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I am using a voltage regulator (AMS1117 3.3V) for my microcontroller with an input voltage of 5V and input max current of 600mA. Can I expect to have more current capability from the output? If yes, what could be the approximate output current in this scenario ? If no, is there a simple solution which would increase the output current ?

Note: I am using 10uF and 22uF capacitors at input and output respectively.

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    \$\begingroup\$ Linear regulators can never provide more current than the supply. \$\endgroup\$ – Ignacio Vazquez-Abrams Mar 16 '18 at 6:55
  • \$\begingroup\$ So my output also can have only 600mA current ? Can you provide some alternate solution which can increase the current ouput ? \$\endgroup\$ – Zac Mar 16 '18 at 7:09
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    \$\begingroup\$ Use a buck converter. \$\endgroup\$ – mkeith Mar 16 '18 at 7:30
  • \$\begingroup\$ @mkeith Buck converter may not be practical in my case due to the additional size. \$\endgroup\$ – Zac Mar 16 '18 at 10:48
  • \$\begingroup\$ Any linear regulator will be limited to less than Iin. If Iin is 600mA max, and you want more than 600 mA out, you will need to use a switchmode DC-DC converter. The buck is the simplest solution. There are small buck converters out there. \$\endgroup\$ – mkeith Mar 16 '18 at 14:49
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You can think of a liner regulator (LDO) like the AMS1117 as a kind of intelligent resistor divider, which adjusts its value to maintain the output voltage stable. It is thus impossible to have more current at the output than at the input. In fact the output current will be slightly smaller by the value given in the LDO datasheets as "quiescent current", which is the current that the LDO needs for itself to operate.

If you need more than the 600mA at the 3.3V side you could switch to a buck converter. The operation of the switching regulators is very different from that of the LDOs and there you should think in terms of power. So, in your case, from the 5V*600mA=3W input power the buck regulator will deliver something like the 95% of it at best case, so 2.85W. This means that the available output current would be around 2.85W/3.3V=0.864A (best case).

I don't see at the moment any other possibility to have more current at your output. Of course you could always try to increase the input current.

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  • \$\begingroup\$ 95% is maybe a bit optimistic. Or at the high end of what is attainable. But good answer. +1. \$\endgroup\$ – mkeith Mar 16 '18 at 14:51
  • \$\begingroup\$ @mkeith It is definitely very optimistic! That's why I said 'best case'. I have seen switching regulators advertising such high efficiencies, although I have never seen them in reality. \$\endgroup\$ – nickagian Mar 16 '18 at 15:04
  • \$\begingroup\$ @nickagian Thanks for you answer. Would be able to recommend some commonly used switching regulator for converting 3.3v to 5v (max current is less than 1A). I would prefer smaller size if available. \$\endgroup\$ – Zac Mar 18 '18 at 18:38
  • \$\begingroup\$ @Zacson It is quite difficult what you are saying! There are literally a couple hundreds buck regulators that fulfill the criteria you have given, with many different options, switching frequency and package. Have a look here. Regarding the small size, I would advise you to opt for a DC/DC regulator and not a controller. The regulators use integrated switching elements (MOSFETs) and thus the external components are minimized, the biggest of which will be the coil. There also exist some modules which include everything that is needed. \$\endgroup\$ – nickagian Mar 19 '18 at 8:18
  • \$\begingroup\$ @nickagian Thanks a lot. My search results mostly shows up "AMS1117", even when I search for '5v to 3.3v dc to dc regulator'. Hence requested your suggestion. I did some filtering in Mouser and found Microchip MIC23050. Is this the correct DC to DC regulator ? \$\endgroup\$ – Zac Mar 19 '18 at 10:24

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