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\$ 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 :

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

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  • 1
    \$\begingroup\$ 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. \$\endgroup\$ – Elliot Alderson Jul 17 '19 at 18:36
  • \$\begingroup\$ yes my error , of course is \$ V_i-V_A \$ \$\endgroup\$ – abbba Jul 18 '19 at 9:03
  • \$\begingroup\$ 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. \$\endgroup\$ – Circuit fantasist Dec 22 '19 at 18:38
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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)

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