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JacobsParts Inc.

JacobsParts Inc. FP5139 100W DC Boost Step-up Converter

SKU CH-F2BN-ANXCJ

Steps up 3–35 V DC input to 3.5–35 V DC at up to 6 A with 96 % efficiency and built-in voltmeter

  • Type100 W
  • Efficient 96% boost conversion reduces energy loss
  • 100W maximum output when input and output exceed 20V
  • Wide 3-35V input range suits diverse power sources
  • Built-in voltmeter simplifies voltage monitoring and adjustment
  • Compact 68x42x14mm module fits tight project spaces

Boost Converter for Low-to-High Voltage Projects

The JacobsParts FP5139 is a dedicated 100 W DC boost step-up converter that raises a lower input voltage to a higher, regulated output. It suits hobbyists and engineers powering 5 V USB gear from a 3.7 V lithium cell, driving LED strips from a 12 V battery, or similar automotive and electronic projects where the supply is always below the load voltage.

Verify Input Range and Polarity Before Wiring

The module accepts 3 to 35 V DC at up to 9 A on the input side and delivers 3.5 to 35 V DC at up to 6 A on the output. Output must always exceed input voltage; reverse polarity will damage the unit. The onboard voltmeter only functions when the input is at least 4 V, so confirm your source meets that threshold before relying on the display.

Efficiency Drops With Wide Voltage Gaps

Peak conversion efficiency reaches 96 percent, yet the 100 W rating applies only when both input and output exceed 20 V; otherwise maximum power falls to 65 W. Low input voltage or a large differential between input and output reduces current capacity and efficiency, so sustained loads should be planned within those derated limits.

Punti di forza

  • Efficient 96% boost conversion reduces energy loss
  • 100W maximum output when input and output exceed 20V
  • Wide 3-35V input range suits diverse power sources
  • Built-in voltmeter simplifies voltage monitoring and adjustment
  • Compact 68x42x14mm module fits tight project spaces

Caratteristiche tecniche

Type100 W

Domande su questo articolo

What comes in the box with the FP5139 100W DC boost converter, and do I need to buy wires or a power source separately?

The module itself is supplied. You must provide a DC source between 3 V and 35 V capable of up to 9 A, plus suitable wiring and a load that needs a higher voltage than the source.

The specs list 100 W and 65 W maximum power — which figure applies to my 12 V battery project?

With a 12 V input the converter cannot reach the 20 V threshold, so the practical limit is 65 W; the 100 W rating only applies when both input and output exceed 20 V.

The headline says 100 W — what does that mean once the board is installed in a real circuit?

It is the peak power available only when the supply and the load are both above 20 V; at lower voltages such as a 12 V battery the continuous capability drops to 65 W, and efficiency falls if the voltage gap is large.

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