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We are trying to solve the following circuit using the Ideal Diode method; however, I'm having a hard time understanding why D1 is always off: circuit - D1 is parallel to R1

The solution sheet states 2 cases:

  1. When Vin is +ve:

    • D1 must be OFF
    • D2 must be OFF
  2. When Vin is -ve:

    • D1 must be OFF
    • D2 must be ON

The conclusions from #2 come from the observation that current will flow from ground to Vin (Vin is -ve)

Although when R2 is moved as in the following circuit:

Circuit 2

We have the following cases:

  1. When Vin is +ve:

    • D1 is ON
    • D2 must be OFF
  2. When Vin is -ve:

    • D1 must be OFF
    • D2 must be ON

Isn't D1 supposed to be ON when current flows from Vin to Vout (Vin is +ve) in the first circuit as well?

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    \$\begingroup\$ There is no load shown on Vout, so no current. \$\endgroup\$ – Peter Bennett Jun 2 '17 at 2:44
  • \$\begingroup\$ @PeterBennett Thank you! I read somewhere that when there is no load/resistance, current should still pass but it would be theoretically infinity (i.e R = 0). Is that wrong? I have edited the question to include the second circuit as well, which has a load but D1 is ON at +ve Vin. \$\endgroup\$ – Jonathan T Jun 2 '17 at 2:57
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    \$\begingroup\$ In the first circuit, with no load shown, the load resistance is infinite, not zero, so there is no load current. With R2 between Vout and Ground, R2 will draw some current, so there will be a voltage drop across R1 - whether that voltage will be enough to turn on D1 will depend on the resistor values. \$\endgroup\$ – Peter Bennett Jun 2 '17 at 3:03
  • \$\begingroup\$ Think like this, as there is no load in electrical sockets in your room, do current flow through that ? This is same. \$\endgroup\$ – Anklon Jun 2 '17 at 3:06
  • \$\begingroup\$ Thank you both very much! I get it now; so in summary, the current was supposed to go from Vin to ground because there is no load on Vout; however, D2 blocked the passage and hence it became an open circuit. \$\endgroup\$ – Jonathan T Jun 2 '17 at 3:08
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For negative Vin , current flows from ground to Vin through R2,D2 and R1.So there is a complete path for current to flow.

But for positive V1 , current can not go through D2 due to reverse biased condition and there is no load at Vout. So there is no path for current to go to the ground. As a result, there is no current through R1 which leads to zero voltage drop across it.

As voltage drop across diode D1 remain zero due to zero current through R1. But there is threshold voltage drop needed for diode to be ON, diode D1 remain OFF

In second image, there is a path fro current to flow through R2 is there in no reverse biased diode across it. So current flow thorough R1 & R2 which cause voltage drop across R1. it created threshold voltage across D1 ,hence it's ON.

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Isn't D1 supposed to be ON when current flows from Vin to Vout (Vin is +ve) in the first circuit as well?

This is actually a badly-formed question, but let's start with the other components.

With Vin positive, it is clear that D2 is reverse-biased, and therefor OFF. This means no current through R2, and therefor no current through R1.

Since there is no current through R1, there is (by Ohm's Law) no voltage across it either. So the voltage across D1 is zero. The question then arises, is an ideal diode with zero voltage across it ON or OFF? Well, that is usually how the ON state is defined, so the answer is as shown.

The thing is, zero current is also the defining condition for the OFF state, so the answer can be argued as OFF.

You need to realize that the condition for both zero voltage and zero current is something of an anomaly, and fits the definition of both states. If you look at an ideal diode as a switch, the 0/0 point satisfies the conditions both for OFF and ON. So, the definition is only clear for diode voltage less than zero OR diode current greater than zero, and a question which addresses the 0/0 point is not a good one.

Just chalk it up to experience: textbook questions and answers are not always well-formed.

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