- Written by Brian Keahl
- Category: Emergency Communications
- Published: 23 July 2020
- Hits: 6069
A recent round-table discussion on portable operation with limited power and its application in EMCOMM resulted in a question about solar power as a sole power source for Summit or Parks on the air operation. The question prompted a whole line of reasoning about not just alternative power, but the many factors involved in selecting and utilizing a variety of power sources.
Volts & Amps & Watts (Oh, my!)
The first thing to determine is the voltage and current demands you have to operate as planned, or the voltage and current limitations you face and must operate your equipment within. In an ideal situation you control the former, in some situations, such as EMCOMM, you may need to operate at the limits of the latter.
Since most amateur radio systems can operate off 12 volts (or something around 12 volts, usually ranging from 12-15V), you will normally be working with voltage in that range. If not, then you'll have to ensure voltage is somewhere within the range your equipment will handle. Handhelds, for instance, will often operate at lower power with as little as 7.2 volts input.
While there is some latitude in operating voltage, the source will generally be around twelve volts. Too much voltage and there will likely be equipment damage and failure as voltage limits are exceeded. However, as the voltage is falls below nominal, it will attempt to pull more current in order to draw the needed total power required. Just like too high a voltage, forcing the equipment to draw too much current may result in poor operation and damage the radio or power supply.
So, for the sake of this presentation, we'll assume your operating voltage is within norms for your equipment. That leaves current, as insufficient current limits power output of your equipment or will at least require you limit the power output for the radio to function properly. The thing about current is that the load (in this case your radio) determines how much current will flow - so a 20A power supply will provide whatever the equipment needs, up to 20A. However, if the equipment requires more than 20A the power supply's ability to provide power will break down. So, in an ideal world the more current your power source can supply, up to what your equipment can use, the better for reliable operation.
A good rule of thumb is your radio needs twice the power in to output the selected power. So, for a 60W output radio to transmit at full power, it will need 2x60W or 120W power in. Let's use 12V as input, and knowing Power (W) = Volts x Amps, we need 120/12 or 10A in. This is an estimate, and the actual draw varies from radio to radio, but this is a good rule of thumb.
In a portable/emergency situation you'll likely be operating at much lower power, as low as 10W or 20W, but the same basic rules apply, it's just that your current requirements will be much lower.
The question about solar panels for portable operation really highlights the challenges of power sources, particularly for portable operation. Let's assume the radio being operated transmits at 10W. Using our power rule above, you'll want to be able to draw twice that much, which is 20W. 20W/12V=1.7A. Getting 1.7A out of solar panels would require a fairly decent size panel even under ideal sunny conditions. But, theoretically, it could work. In this case, a better option would be to have a battery capable of sourcing at least 1.7A and using the panels to charge the battery. A big positive of using this configuration is that the radio pulls very little current on receive, and the excess produced by the solar panels can charge the battery during receive, ensuring the battery can provide the necessary current, in conjunction with the solar panel, for transmit. There is the additional benefit that if clouds pass over, the battery allows continued operation.
Maximum Power Output & Amp-Hours
Whatever power source you are using, solar, battery, or A/C powered supplies, there will be a "Maximum Current" rating, indicating the most current the source can provide reliably for extended periods of time. Naturally, if the source is not a power supply, the period of time maximum current can be provided will not be indefinite. But whatever your power source, ensure your maximum current is at least your anticipated draw, but ideally significantly higher. The reasons for this vary, but the rule applies across the board.
Power supplies, usually connected to commercial power, are the most reliable, assuming you have AC power available, of course! However, running the supply at maximum current puts a heavy load on the components. Just like running your automobile at maximum RPMs is not good for your car, running a typical power supply routinely at maximum output is bad for the power supply. Also, in most cases, as the power supply reaches maximum current ripple and noise will appear on the supply output.
In the case of solar power, regardless of the rated maximum output, you are at the mercy of the weather, with maximum current being theoretical, and often being less than published due to available light to the solar panels. The ideal configuration for solar is to have batteries charging from the panel(s) providing power to the equipment.
Batteries have an additional rating besides voltage and max current, called Amp Hours, usually identified as AH or mAH if the rating is in milliamp-hours (1000ma = 1 amp). The rating is actually a measure of the amount of energy the battery can provide over 20 hours, condensed down into an hour. For instance, if a battery can source 1 Amp (1A) for 20 hours before being considered discharged, it would be rated at 20 Amp-Hours (or 20AH). Amp-Hours are often misunderstood, because it is assumed that the amount of current in the rating could be sourced for an hour. Using that reasoning, our hypothetical 20AH battery would source 20A for an hour, or 10A for two hours.. Unfortunately, batteries don't work that way.
Basically, the more current you pull the less you'll get out of the battery. As mentioned, the AH rating of a battery is generally taken by determining how many amps the battery can draw over a period of 20 hours, actually multiplied by 20. As you increase your draw above AH/20 the number of hours the battery will be able to supply it decreases by a disproportionate amount, which varies by battery type. Assuming our hypothetical 20AH battery could actually source 20A, you would likely get anywhere from 20 to 40 minutes, depending upon battery chemistry.
In addition to the variables above, there is the additional fact that our radios don't consistently draw a fixed amount of current. While transmitting at 100 watts (100W) may draw 20A to the radio, that same radio will typically draw under an amp while receiving. So during a portion of the time the battery will likely be asked to provide well above the current used to calculate its AH rating, much of the rest of the time it will be drawing at or below the rate used for the calculation. We already know the discharge curve is not linear, so this is yet another complication.
The best you can do to calculate your possible needs is to figure out current draw transmitting and current draw receiving and the percentage of time you are engaging in each. So, lets say your radio draws 3A transmitting and 1A receiving and you expect to be transmitting 30% of the time. That means your average current draw per hour would be 3x0.30 + 1x0.70 (3 amps at 30%, 1 amps at 70%). So your draw is 1.6AH. Multiply your draw times number of hours you want to operate and you have a rough idea of what you need to get out of your battery. The problem is, the higher transmit current will decrease the actual AH you'll get out of your battery. From my own experience, if you increase that battery capacity by the percentage of time you will be transmitting and you'll be in the ballpark.
As you can see, accurately determining proper battery ratings is more an art than a science. However, most manufacturers publish the charge and discharge characteristics for their batteries, so with a little work and a little math, you can approximate the appropriate sized battery for your application.