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I'm trying to analysis circuit with parallel circuit of diode and resistor. enter image description here
the diode is ideal diode with forward voltage of 0.8V, and I'm trying to calculate voltage across D2 and R2 (V_out)
For positive V1, D2 should be act like open switch, so V_out is voltage drop across parallel circuit (R1, D1 in parallel) plus V1.
For 0V < V1 < 0.8V, current will flow only through R1 as D1 is open switch yet, so V_out should be V1 - voltage drop across R1, which is 0V. And for V1 > 0.8V, voltage across parallel circuit should be fixed to 0.8V, so V_out = V1-0.8. Then it gives me nice continuous V1-V_out graph.
And to confirm my thought, I simulated with pspice and it turns out that V_out for V1 positive is V1, and voltage across parallel circuit is zero (not exactly zero, but I'll say zero for it is ideal). I can't understand this result. I also edited Vj of dbreak (forward voltage as far as i know) to 0.8V
enter image description here
Why the voltage across parallel circuit is zero?

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  • \$\begingroup\$ First do a .DC analysis on the diode, alone, to make sure it has the required transfer function. Otherwise this is a quick test in LTspice with an ideal diode with forward drop 1 V and reverse 2 V. \$\endgroup\$ Apr 15 at 10:47
  • \$\begingroup\$ @aconcernedcitizen thank you I'll try! \$\endgroup\$ Apr 15 at 11:18
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Voltage across parallel circuit (R1, D1 in parallel) equals to voltage drop across R1. If the voltage drop gets to 0.8V, current will start flowing through D1 so the voltage drop across R1 will not increase anymore.

So Vin-Vout is either voltage drop across R1 or 0.8V maximum. But this is V1 minus Vout, not Vout itself.

If D2 is open-circuit, there is no current flowing through the R1 and D1 in parallel). Voltage drop across R1 is zero, so Vout equals V1.

If for example, V1 is 10V, Vout will also be 10V (there is no current flowing through R1, so there can not be any voltage difference). Voltage across D1 is still zero.

The "not exactly zero" from your simulation is probably because there is some small but nonzero reverse bias current flowing through D2, that also flows through R1 and creates a small voltage drop.

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  • \$\begingroup\$ Thank you very much! I thought that voltage measure pin includes some kind of ground or something. Now I can understand why voltage drop is zero. Thank you! \$\endgroup\$ Apr 15 at 10:59

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