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3.3V 800mA Linear Voltage Regulator - LD1117-3.3 TO-220

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SEK 19,00


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3.3V 800mA Linear Voltage Regulator - LD1117-3.3 TO-220

https://www.m.nu/web/image/product.template/41948/image_1920?unique=5e2c5cc
Ah the esteemed LD1117, who amongst us has not used this popular low drop voltage regulator? This big chunky regulator will help you get your 4-15V battery or wall adapter down to a nice clean 3.3V with 1% regulation. Perfect for just about all...

SEK 19,00
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Ah the esteemed LD1117, who amongst us has not used this popular low drop voltage regulator? This big chunky regulator will help you get your 4-15V battery or wall adapter down to a nice clean 3.3V with 1% regulation. Perfect for just about all electronics!  This is the TO-220 version, with up to 800mA current capability, and has internal current limiting + thermal shut-down protection which makes it sturdy and pretty much indestructible - at least electronics-wise (we're pretty sure a hammer might work...)

This regulator has a ~1V linear drop-out, better than the 780X series' 2V. That means you must give it at least 4.3V to get a clean 3.3V out. This regulator is often used to get a 5V power supply to a a clean 3.3V. There is a constant 'quiescent' current draw of 5mA.

This regulator can provide up to 800 mA as long as it has proper heat-sinking. The higher your input voltage and output current, the more heat it will generate. Without an extra heatsink, you can burn off up to 2W. We like this calculator for determining your heat sink requirements It's a TO-220 package, so use 62.5°C/Watt junction thermal resistance. The wattage of your set up is = (InputVoltage - 3.3V) * AverageCurrentInAmps. E.g. a 9V power plug and 0.5 Amp of average output current means the regulator is burning off (9 - 3.3)*0.5 = 2.85 Watts! This setup would need a heat sink Or you could use a 5V power supply for (5-3.3)*0.5 = 0.85W which would not require a heatsink.

This regulator requires at least 10uF electrolytic capacitors on both input and output for stability