Thanks, how can I prove that method work? I think it will have big difference if we remove diode.If you haven't had Thevenin's Theorem yet (or even after you have replaced the circuit as seen by the diode by the Thevenin's equivalent), you want to first determine if the diode is forward biased or not. Perhaps the simplest way to do this is to remove the diode from the circuit and analyze the modified circuit to find the voltage across the nodes where the diode was. If this voltage would result not result in the diode being forward biased, then you are done since putting the diode back in the circuit would not significantly change anything. But if that voltage would result in the diode being forward biased, then replace the diode in the circuit with a 0.7V battery to hold the voltage between the two nodes equal to the forward diode drop (from the figure I am assuming that the diode model you are using assumes that, when forward biased, the diode will have 0.7V across it regardless of the current through it). Then reanalyze the circuit and whatever current is flowing in the battery is the same current that would be flowing in the diode if you put the diode back into the circuit (in place of the 0.7V battery).
I mean, for example, if I remove the diode and then I recaculate the voltage between anode and cathode in the circuit. If the voltage is Vd=0.3 V, how can I know that if I insert the diode into circuit, the diode will be reverse-biased?I don't understand the question. Are you asking how to prove that the analysis method I described will lead to a valid result?
If the diode is reverse-biased, then removing it will have no effect. If it is forward-biased, then removing it will have a significant effect. The point of analyzing the circuit with the diode removed is only to determine if the diode will be forward biased or not when it is inserted back into the circuit.
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