LM2596 Minimun Input Voltage

Ricardo Ra

Other Parts Discussed in Thread: LM2596

Hello,

I was reading the Datasheet of the LM2596 but i didnt find anything about wich is the minimum voltage that i should have as input in order to obtain a fixed output, is to say, if i would like to have 3.6V on the output, which is the minimum voltage that i can put on the input to take this?

Thank you for your help...

over 11 years ago

0 Denislav Petkov over 11 years ago

TI__Mastermind 19891 points

Hello Ricardo,

Please take a look at figure 8 (page 6) of the datasheet.

Regards,
Denislav

0 Ricardo Ra over 11 years ago in reply to Denislav Petkov

Intellectual 625 points

Hello Denislav,

Thanks for your reply,

According to the figure i could say that at higher load current the input-output differential viares between 0.2 - 01 V aproximmatly? So i could have a input voltage of 3.8V to a output voltage of 3.6?

Regards,

Ricardo

0 Denislav Petkov over 11 years ago in reply to Ricardo Ra

TI__Mastermind 19891 points

Hello Ricardo,

May be we are not looking at the same figure.

Here is the plot I am referring to:

This plot suggest that if you have 3A load and your inductor has 40mohms of DCR, at 25C junction the dropout voltage will be about 1.3V. This means that for a 3.6V output, the edge of regulation will happen when your input voltage reaches 4.9V. The dropout voltage will be less with less load.

If you inductor has higher DCR, then there will be additional voltage drop on top of the voltage in the plot.

Let me know if this makes sense.

Regards,
Denislav

0 Ricardo Ra over 11 years ago in reply to Denislav Petkov

Intellectual 625 points

Thanks Denislav,

According you tell me, if i ensure that the input voltage be Vin=1.3V+Vout I could have the correctly behavior of the Regulator?

Best regards,

Ricardo

0 Denislav Petkov over 11 years ago in reply to Ricardo Ra

TI__Mastermind 19891 points

For the condition I mentioned, yes that is what the plot is suggesting.

If you have inductor with higher DCR, then you may have to provide higher voltage because of the extra drop. See the conditions in the top right corner of the plot. The voltage differential in the plot is for Vout - 50mV, so right when Vout is starting to drop.

Regards,
Denislav

0 Ricardo Ra over 11 years ago in reply to Denislav Petkov

Intellectual 625 points

Thanks Denislav,

According you told me about at higher DCR ill have to provide gihger voltage, there is not a relation which i can use to calculate the dropout? This is because my application is about to work with a Li Ion Battery, so i always ill need to have 3.6V at the output, so i need to know which is the correct value choose the battery....

Best Regards,

Ricardo

0 Denislav Petkov over 11 years ago in reply to Ricardo Ra

TI__Mastermind 19891 points

Hi Ricardo,

If you use an inductor with higher DCR and you know your load current, you can calculate the additional voltage drop on top of the voltage in the graph. The plot is done with an inductor with 40mohm DCR.

So for example, if your inductor has 140mohms of DCR you will have additional 100mohm of resistance in the current path. If your load is 1A, then you will need to have 100mohm x 1A = 0.1V of additional input voltage on top of the voltage in the graph.

Check the inductor specifications and pick one with lower DCR.

Regards,
Denislav

0 Ricardo Ra over 11 years ago in reply to Denislav Petkov

Intellectual 625 points

Ok Denislav,

Now everything is clear.... Thanks for your huge Help!

Best Regards,

Ricardo

0 Denislav Petkov over 11 years ago in reply to Ricardo Ra

TI__Mastermind 19891 points

No problem. Thanks for coming to the forum.

Cheers,
Denislav

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LM2596 Minimun Input Voltage