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What to do when your power source is higher than your power requirements

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GeorgeOfTheJungle:
Thought so too, but double the voltage would be 4x the power => 4*0.78 = 3.12W, still not 2.2W  :-//

dazz:

--- Quote from: GeorgeOfTheJungle on August 30, 2018, 01:01:10 pm ---Thought so too, but double the voltage would be 4x the power => 4*0.78 = 3.12W, still not 2.2W  :-//

--- End quote ---

Well, it's not 100% efficient. According to the datasheet it can do 2.5W at 5V 4 ohms, 1% THD. So about an 80% efficiency which seems about right

GeorgeOfTheJungle:
Oh, I see, that must be, it's a switching amplifier and the figure comes from the datasheet, lol. Thanks!

tooki:
It is a class D amp. In my testing of that chip, I found that (at around maximum volume without clipping, into 8 ohm speakers), its current had 8 amp transient peaks. (Average current about 800mA.) So capacitors are essential for operation on power supplies that provide less than that peak current.

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