Adjustable Power Supply Design
First, it should be noted that there's a comparator inside the LM2575 and its purpose is to compare a portion of output voltage (the voltage on FB pin) to a reference voltage (1.23V for LM2575) and generate a signal for the control circuitry. So, what the LM2575 does is to keep the voltage on FB pin at 1.23V. That's why minimum output voltage is given as 1.23V.

Consider the following schematic:







simulate this circuit &ndash; Schematic created using CircuitLab

The key point here is the current into/out of the resistors connected to FB pin. And we have 3 equations: 

1) \$I_{R2}\$ is constant: \$I_{R2} = I_{R3} + I_{R1} = V_{ref}/R2\$. 

2) \$I_{R3}\$ varies with DAC output voltage, \$V_{DAC}\$: \$I_{R3} = (V_{DAC} - V_{ref})/R3\$. 

3) And finally, \$V_o = V_{R1} + 1.23V\$, where \$V_{R1} = I_{R1} \cdot R1 = (I_{R2} - I_{R3})\cdot R1\$. If we make \$I_{R1}\$ negative then we can get output voltages lower than 1.23V.

But selecting resistors needs a two-unknown equation to be solved.



For your needs: \$V_{o-min} = 0.5VDC\$ and \$V_{o-max} = 12VDC\$

Let's select R2 = 1k2.

\$I_{R2} = 1mA\$

For \$V_{DAC} = 0V\$ (zero code);


\$I_{R3} = (0 - 1.23)/R3 = -1.23/R3\$
\$I_{R1} = I_{R2} - I_{R3} = 1 - (-1.23/R3) = 1 + 1.23/R3\$
\$V_o = 1.23 + I_{R1}\cdot R1 = 12V \Rightarrow R1 = \frac{10.8}{1+\frac{1.2}{R3}}\$


For \$V_{DAC} = 5V\$ (full-scale code);


\$I_{R3} = (5 - 1.23)/R3 = 3.77/R3\$
\$I_{R1} = I_{R2} - I_{R3} = 1 - (3.77/R3)\$
\$V_o = 1.23 + I_{R1}\cdot R1 = 0.5V \Rightarrow R1 = \frac{0.73}{\frac{3.8}{R3} - 1}\$


From \$R1=\frac{10.8}{1+\frac{1.2}{R3}} = \frac{0.73}{\frac{3.8}{R3} - 1}\$, you'll get R3 = 3k5 and R1 = 8k.

If you crosscheck, you'll see that the minimum output voltage (with full-scale DAC output) will be 0.51V and maximum output voltage (with zero-scale DAC output) will be 11.94V.



With this method, any output voltage range can be obtained.