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1

Although, I don't understand where an increased power-consumption would come from when the INPUT pin is unconnected and no pull-up is active. This part of your question already has 2 answers, so I'll just answer the "power saving part". The data sheet says "If some pins are unused, it is recommended to ensure that these pins have a defined ...


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The theory here is that an pin without a pull-up will probably float somewhere around the switching voltage. A bit of noise from the surroundings or whatever might push the pin up or down slightly, which will cause the internal circuitry to "flip" from high to low (or vice versa). All that flipping consumes quite a lot of power. The little tiny ...


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Current is consumed when an input pin switches from HIGH to LOW or from LOW to HIGH. An unconnected "floating" input pin can change randomly depending on the local environment (EMI, noise, etc). That is unpredictable. The "it will consume more" is erroneous - it should be "it could consume more". The could is important, as I ...


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In regards with your schematics Are you sure you want your stuff working only when both D3 and D5 are up? Because it seems to me you connected them in series, so if one transistor is down, nothing's gonna work. check the power rating of the ultrasonic transducer, if you get your power source from the computer USB port, you will get at best about 2.2W from ...


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About the ultrasonic fogger: The product description is misleading. This is a pure transducer. That means it does not generate its own ultrasonic frequency and is thus not DC. You need to provide an AC voltage at the frequency of 113kHz (though one reviewer states to have received products with a resonance frequency of 108kHz. Maybe just try out what you ...


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