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Question about UBECs


G+_Ben Reese
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Thanks for the replies. My thought is that a transformer will change voltage, but have little impact on the overall wattage (I could be wrong on this assumption); whereas a voltage divider using resistors would reduce the voltage and available wattage at the same time - possibly even impacting amperage more than voltage.

 

My initial assumption is that the UBEC may not be able to boost the amperage as it reduces the voltage.

 

Reason for the question... I am wanting to install some WS2812 LED strips. Over the years I've collected several old laptop power supplies, so I'm hoping I can use a 5V 3A UBEC coupled with a 19v 2A power supply to power things. For example, here's a 3A that will take up to 26V as the source: https://www.banggood.com/3A5A7A15A-BEC-Brushless-UBEC-For-FPV-Receiver-p-1007444.html

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Haven't used a UBEC module myself, but it is most likely a form of a buck converter. The trick with this is to store pulses of energy in a magnetic field, and feed it slowly to the load. The 12 volt input voltage is fed to a large capacitor. This capacitor allows a high (3+ amps) current to be switched into an inductor. The output of this inductor charges another large capacitor. When the voltage on this capacitor reaches 5 volts, the input current to the inductor is switched off. A diode from ground to the input of the inductor allows it to continue to dump energy into the output capacitor, maintaining the 5 volt output (it's actually a bit more complex than this, but close enough to get an idea of how it works).

 

This switching is a form of Pulse Width Modulation (PWM) and happens at a fairly high frequency. The high frequency allows the use of a smaller inductor, as it doesn't have to store very much energy at any given point in time. So, because of this switching action and the input capacitor, the input current draw of the converter can average around two amps, even though the inductor (and thus the output) current is up around three amps. The inductor is effectively transforming a higher input voltage to a lower output voltage, As a side effect, it is also transforming a lower input current to a higher output current.

 

Pin = Pout + some losses

- or -

Vin x Iin = Vout x Iout + some losses

 

See https://en.wikipedia.org/wiki/Buck_converter for a painfully detailed explanation. (Note: doesn't show the input capacitor, for some reason.)

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