# Diode and op amp exercise

$$\ R_1=1.2 k\Omega\$$, $$\R_2=2k\Omega\$$, $$\R_3=8.2k\Omega\$$, $$\V_R=15V\$$

Exercise:

1. For which value of $$\V_i\$$ the diode is conducting?
2. Find $$\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 :

$$\begin {cases}0 & \text{if}\ V_i<2.2V \\-V_iR_2/R_1+15VR_2/R_3\ \text{if} \ V_i>2.2V \end{cases}$$

• A hint: the voltage across $R_1$ is not $V_i+V_A$. The voltage across this resistor is either $V_i-V_A$ or $V_A-V_i$, depending upon the assumed direction of the current...which you have not indicated. – Elliot Alderson Jul 17 '19 at 18:36
• yes my error , of course is $V_i-V_A$ – abbba Jul 18 '19 at 9:03
• Another hint: R1 and R3 form a summing circuit. While the diode is off, according to the superposition principle, VA = Vi.R3/(R1 + R3) + VR.R1/(R1 + R3). After the diode is on, VA is fixed at 0.7 V and the whole circuit is an op-amp inverting summing amplifier. Roughly, Vo = -Vi.R2/R1 - VR.R2/R3. – Circuit fantasist Dec 22 '19 at 18:38