-Hours = 2.875 Watt-Hours.

How much runtime?

If you had to run your 10 Watt device for one hour, you need a battery with 10 Watts * 1 Hour = 10 Watt-Hours of power in it. Given that a single C battery is only 2.8 Watt-Hours, this means that you need at least four C size batteries to run for one hour.

If you want to run your 10 Watt device for eight hours, you need a battery with 10 Watts * 8 Hours = 80 Watt-Hours of power in it. Given that a single NiCd C battery is only 2.8 Watt-Hours, this means that you need at least 29 C size batteries!

How much weight?

This is fairly straight forward: once you know how many batteries you need, you just multiply the number of batteries by the weight of each battery. The C size batteries listed above weigh 2.75 ounces. Four batteries (2.75 oz * 4 batteries) weigh 11 ounces, less than an pound (16 oz). 29 batteries (29 batteries * 2.75 oz) weigh a hefty 79.75 ounces, or 4.98 pounds.

For those who prefer metric, the batteries weigh 78 grams, so four batteries weigh 312 grams and 29 batteries weigh 2262 grams!

Naturally, you can reduce your weight load by spending more money. NiMh batteries are more expensive, but they weigh less and contain more power. Li+ batteries push this envelope even further.

How much power density?

Power density is a measure of Power per Pound (or your favourite weight unit).

Compare the various NiCd batteries:

AA Cell = 1.25V * 1000mAHr / 23 grams = 54 milliWattHours per gram C Cell = 1.25V * 2300mAHr / 78 grams = 37 milliWattHours per gram D Cell = 1.25V * 5000mAHr / 145 grams = 43 milliWattHours per gram

Compare various NiMh batteries:

• AA Cell = 1.25V * 1500mAHr / 26 grams = 72 milliWattHours per gram
• SC Cell = 1.25V * 3000mAHr / 55 grams = 68 milliWattHours per gram
• D Cell = 1.25V * 6500mAHr / 170 grams = 48 milliWattHours per gram

NOTE: SC stands for Short-C, which is a physically different size (it's shorter, duh!) than a standard C size battery.

And further calculations can tell you cost per Watt-Hour

• NiCd AA = $3.16 / 54 mWHr = $3.16 / 0.054 WHr = $58 per WHr
• NiCd C = $7.84 / 37 mWHr = $7.84 / 0.037 WHr = $211 per WHr
• NiCd D = $12.93 / 43 mWHr = $12.93 / 0.043 WHr = $301 per WHr
• NiMh AA = $3.27 / 72 mWHr = $3.27 / 0.072 WHr = $45 per WHr
• NiMh SC = $9.30 / 68 mWHr = $9.30 / 0.068 WHr = $137 per WHr
• NiMh D = $18.22 / 48 mWHr = $18.22 / 0.048 WHr = $379 per WHr

Of course, you can get MUCH better prices for these batteries than at Digikey. I leave that exercise upto the Savvy Internet Shopper.

-- KevinWang - 01 Feb 2002