\$ 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 :
$$ \begin {cases}0 & \text{if}\ V_i<2.2V \\-V_iR_2/R_1+15VR_2/R_3\ \text{if} \ V_i>2.2V
\end{cases}
$$
Best Answer
Kirchoffs current law - make the assumption of whether currents going into VA is positive or negative. Remember the voltage source VR is flipped. Assume the diode is forward biased, subtract the 0.7 voltage drop, calculate the total input resistance to the op amp, calculate the gain of the opamp (-R2/Rin)