When the power fails: designing for a smart meter’s last gasp
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Smart-meter designers have an unusual predicament: The meter receives its power from the same bus that the meter is monitoring. When the meter loses power, it must record state information to flash memory or send out a wireless signal—the meter’s “last gasp.” Some utilities also disconnect subscribers from the grid during power outages to minimize the inrush demands when power is ultimately restored. Disconnecting a subscriber after loss of power also requires stored energy in either electrical or mechanical form.
The problem of efficiently and cost-effectively providing holdup energy typically falls to power-supply designers, who have various options for solving this problem. Here we evaluate the benefits and costs of these options in a flyback switch-mode power supply.
Figure 1 shows a basic offline flyback circuit. The supply accepts 85 to 265V ac and generates a 3.3V-dc, 5W output. The holdup requirement of the load is 50% power, or 2.5W, for 0.5 sec, or 1.25J. The figure highlights three sections for energy storage. Option A stores the holdup energy in the high-voltage capacitor, CBUS. Option B stores the energy in a 20V intermediate-voltage capacitor with a downstream dc/dc buck regulator that steps down the voltage to the load’s working voltage at 3.3V. Option C is simpler and stores energy in a large capacitor at the output.
Because all of the options involve storing energy as electric potential in a capacitor, you should review the relationship of voltage, capacitance, and potential energy, as the following equation shows:
The first option is to increase the capacitance of the high-voltage bulk electrolytic capacitor, CBUS, on the primary side. This capacitor typically stores just enough energy to continue power conversion during the ac cycle valleys—in this case, 1/120 sec, or 8.3 msec, for a full-wave rectified input. When voltage from the line disappears, the converter continues to run and consume energy from CBUS. The voltage on CBUS eventually reaches a point at which the voltage cannot ramp sufficient current through the primary-side inductor to sustain the output. The voltage on the auxiliary output, from which the controller draws its power, also drops to the controller’s undervoltage-lockout level. After this drop, the controller shuts off, and the output falls to 0V (Figure 2).
The supply must hold up the output under all conditions, so we will focus on the worst case: when source power disconnects shortly after the ac zero-crossing at low line input. Because the supply draws half-power during holdup, the supply operates normally. It has enough voltage to impose across the inductor to reach the required primary-side peak current—down to 70% of the minimum input voltage requirement. You can therefore estimate that the supply can operate down to 70V on CBUS. Solve for the capacitance to provide the 1.25J of holdup energy and set V0 to the minimum rectified ac line voltage because the bus-ripple voltage is negligible when the bulk capacitance is large:
This capacitance is large, and these capacitors must have a minimum voltage of 400V. One benefit of using the primary side is that it can store a large amount of energy if the minimum line voltage is fairly high. For instance, if the minimum line voltage were 190V ac, then the same holdup would require only 68F. Figure 3 shows the holdup-time curve for this supply.
Storing energy in the primary side can be expensive. Large-value, high-voltage capacitors can get pricey, and the leakage current through them increases with voltage and value. Most designs try to minimize the primary-side loop to minimize electromagnetic-compliance problems. Using capacitors with diameters of 30 mm may make this task difficult.
You must also take into consideration the efficiency of the power supply. Energy stored in the primary side must be processed through the converter, which includes the switch, the inductor, and the secondary-side diodes, and the efficiency of the converter will decrease this energy. Many flyback converters are 75 to 85% efficient, so the primary-side energy must increase by 20%.