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schematic

simulate this circuit – Schematic created using CircuitLab

I'm really unsure how to start with this. The instructions said to find Vin such that Vout will be 0.5Vin. The only given is Vd = 0.7 and Rd = 5 ohms. I tried to do a KVL equation but then I don't know what the current is since Vin is not given.

Additional info: Diode model that is used is the constant voltage drop model with resistance of Rd=5 ohms and forward voltage of 0.7

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    \$\begingroup\$ Compared to your previous question, you've now clearly left the concept of the "ideal diode", and are now working with a diode that has a more complex mathematical model. But you forget to specify your diode's model – is it maybe the Shockley equation? No matter the model, you've also forgotten to specify the parameters of your diode (something like forward voltage in saturation, or saturation current). \$\endgroup\$ Oct 8, 2020 at 13:06
  • \$\begingroup\$ @MarcusMüller I should have mentioned that Vd = 0.7 meant that its the forward voltage of the diode. Hence, ill be using the constant voltage drop model here. \$\endgroup\$
    – user263783
    Oct 8, 2020 at 13:11
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    \$\begingroup\$ Oh, Ok, that makes it... super easy? Then Vout = 0.7 V by definition, then 2Vout = Vin = 1.4 V by the problem statement? \$\endgroup\$ Oct 8, 2020 at 13:14
  • \$\begingroup\$ Thats what I was thinking as well. But then what im doubting about is the existence of the resistor Rd. Would it not affect Vout? \$\endgroup\$
    – user263783
    Oct 8, 2020 at 13:16
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    \$\begingroup\$ that doesn't matter. Your problem statement says Vin = 2 Vout, and Vout=Vd is given. so, solved. (Unless your diode model is more complex, AND Vd is misleadingly named; but again, you're still not stating that model! We can't know what's in your head. Please define what a diode looks like for you, and what Rd is in case of your diode model.) \$\endgroup\$ Oct 8, 2020 at 13:34

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An explicit model to be used for the diode is missing in the original question, however the diode parameters provided (\$V_d\$ and \$R_d\$) point towards using a piece-wise linear model for the diode. Therefore, your circuit should be redrawn as follows.

schematic

simulate this circuit – Schematic created using CircuitLab

Assuming \$V_{out} > V_d\$, the ideal diode has a forward bias and acts as a short-circuit. Voltage equations: $$ \frac{V_{out}-V_i}{R_1}+\frac{V_{out}-V_d}{R_d}=0 $$ With the additional constraint \$V_{out} = 0.5V_i\$: $$ \frac{-0.5V_i}{R_1}+\frac{0.5V_i-V_d}{R_d}=0 $$ $$ 0.5\left(\frac{1}{R_d}-\frac{1}{R_1}\right)V_i=\frac{V_d}{R_d} $$ $$ V_i=2\frac{V_d}{R_d\left(\frac{1}{R_d}-\frac{1}{R_1}\right)} $$ Checking numerically if the forward bias assumption holds true: $$ V_i=\frac{1.4}{5(0.2-0.1)}=\frac{1.4}{0.5}=2.8 $$ $$ V_{out}=0.5V_i=1.4 > V_d $$

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