EDN
Some battery-powered devices require large amounts of current in a short period of time but spend most of the time in sleep (power-down) mode. The momentary large-load current demands large batteries to meet the time requirement, even though the average current consumption is low. For instance, a system operates for 1.5 sec every 10 hours and needs 500 mA at 3.3 V during the operation. Although the average current is only 21 µA, small, “coin” batteries cannot drive such a heavy load.
To eliminate the need for larger batteries, the circuit in Figure 1 solves the problem by gradually building up energy in a supercapacitor. The device releases the energy when it is needed. Because the supercapacitor has low internal impedance, the momentary current can easily exceed several amperes.
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| Figure 1. | A supercapacitor helps a small battery to deliver large pulses of energy. | |
Because a coin-type lithium battery delivers 3 V and the supercapacitor's rated voltage is 2.5 V, the circuit uses a voltage-controlled switch to cut off the battery once the voltage on the super-capacitor reaches 2.2 V. This design uses a 1.5 F, 2.5 V supercapacitor from PowerStor, model A1030-2R5155. IC1, a micropower voltage comparator with built-in 1.245 V reference, senses the voltage on the supercapacitor. R3 provides 0.5 V hysteresis to the comparator. When the voltage is lower than 1.7 V, the comparator's output is low and thus turns on the p-channel MOSFET, Q1. The battery charges the supercapacitor. Once the voltage on the supercapacitor reaches 2.2 V, the comparator switches high to shut off Q1. You could use this low-to-high transition at Point A as a battery-charge-complete indicator or to trigger another device, such as a microcontroller's interrupt line.
Q2, another p-channel MOSFET, controls the discharge of the supercapacitor. When Point B is floating, the switch is off. When an open-drain or open-collector device pulls down Point B, the switch is on. Because the voltage on the supercapacitor continuously drops when the switch is on, you can use a boost dc/dc converter to generate a constant output voltage. Select a boost converter with the lowest possible working input voltage to obtain the maximum energy from the supercapacitor. For example, you can use an LTC3402 to generate a stable 3.3 V output. Once it starts, the LTC3402 can work with input voltages as low as 0.5 V. The energy from the supercapacitor is
1/2V2 C,
or
1/2[(2.2V)2 ×1.5F–(0.5V)2 ×1.5F]=3.4J.
