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Select The Right Ultracapacitor Solution

Nov. 5, 2009
Alex Pastos tells you how to choose the ultimate ultracapacitor for your design.

Like all capacitors, ultracapacitors have a high power density. Yet unlike their traditional counterparts, electrolytic capacitors, ultracapacitors offer high energy density, allowing them to store a vast amount of energy in a small package. The capacitors that most design engineers are familiar with have very short time constants, which means their voltage cycles quickly. Ultracapacitor arrays, though, have time constants on the order of tens of seconds to minutes.

The large capacitance and extremely low-frequency time constants enable ultracapacitors to be used in applications that have not been practical or economical for other types of capacitors. Since ultracapacitors are still rather new to the electronics industry, few people are aware of their existence, much less how to use them.

While ultracapacitors store a large amount of charge, they are still well below the energy density of storage batteries. Batteries in general will have 10 to 30 times the energy storage of ultracapacitors of comparable masses. The Ragone chart illustrates the relative power and energy densities of various energy storage devices (Fig. 1). There aren’t many situations in which an ultracapacitor solution can replace a battery outright.1

But since ultracapacitors have a much lower internal resistance and much faster charge rate than batteries, they can make a battery-powered system run much more efficiently. An array of ultracapacitor cells in series coupled to a load in parallel with a storage battery creates a hybrid power source with higher power and energy density than either device in a standalone configuration.

ULTRACAPS ON THE ROAD
One practical application for ultracapacitors pairs them with the starter battery for an internal combustion engine (ICE). Starter batteries are typically sized for starting ICEs in 0°F (–18°C) conditions. Yet in many instances, automotive starters are required to operate well below 0°F, as anyone in northern climes is well aware.

Unfortunately, a flooded lead-acid battery can’t handle temperatures much below 0°F and maintain an appreciable power output. The power output from starter batteries decreases as temperature decreases due to an increase in internal resistance. On the other hand, an ultracapacitor can work down to –40°F (–40°C) with very little increase in its internal resistance, making the combination of battery and ultracapacitor a robust power source even at very low temperatures.

Suppose an engineer wanted to reduce the starter battery size, but still wanted the ICE to start reliably in below freezing temperatures. To do so, it’s necessary to determine the power required to crank the engine and determine the energy consumed in the process of starting the engine.

As an example, assume that under normal conditions the load for the starter motor is 0.025 Ω and that the internal resistance of the battery is 0.015 Ω. The current draw from the battery under these conditions will be 12 V/0.04 Ω = 300 A. In this instance, a large amount of the power is dissipated as heat due to the internal resistance of the battery causing an I2R loss.

Using these figures, the power delivered to the starter motor is 2250 W. An additional 1350 W is dissipated as heat due to the battery’s internal resistance. If it is further assumed the starting process takes 3 seconds, the total energy consumed is 10.8 kJ (3 Wh); 6.75 kJ to the starter motor and the balance, and 4.05 kJ dissipated as internal heating in the battery.

In subfreezing conditions, the battery’s internal resistance will increase. Suppose it doubles for this illustration, which reduces the total current that can be drawn from the battery to 218 A, or nearly a 30% reduction. The power that can be delivered to the starter motor under these conditions is reduced to 1190 W or about 50%. Based on this example, it will be more difficult to start the engine since the battery can only supply half the power it could in warmer conditions.

Now pair the battery with an ultracapacitor device to produce a more robust power supply to be used in cold weather conditions. Using the figures above, the new power source must supply 2250 W for 3 seconds and have sufficient power to start the ICE and to supply power to the other electrical systems. It is desirable to use ultracapacitors to provide the power for engine start and the battery to provide low-power energy for accessories when the engine is off.

Under this scenario, the energy required to start the ICE is given as 6.75 kJ. The equation for energy stored in the capacied tor is ½ CV2. Using this relationship and setting a requirement that sufficient energy to start the ICE must be stored in the ultracapacitor bank without dropping below 9 V, the capacitance must be at least 215 F. A 15% cushion is added to be safe, bringing the required capacitance to 250 F.2

WORKING IN SERIES
Ultracapacitors have a maximum cell voltage of 2.7 V, so they must be connected in series to reach the required working voltage. With any identical capacitors, the capacitance of a series array goes down as the capacitors are connected in series, but the working voltage increases by the rated voltage of each additional cell.

A six-cell lead-acid battery requires six ultracapacitors, because the maximum voltage a 12-V battery can be charged to is 14.4 V. With five ultracapacitors, the maximum voltage across each cell would be 14.4 V/5 = 2.88 V, which would cause premature failure of the cells. At highervoltage battery configurations, it’s possible to have slightly fewer ultracapacitor cells than lead-acid cells. In general, though, the cells are equal to the number of lead-acid cells when directly connected in parallel with the battery.

Since a minimum of six cells is required and 250 F was the minimum capacitance, the cell capacitance has to be at least 6 × 250 F or about 1500 F. Several manufacturers offer different sizes of ultracapacitors close to this capacitance. This example will use 2000 F prismatic manufactured by Ioxus (Fig. 2). The equivalent series resistance (ESR) specified for these cells is 0.0006, resulting in a total ESR of 0.0036 .

As can be shown when two power sources are connected in parallel, the current draw from each will be inversely proportional to the internal resistance of the devices. In this case, theory predicts that at the instant the ignition key is engaged, the ultracapacitor bank will provide most of the current and will supply approximately 80% initially. The battery will gradually pick up load as the capacitor discharges. However, the initial current supplied by the ultracapacitor bank will start the ICE and reduce the current drain on the battery.

The parallel configuration will present a lower internal resistance as seen by the load. Again using the numbers in this example, the internal resistance of the power source is about 3 m. When the starter motor load is connected to the power source, it will draw about 415 A, providing about 4300 W, which is more than enough to start the ICE.3 Because the internal resistance of the ultracapacitor bank is so much smaller than that of the battery, the maximum I2R internal loss is on the order of 600 W, and the smaller current drawn from the battery reduces the initial I2R loss of battery to about 150 W.

In general, the use of an ultracapacitor in combination with a battery is an excellent way to increase the overall power density of the power source and decrease the strain on the battery. As illustrated in this example, a smaller battery could be used since the available power of the hybrid power source is so much more than required. In any case, where energy storage with highpeak power is required, it is very likely an ultracapacitor device will be useful.

References
1. For volatile memory backup and clock backup power in computers and other electronic devices with small energy consumption, an ultracapacitor can be used without a battery.

2. The energy required is 6750 J = ½ C(V2upper – V2lower); C = 2 × 6750/ (12.02upper – 9.02lower) = 214 F

3. The currents from the battery and ultracapacitor bank can be calculated more precisely, and the method may found at www.ioxus.com in an application note.

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