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I have a 5V power supply, I also have a chip that its VCC input is 3.3V, the chip datasheet doesn't specify any information about the current needed in the VCC pin, I want to use a voltage regulator so I can use my power supply to power up this chip, The voltage regulator specifies current limit, I'm not sure what it means because I only need voltage to power up the chip (or do I?), Does current mean anything in this situation? In what situation the current of a voltage regulator will mean something?

Thanks.

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    \$\begingroup\$ the 'chip' will have an identification number on it - find the datasheet and it will give you its specs. \$\endgroup\$ – JIm Dearden Mar 7 '17 at 19:25
  • \$\begingroup\$ Is that it "Active Power Supply Current"? \$\endgroup\$ – Kikapi Mar 7 '17 at 19:29
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You 5 V regulator has a maximum current limit (probably printed on it, and probably many 100's of mA), Your 3.3 V regulator is limited to 100 mA (if you go above this it will shutdown, but it will not fail). Your chip will have a VCC or VDD current specified in the datasheet.
Look at the device (chip) datasheet and ensure the current required is less than 100 mA under all operating conditions.
You should also put a capacitor on the output of the 3.3 V regulator, anything from .1 - 10 uf should do.

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  • \$\begingroup\$ What is the purpose of the capacitor in this configuration? \$\endgroup\$ – Kikapi Mar 7 '17 at 20:43
  • \$\begingroup\$ Stability, to lower noise if your 'chip' is digital. \$\endgroup\$ – Jack Creasey Mar 7 '17 at 21:07
  • \$\begingroup\$ Thanks, If you can, please tell me how did you calculate the capacitor capacitance. \$\endgroup\$ – Kikapi Mar 7 '17 at 21:42
  • \$\begingroup\$ I didn't calculate it. The capacitance varies with the regulator, here's one: ti.com/lit/ds/symlink/lp2951.pdf ....read section 8.2.2 onward to get the recommendation for input and output capacitor values. This is very typical of most linear regulators. \$\endgroup\$ – Jack Creasey Mar 7 '17 at 23:11
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If the chip tries to draw to much power (current) from the regulator, the regulator will fail. If you have just one chip, it's current draw is likely minimal, but as suggested you can look up how much current it needs at 3.3V.

You just need to be sure the voltage regulator can source more than the regulator needs to sink.

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  • \$\begingroup\$ I think the current of the Vcc is <100mA, will this 78L33 (pdf1.alldatasheet.com/datasheet-pdf/view/22618/…) work? it has dropout voltage of 1.7V and that is exactly 5V. \$\endgroup\$ – Kikapi Mar 7 '17 at 19:39
  • \$\begingroup\$ It's close, but if you're right on the edge, you might be better off including some margin. If you have that one on hand, there's no reason not to give it a go. If you notice it overheating, then be sure to disconnect it. \$\endgroup\$ – Brian Dohler Mar 7 '17 at 23:33

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