\$ R_1=1.2 k\Omega\$, \$R_2=2k\Omega\$, \$R_3=8.2k\Omega\$, \$V_R=15V\$
Exercise:
- For which value of \$V_i\$ the diode is conducting?
- Find \$V_o\$ as a function of \$V_i\$.
Attempt solution:
First I suppose the diode is not conducting and I find: \$ \frac{V_i-V_A}{R_1}=\frac{V_A+15V}{R_3}\$, from this I get \$ V_A=\frac{R_3V_i-15VR_1}{R_1+R_3} \$, but since 2 is at \$ 0 V \$ the diode to conduct must be \$ V_A > 0\$, and I find \$ V_i>15VR_1/R_3\approx 2.2 V \$. This should give me the first answer.
Then I suppose the diode is conducting, and we have \$ V_A=0V\$, then \$ \frac{V_i}{R_1}=\frac{15V}{R_3}-\frac{V_o}{R_2} \$ this gives me : \$ V_0=-V_iR_2/R_1+15VR_2/R_3 \$.
So the second answer should be :
$$ \begin {cases}0 & \text{if}\ V_i<2.2V \\-V_iR_2/R_1+15VR_2/R_3\ \text{if} \ V_i>2.2V \end{cases} $$