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The LM2575 is a step-down voltage regulator. I want use it to step down from 9V to 5V in order to power my Arduino board and Bluetooth module.

5 ON/OFF, 4 Feedback, 3 Ground, 2 Output, 1 Vin

However, I'm not sure how to wire this into my breadboard:

  • Where does the feedback pin go?
  • Where does the ON/OFF pin go?

Should I be using capacitors? Finally, will the RF interference generated be a significant problem with my Bluetooth module?

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  • \$\begingroup\$ See the datasheet, p. 2, figure "Typical Application (Fixed Output Voltage Versions)" \$\endgroup\$
    – diverger
    Commented Nov 20, 2014 at 4:00
  • \$\begingroup\$ @diverger Interesting. I don't own a 330uH inductor, diode, or a 330uF capacitor, though. I think I'll just use a linear regulator instead. \$\endgroup\$ Commented Nov 20, 2014 at 4:18
  • \$\begingroup\$ If I wanted to use the LM7805, would I need to buy capacitors? I currently have 3x (100pF, 100nF, 100uF). \$\endgroup\$ Commented Nov 20, 2014 at 4:21

2 Answers 2

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LM2575 is a switching regulator. See the datasheet, p. 2, figure "Typical Application (Fixed Output Voltage Versions)" for it's typical wiring.

If you only need to step down from 9V to 5V, and don't need high efficiency. You may choose linear regulator. With it you don't need to worry the switching noise with switching regulators.

You should use bypass capacitors to make your power rail more quiet. Generally, the regulator's datasheet will give you the typical application circuits, like this LM7805, p. 18. If you don't have so many capacitors with different values, you can parallel small capacitors to get a bigger one, such as if you have many 100nF capacitors, you can parallel ten to get 1uF.

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  • \$\begingroup\$ For LM7805, the datasheet recommends 0.33uF and 0.1uF for C_in and C_out. This source claims I can use my 100uF for C_in instead. Is this correct? \$\endgroup\$ Commented Nov 20, 2014 at 5:39
  • \$\begingroup\$ Generally, the bigger the better. Bigger capacitor has better filterring capability. \$\endgroup\$
    – diverger
    Commented Nov 20, 2014 at 5:42
  • \$\begingroup\$ Does the coil inductance matter? How much wiggle room is there for it? \$\endgroup\$ Commented Apr 16, 2019 at 18:54
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Try the circuit as shown below

enter image description here

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    \$\begingroup\$ It would be good if you could expand this answer a little to explain where you got it and why you think it is the correct solution. \$\endgroup\$
    – David
    Commented Nov 20, 2014 at 6:35

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