Soft Clipping with Diode
I think that when \$V_i &lt; V_d\$, the diode does not conduct, so the circuit behaves live a potential divider, with \$V_o=V_i\dfrac{R2}{R1 + R2}\$. Otherwise, the diode behaves like a short circuit, so \$V_o=V_i−V_d\$.


One problem (maybe the only one) with your analysis, is that the diode behavior doesn't depend on \$V_i\$ but on the voltage across the diode itself, which is not the same thing in this circuit.

If you consider that the diode is "on" when \$V_i\dfrac{R_1}{R_1+R_2} &gt; 0.7 \mathrm{V}\$ instead of just \$V_i &gt; 0.7\mathrm{V}\$, the behavior in the "on" and "off" states will match up better at the transition.

Also, for this kind of simple modeling, when the diode is "on" we don't consider it as a short circuit, but as a fixed voltage of about 0.7 V.