What would be the least expensive way to boost a DC voltage?
The aim is to convert 1.2 V/1.5 V (from an AA/AAA cell) to 3.3 V to power a small 8-bit microprocessor, like Atmel ATtiny45 or ATtiny2313, and also (if possible) 6 V to power a buzzer.
Also, what would be the maximum current one could draw safely from an alkaline battery, after boosting it to 3.3 V/6 V?
Finally, how I could compute the duration for which the alkaline battery would last, given a certain consumption?
Answer
There's a technique called a charge pump with which you can make a voltage doubler, but that will only give you 3V from a 1.5V cell, and even less from the 1.2V cell. I'm still mentioning it because several microcontrollers these days will work with voltages down to 2V. A charge pump can only supply limited current, enough to power the microcontroller, but extra power devices like motors or relays are out. The voltage will also drop under load. So not ideal. The LM2660 is a switched capacitor charge pump.
The better solution is a switching regulator. These exist in two major topologies: "buck" to go from higher to lower voltage, and "boost" to go from lower voltage to higher. So you want a boost regulator. Major manufacturers include Linear Technologies (more expensive) and National Semiconductor (recently acquired by Texas Instruments). The LM2623 can operate on input voltages as low as 0.8V.
About current and battery life. I'll assume you're working with 1.5V batteries. The ones here on my table are rated for 2300mAh, so let's use that value. Also let's say your microcontroller plus extras need 100mA at 3.3V. That's 330mW. If the switcher is 85% efficient that means it draws 330mW/0.85 = 390mW from the battery. That's at 1.5V, so you'll draw 260mA from the battery. The battery is rated at 2300mAh, then your device can run for 2300mAh/260mA = 9 hour on one charge.
If you plan to load the battery rather heavily, I would remain below 2300mA, which will drain it in 1 hour.
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