MBCC breakout carrier board problem

Question

Hello

I have checked the MBCC Breakout carrier board's datasheet for MicroZed. I see a problem!

we can calculate the output voltage for the LMR10510 as follow:

Vout=0.6*(1+R1/R2)

you can find above folrmula at page 12 of the below datasheet:

http://www.ti.com/lit/ds/symlink/lmr10510.pdf

but in your design:

http://s8.picofile.com/file/8316564600/LMR10.JPG

as your design to produce 2.5V we have:

Vout=0.6(1+107k/10k)=7.02 !!!

could you please clear this problem?

for other output voltage there is the same problem.

Thanks

jafoste4 · Accepted Answer

Hello Xzsawq21,

Excellent question! If you refer to the schematic you will notice that the 45.3k Ohm resistor is shorted. This is done so when no Jumper is placed 3.3V VADJ is present by default.

So when jumper position is 3-4 is placed you have a 107k and a 45.3k Ohm resistor placed in parrallel. So to find the resistance of the parrallel resistors you do 1/(1/107k+1/45.3k) = 31.82k Ohm.

Now we plug that into the forumla from the TI datasheet you provided we get

Vout = .6(1+31.82k/10k) = 2.509V which is correct.

This is similar to how the 1.8V VADJ was calculated.

Let me know if you need a further explination.

Thanks,

Josh

MBCC breakout carrier board problem

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