SMPS and two power voltages output using N channel mosfet

[newtekuser]

Taking the attached schematic from a STM SMPS design which is supposed to provide selectable voltages using an n-channel mosfet with a third resistor added in parallel that controls the voltage ratio. Under "normal" operation, the output voltage is 0.81V, otherwise, when controlled by the MOSFET, the output voltage is 0.89V.
I understand that when the mosfet is off, the circuit acts as a traditional resistor divider network so the output voltage becomes 0.81V using the formula from the SMPS IC datasheet R1=R2((Vout/Fb)-1), where R2 is chosen to be 56K. After solving, R1 becomes 160K.
What I do not understand is how the value for R3 is calculated such that the output voltage becomes 0.89V when R3 is turned on by the mosfet.

[PGPG]

using the formula from the SMPS IC datasheet R1=R2((Vout/Fb)-1),

Since you did not provide Fb value, I try to find it.
R1=R2((Vout/Fb)-1), so
R1=R2(Vout/Fb)-R2, so
R1+R2=R2(Vout/Fb), so
(R1+R2)/R2=Vout/Fb or R2/(R1+R2)=Fb/Vout

where R2 is chosen to be 56K. After solving, R1 becomes 160K.

From equation I got, it looks that not R2 but R1 was chosen to be 56k.
So, in contrast to what you wrote I assume R1=56k, R2=160k and I can find Fb to be 0.81*160/216 = 0.6V.

Now we need to find R2 for 0.89V.
From your first equation R2=R1/((Vout/Fb)-1) so R2=56/(0.89/0.6 -1)=56/0.483333=116k.

One of first things I have learned about electronic was how to calculate resistance of connected in parallel resistors R=R1*R2/(R1+R2). Later, when I learned how to transform equations I found that when we know the paralel result and one of resistors than the lacking one we can get from R2=R*R1/(R1-R). So to get 116 from 160 || with something we need 116*160/(160-116)=422k.

You have to be very young as these calculations are at primary school level. If you seriously plan to play with electronic assume math is the most important subject at school.

[newtekuser]

using the formula from the SMPS IC datasheet R1=R2((Vout/Fb)-1),

Since you did not provide Fb value, I try to find it.
R1=R2((Vout/Fb)-1), so
R1=R2(Vout/Fb)-R2, so
R1+R2=R2(Vout/Fb), so
(R1+R2)/R2=Vout/Fb or R2/(R1+R2)=Fb/Vout

where R2 is chosen to be 56K. After solving, R1 becomes 160K.

From equation I got, it looks that not R2 but R1 was chosen to be 56k.
So, in contrast to what you wrote I assume R1=56k, R2=160k and I can find Fb to be 0.81*160/216 = 0.6V.

Now we need to find R2 for 0.89V.
From your first equation R2=R1/((Vout/Fb)-1) so R2=56/(0.89/0.6 -1)=56/0.483333=116k.

One of first things I have learned about electronic was how to calculate resistance of connected in parallel resistors R=R1*R2/(R1+R2). Later, when I learned how to transform equations I found that when we know the paralel result and one of resistors than the lacking one we can get from R2=R*R1/(R1-R). So to get 116 from 160 || with something we need 116*160/(160-116)=422k.

You have to be very young as these calculations are at primary school level. If you seriously plan to play with electronic assume math is the most important subject at school.

Thank you my friend!