Setting the output voltage of SMPS using TPS62088 IC

[newtekuser]

I'm not clear what to set R1 and R2 to if I follow the datasheet. They say choose R2 that is 100K or less, ok, I get that. But in their example, if Vfb is set to 0.6 (taken as typical value from the table in the datasheet) and I want an output voltage of 0.6V, then R2 becomes 0. Is that correct?

[benj38]

I'm not clear what to set R1 and R2 to if I follow the datasheet. They say choose R2 that is 100K or less, ok, I get that. But in their example, why is Vfb set to 0.6? (I take they chose this as typical value from the table in the datasheet). But if they want to achieve an output voltage of 0.6V, setting Vout also to 0.6 then the entire equation becomes 0 as R2 is multiplied by 0. (i.e: (0.6/0.6 - 1) = 0)

Vfb is not exactly "set" to 0.6V. Inside the IC there is a reference voltage that is 0.6V, and the IC compares the value at the FB pin to this reference voltage. The IC increases/decreases the output voltage to make the Vfb equal to the internal reference. Hence, if you want an output of 0.6V, you need Vfb to be equal to the output voltage, i.e., an R2 of zero ohms.

There is nothing wrong with the equation in the datasheet: if you want 0.6V output, R2 is simply a short; lower output voltages are not possible, and for higher output voltages increase R2 according to the equation (e.g., for Vout=1.2V, R2=R1).

[newtekuser]

I'm not clear what to set R1 and R2 to if I follow the datasheet. They say choose R2 that is 100K or less, ok, I get that. But in their example, why is Vfb set to 0.6? (I take they chose this as typical value from the table in the datasheet). But if they want to achieve an output voltage of 0.6V, setting Vout also to 0.6 then the entire equation becomes 0 as R2 is multiplied by 0. (i.e: (0.6/0.6 - 1) = 0)

Vfb is not exactly "set" to 0.6V. Inside the IC there is a reference voltage that is 0.6V, and the IC compares the value at the FB pin to this reference voltage. The IC increases/decreases the output voltage to make the Vfb equal to the internal reference. Hence, if you want an output of 0.6V, you need Vfb to be equal to the output voltage, i.e., an R2 of zero ohms.

There is nothing wrong with the equation in the datasheet: if you want 0.6V output, R2 is simply a short; lower output voltages are not possible, and for higher output voltages increase R2 according to the equation (e.g., for Vout=1.2V, R2=R1).

Thanks!  Didn't know that was possible (i.e.: having zero ohms in a resistor divider network, or not having an R2).