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I have a trouble in designing power supply for my device. I've shifted 3.3V up to 5v by using LMR 62421 manufactured by Texas Instruments. I'm planning to use the 5v rail to power my devices(module, sensor, etc...) Everything is fine except that I can't find out how much load(output) current LMR62421 allows in maximum. Datasheet only says it has minimum switch current limit of 2.1A. I don't think this is equal to Maximun load current. There are some equations on datasheet but they start from knowing VIN VOUT Ioutmax. My devices which are connected to 5V rail will consume approximately 1.6A in sum. I expect that LMR 62421 will be fried if I do not take careful consideration on its maximum allowable current value. Can anyone help me find it out?

I refered to this diagram on datasheet(same inductor value, a little bit different capacitance and resistors for voltage divider

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    \$\begingroup\$ On a tangent: the input voltage strikes me as a bit odd - a 3.3V supply is often a regulated supply rail for 3.3V logic devices. If you're designing a component for a larger system, be careful about feeding back switching noise into that supply, as that could cause problems elsewhere. \$\endgroup\$ – Mels Jul 19 '17 at 9:39
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    \$\begingroup\$ If at all possible, using an unregulated input rail for your 5V supply would almost certainly give better efficiency - when regulation stages are stacked, the losses add up. It will also be easier to ensure stability as the supply interactions are simpler to model. \$\endgroup\$ – Mels Jul 19 '17 at 9:51
  • \$\begingroup\$ Depending on switching frequency and inductor size, the output current will vary even when the peak current is kept at 2.1 A. Do yourself a favor, calculate the duty cycle, assume no trapezoid current waveform, substitute with square wave and just cumpute your average for that duty cycle. You will both learn something and get close to the real value. \$\endgroup\$ – winny Jul 19 '17 at 9:51
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    \$\begingroup\$ Basically, the answer is it depends. This is a boost converter so you will have a higher peak current than average output current. Peak switch and inductor current are equal. And this depends on inductor size and frequency as well. Take a look here for the equations ti.com/lit/an/slva372c/slva372c.pdf Also make sure to calculate for worst case (highest load, biggest output voltage, lowest input voltage). Also you need to make sure the inductor will not saturate. Moreover, make sure your transients don't exceed 2.1A in the switch \$\endgroup\$ – Andrés Jul 19 '17 at 9:51
  • \$\begingroup\$ Mels : Thanks for your advice. I'll keep that in mind \$\endgroup\$ – 김현일 Jul 19 '17 at 23:32
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From page 1 of the data sheet you see these two graphs (I have added max current figures in red just to emphasize things): -

enter image description here

So, when outputting 20 volts, the graph informs you that the maximum sustainable output current is 200 mA. It tells you this because the curve on the graph ends at 200 mA. Now, this is page 1 of the data sheet and if it could supply 300 mA (for instance) it would definitely have that curve going to 300 mA. Page 1 tries to promise as much as possible!

OK that's a power output of 20 volts x 0.2 amps = 4 watts.

Similar story for the 12 volt output, max current is about 330 mA and at 12 volts this means it can supply an output power of 3.96 watts.

What current are you going to get at 5 volts - you could say that maximum output power is 4 watts and this predicts an output current of only 0.8 amps.

You won't get 1.6 amps at 5 volts. Maybe try one of these: -

enter image description here

And yes, they are a little expensive so maybe try again on TI's site and narrow down your selection using their search engine.

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  • \$\begingroup\$ Thank you! You not only teach me some basics but also give me a example. It would be great help to me \$\endgroup\$ – 김현일 Jul 20 '17 at 0:02
  • \$\begingroup\$ Oops I forgot it. Sorry for late accepting. It really helped me a lot. \$\endgroup\$ – 김현일 Jul 21 '17 at 7:45
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If you run TI's WEBENCH tool here.

enter image description here

Now with 1.6 A output current the design cannot be created because "Current exceeds limit". If you run the tool with output current set to 2 A, then an additional explanation will be shown.

NOTE: The current ratings for boost parts are the peak switch current ratings, not the output current. The switch current is dependent both on the output current and also on the amount of boost in voltage that is required in the design. So for designs which require a large increase in voltage, the switch current is much greater than the output current. Energy must be conserved by the following equation:

\$ P_{out} = P_{in} \times Efficiency \$

or

\$ I_{in} = \large \frac{(V_{out} \times I_{out})}{(V_{in} \times Efficiency)} \$ which approximately = \$ I_{switch} \$

For this design if we assume 85% efficiency:

\$ I_{switch} = \large \frac{(5 \times 2)}{(3.3 \times .85)} = 3.57 A \$

For \$ I_{out} = 1.6 A \$ it would be \$ I_{switch} = \large \frac{(5 \times 1.6)}{(3.3 \times .85)} = 2.85 A \$ which still exceeds the maximum 2.1 A switch current.

With \$ I_{out} \$ unknown and \$ I_{switch} \$ 2.1 A, the maximum output current will be ~ 1.17 A.

The tool could generate the design with 1 A output current.

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    \$\begingroup\$ Agreed. Just wanted to say, one could also use the formula (7) of the datasheet, giving the peak current at the inductor. They also say there, that this current should not exceed the minimum current limit of 2.1A. \$\endgroup\$ – nickagian Jul 19 '17 at 10:52
  • \$\begingroup\$ @nickagian Yes, using the datasheet is always good. Their tool is only a convenient solution which probably uses the same equations. \$\endgroup\$ – Bence Kaulics Jul 19 '17 at 10:55
  • \$\begingroup\$ Really thank you for your clear explanation and introducing me very useful tool! \$\endgroup\$ – 김현일 Jul 20 '17 at 0:00

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