aircraft electrical power system holdup … · atrenne integrated solutions 10 mupac drive...

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AIRCRAFT ELECTRICAL POWERSYSTEM HOLDUP REQUIREMENTS


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Atrenne Integrated Solutions | 10 Mupac Drive | Brockton, MA 02301 | 508-588-6110 | www.atrenne.comAtrenne Integrated Solutions | 10 Mupac Drive | Brockton, MA 02301 | 508-588-6110 | www.atrenne.com

AIRCRAFT ELECTRICAL POWER SYSTEM HOLDUP REQUIREMENTS PART 2 OF 2

PART 2 – DETERMINING THE PHYSICAL VOLUME AND WEIGHT REQUIREMENTS

This white paper is the second in a two-part series exploring the standards and design considerations related to MIL-STD-704 and DO-160 power supply hold-up.

A Transfer Operation as defined in MIL-STD-704 is aswitching operation that transfers the aircraft’s electrical power feed from one source to another. In the process of executing the power transfer, there will be a momentary interruption in electrical power supplied to utilization equipment. It is sometimes a requirement that utilization equipment continue to operate during the resultant power interruption. This capability is defined as a hold-up function. Simply stated, the utilization equipment must ride-through a power interrupt without any interruption of its intended function. In such systems, the power supply within the equipment must have sufficient energy stored in reserve to draw upon during these dropout events. A hold-up requirement can be applicable to systems of various power levels, and hold-up time durations can range from 50 mS to up to 1 full second.

OVERVIEW Part 1 of this series reviewed the standards and specifications that define airborne electrical power hold-up requirements. The system designer was guided through the process of determining if there was a need to specify a hold-up capability within the power supply system, and what standards defined and bound the technical requirements. The goal was to provide the system designer with the knowledge so that he or she will have been able to determine if their proposed Utilization Equipment needs to comply to MIL-STD-704, RTCA DO-160, (or both) as it pertains to Transfer Operation hold-up. It is hoped that he or she will have been able to determine whether some or all of the system functions have a critical need to operate through a Transfer Operation power interrupt. Ultimately the designer will have gained the knowledge to understand the system power delivery requirements during the interrupt, and will have determined the required duration of the hold-up.

In any airborne application, physical volume and weight requirements must be considered. This installment will help the system designer to understand and quantify the volume and weight penalty associated with the hold-up performance parameters that have been chosen. This second installment walks the reader through a top level design example with the goal of providing him or her with the knowledge to develop their own estimates of the volume and weight requirements associated with the hold-up function within the power supply subassembly of the airborne Utilization Equipment.

Storing energy takes physical space. For systems that draw hundreds (or thousands) of watts, providing power hold-up for even a 50mS Transfer Operation interrupt can more than double the volume, weight, and cost of the power supply subassembly. Some applications require extended hold-up times of 200mS, or even a full second of time. Such requirements will greatly add to this penalty. The following text will first provide an overview of how a basic hold-up function is implemented. It will then detail a more efficient architecture and provide methods to estimate the volumetric impact of such a design. The calculations are simple to execute once the requirements are understood. Performing these calculations early in the design process will give the system designer insight into the potential volume and weight costs to be incurred.

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This example will present a typical MIL-STD-704F 28VDC (22V to 29V) application. It is assumed that the system requirements are to deliver 500W for a minimum 50mS duration. Electrical power hold-up in most systems is accomplished by maintaining a capacitor bank charged and ready to provide energy to a downstream voltage regulator (or regulators) when needed. In it’s simplest implementation (see diagram above), a large value capacitor bank is electrically connected across the power bus to store reserve energy for use by the downstream regulator(s) during power interrupts. Note that all voltage regulators have an input voltage range of operation, and will only function within their specified range. In this example, the regulator design will accept a DC input voltage as low as 10V. Below this range the regulator will cease to deliver power to the load.

This simple architecture has a couple of significant drawbacks. First, the capacitor only has the range from the nominal 28V input down to the 10V regulator lower threshold to provide energy to the load. In this example, to accommodate a 50mS power interrupt in a 500W system, the capacitor size will need to be quite large ( >74,000 uF). If a power interrupt occurs, the capacitor will source power to the regulator until its charge drops from nominal down to the 10V threshold of the converter(s). If at the end of the 50mS period the input power is not restored, the regulator will shut down. Note that the system will have left the capacitor charge voltage at approximately 10V. In this simple example, this leaves 12.7% of the stored energy in the capacitor. This

HOLD-UP BASICS


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DC/DC Converter28VDCOutput500W

With Or Without Output Converter(Depending upon Requirements)

+Vin

-Vin

+Vout

-Vout

28VDCInput

EMI Filter & TransientSuppresion

74,000 uF Holdup Capacitor Bank

+Vin

-Vin

+Vout

-Vout

remaining energy will not be harvested. Secondly, if the 28V input bus happens to be running near to the MIL-STD-704F Normal Low Steady State (NLSS) voltage of 22V when the interrupt occurs, there will be less than 30mS hold-up time available before the capacitor discharges to the 10V level where the regulator shuts down. Increasing the capacitor value to >130,000 uF would increase the hold-up back to 50mS, but at a very significant volume and weight penalty.

A more efficient method would be to implement a system where the downstream regulator(s) are operated at a much wider input voltage range between say 60V down to 10V. In this architecture, a voltage boost circuit would be utilized to charge and maintain a storage capacitor bank at a 60V level. Such a boost circuit can generate this higher voltage with very high efficiency. During a voltage interrupt event, the capacitor will begin supplying current as before, but now starting from a 60V charge. During normal operation, the 60V will be maintained across the capacitor even if the input voltage were to drop to the 22V NLSS value. In the event of a power interrupt, the regulator will continue to provide output until the 60V capacitor charge drops below the regulator’s lower 10V input threshold. As long as the input power resumes nominal voltage before the lower threshold is reached, the system will function throughout the Transfer Operation event. This architecture will harvest over 97.2% of the capacitor’s stored energy, thus significantly improving volumetric efficiency. This configuration requires a significantly smaller capacitor value (approx. 14,300 uF),


and less than 2.8% of the stored energy is left in the capacitor after the discharge cycle. This architecture maintains the same energy storage as long as the input voltage is within normal parameters, and significantly more “usable” energy will be harvested from the storage capacitor. Also storing energy in a capacitor at a higher voltage allows for more energy storage for a given capacitor physical volume, further improving volumetric efficiency of the hold-up function. This configuration will be reviewed in more detail in the following section.

As illustrated above, meeting hold-up requirements is an energy storage challenge, and physical volume (space & weight) is the cost incurred.

CALCULATING SYSTEM STORAGE REQUIREMENTSFor the following exercise, it is assumed that the following three pieces of information have been determined.

• Which specific system functions will require hold-up.

• The actual power consumption requirements of the above functions needing hold-up.

• The hold-up duration.

The joule is a unit of measure of energy in the International System, and in electrical terms is equal to the energy it takes to deliver one watt of power for one second. For this example, the designer will need to determine how many usable joules of energy will need to be stored in reserve to support the hold-up requirement. We say “usable” as a reminder that some energy will remain in the capacitor


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+Vin

-Vin

+Vout

-Vout

28VDC Input(Cont. 22V to 32V)

EMI Filter & TransientSuppresion

+Vin

-Vin

+Vout

-Vout

Sequencing, Inrush, &Reverse Blocking

InputRange

10 VDCto

60VDC

+Vin

-Vin

+Vout

-Vout

60VDC Boost Driver 60 VDC

16,000 uF Holdup Capacitor Bank

ThresholdDetect,

Control, &Switching

+Vin

-Vin

+Vout

-Vout

DC/DC Converter28VDCOutput500W

USEFUL FORMULAS

Power / Energy / TimePW = E(J) / t(s)E(J) = PW * t(s)

To calculate required C when you know Power, Time, Vu and VL

C = 2*(P*t) / ((VU)2 - (VL)2)

To calculate energy (joules) stored in capacitor

E(J) = 1/2 * C * V2

PW = Power in Watts

V = Volts

E(J) = Energy in Joules

t(s) = Time in Seconds

C = Capacitance in uF

VU = Max charge Volts

VL = Min charge Volts


unharvested. Lets assume it has been determined that the system requirements are to deliver 500W for a minimum 50mS duration.

Power available in watts is equal to the energy in joules (J) divided by the time period in seconds.

PW = E(J) / t(s)

Rewriting the above formula and plugging in the above numbers yields the minimum number of joules of energy that are required to be delivered to the load for a 50mS period.

PW * t(s) = E(J)

500W * 50mS = 25 joules

This calculation shows that a minimum of 25 joules of stored energy are required to source 500W of power for 50mS. Note that this example is assuming 100% efficiency. Real world physical constraints create inefficiencies in both the energy transfer process (dissipation losses), as well as in the energy storage process (dielectric losses). The power system designer will add margin to cover these conversion and dielectric losses.

As discussed earlier, hold-up is accomplished by keeping a capacitor bank charged and ready to provide energy to downstream converters when a power interrupt occurs. As it provides this energy, the capacitor voltage will discharge from an initial full charge value. Such a system can only harvest the usable energy between the upper capacitor charge voltage, and a lower voltage threshold where the downstream converter(s) will cease to function.

The setting of the 60V upper charge voltage (VU) is based on practical boost circuit limitations, as well as upper input limits of typical regulators. As discussed in the previous section, the 10V lower threshold (VL) is a typical lower input limit of wide input range downstream converter(s). As described above, it should be clear that no further energy can be harvested from the charged capacitors once the charge falls below the lower threshold of the converters. With the information derived from the calculation above, it is clear that this system must be designed to maintain the upper charge on the capacitor to be at least 25 joules higher than what would remain in the capacitors when discharged to the lower voltage threshold.

Knowing the need to deliver 500W for at least 50mS, the system designer can now proceed with some calculations to determine the capacitor requirements.

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The next step is to determine how much capacitance will be required to deliver 500W for 50mS, and have better than 10V of charge remaining at the end of the period. Remember that the system will leave some energy in the capacitor even after a full 50mS discharge.

The following formula will calculate the capacitance value for a hold-up function as described above:

C = 2*(P * t) / ((VU)2 - (VL)2)

Capacitance (C) is in microfarads. Plugging in the values for power (P), time (t), upper charge volts (VU) and lower charge volts (VL), the system designer will arrive at the following conclusion:

2*(500*0.05) / ((60)2 – (10)2) = 14,286 uF

To meet the hold-up requirements, the system will require a minimum of 14,286 uF of capacitance. Remember the designer will need to increase the capacitance by some percentage in order to compensate for losses and to provide margin for operating temperature, dielectric losses, etc.

To prove out the above calculations, and to understand the total energy (in joules) stored in the capacitor, the system designer can use the following formula:

E(J) = ½ * C * V2

At the maximum charge of 60V: ½ * 14,286 uF * 602 = 25.71 joules

At the minimum charge of 10V: ½ * 14,286 uF * 102 = 0.71 joules

At a 60V charge there will be 25.71 joules of energy stored. Since 25 joules are required to supply the energy for the hold-up event. At the end of the discharge, 0.71 joules will remain. It can be seen that during the hold-up event, this architecture will harvest over 97% of the energy stored in the capacitor leaving less than 3% remaining. This is significantly better energy storage utilization when compared to the first example discussed earlier in this paper where 12.7% of the charge was left unused. From a capacitor volume and weight perspective, using an efficient boost circuit to charge to a higher voltage is a better solution. There are more exotic circuit topologies in use that are designed to squeeze even more efficiency into the hold-up function, however this is beyond the intended scope of this paper.


CAPACITOR BANK CONFIGURATION:Once the energy storage requirements are known, the system designer can begin to investigate suitable capacitors. These must be sized properly to meet the requirements with proper energy storage margins for efficiency, operating temperature range, and voltage. These margins depend on his or her system requirements and are left up to the designer. Once completed, the designer will have a better idea of the physical space and weight penalty associated with the hold-up function.

There are several voltage and capacitor combinations that could be chosen for this application. A common choice for such systems is to use Cornell-Dubilier flat pack capacitors as shown on the right. For our example, Part Number MLP332M080EB0A is a 3300 uF device with a voltage rating of 80VDC (100VDC Surge). These parts are quite stable over a wide temperature range. They have a long operating life, and are ideal for airborne applications. An array of five 3300 uF capacitors would provide 16,500 uF of capacitance.

In this application the capacitor bank is simply an energy storage block, and follows the simple formula based on joules of energy storage. Any reasonable requirement can be supported using different combinations of devices. A common alternate capacitor choice to the flat-packs is to design a printed wiring board to use banks of radial lead cylindrical electrolytic capacitors.

VOLUME, WEIGHT & COSTWhether the system is designed using flat pack, cylindrical, or other capacitor configurations, physical space will be consumed. Referring to the 28V/28V Hold-up w/Boost block Diagram presented earlier in this paper, note the circuit blocks shaded in gray. These blocks together comprise the additional hold-up function added to a basic power supply subassembly. The boost driver, threshold detect, control and switching blocks will consume no more than a 3” x 5” area of a printed wiring board, for a volume of approximately 11.3 cu. in. It will weigh less than 1.5 pounds. The five flat-pack capacitors will consume a total volume of 13.2 cu. in. The five capacitors will weigh approximately 0.8 pounds. Be sure to take into account mounting and packaging of this additional hardware within the power supply system. It can be seen in the above tables that including the hold-up capability into this system will consume approximately 24.5 cu. in., and will have a weight of approximately 2.3 pounds.

Note: The first design example would have required a quantity of twelve 11,000 uF capacitors of the same physical size. The capacitor volume would have increased from 13.2 to 31.5 cu. in., and the capacitor bank weight would have increased from 0.8 to approximately 1.75 pounds.

While at first glance the above volume and weight estimates may not seem significant, it can quickly prove to be a challenge when trying to include this functionality into a 3U VPX airborne chassis. Another option to achieving very high storage density is to use hybrid capacitors such as those shown below.

These devices are a tantalum hybrid manufactured by Evans Capacitor Co. Tantalum hybrid capacitors are being used in high reliability applications (usually airborne) where volume and weight constraints are paramount. They achieve a storage density per unit volume that can be 4 times greater than the flat-pack electrolytic devices described above.

SUMMARYThe system power supply will be required regardless of whether there is a hold-up requirement. The increased volume & weight required by the hold-up function can be estimated using the techniques described above. The information in this paper will help the system designer to understand the volume, weight, and cost impacts that the hold-up function adds to the overall system power supply.

AUTHORDavid R. Masucci is a Program Manager at Atrenne Computing Solutions in Littleton, MA.

CONTACT [email protected] or 800.926.8722

The information in this document is subject to change without notice and should not be construed as a commitment by Atrenne Integrated Solutions. While reasonable precautions have been taken, Atrenne Integrated Solutions assumes no responsibility for any errors that may appear in this document. All products shown or mentioned are trademarks or registered trademarks of their respective owners.

Cornell-Dubilier Flat Pack Capacitors


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