What do you mean by open circuit and close circuit pn junction?
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Whenever, there is a concentration gradient, a junction is expected and thus a contact potential develops. In a p-n junction diode, there is a concentration gradient across the junction and as a result a junction potential develops across the junction. In order to connect this junction diode to the external circuitry, two conducting wires are attached to both ends of the semiconductor diode - one at each end of the diode. One can assume a conductor to be an n-type material with extremely high electron concentration. Thus two metal-semiconductor junctions are formed at these two ends : a metal-p-type-semiconductor junction at one end and a metal-n-type-semiconductor junction at the other end. These additional junctions are referred to as “Ohmic Contacts” and naturally a potential, called as Ohmic Contact Potential, is developed across these contacts.

Now, if we short circuit the terminals of the p-n junction, the junction potential must be exactly compensated by the metal-to-semiconductor Ohmic Contact potentials at the Ohmic contacts because under short-circuit-conditions, the some of the voltages around the close loop must be zero. Hence the current would be zero.
If you are not convinced enough till now, then, let us assume the current to be non zero. Then metal wire gets heated and it dissipates energy in the form of heat. Due to conservation of energy and non-availability of any external sources in the circuit, and since the wire dissipates energy, the semiconductor diode should be cooled since it has to lose equal amount of energy. This violates the principle of thermal equilibrium as simultaneous heating of one part (metal wire) and simultaneous cooling of the other part (semiconductor bar) of the same physical system without the presence of any external energy generating source is not possible. So there should not be any current flowing in the short circuit.
Hope its help you
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Now, if we short circuit the terminals of the p-n junction, the junction potential must be exactly compensated by the metal-to-semiconductor Ohmic Contact potentials at the Ohmic contacts because under short-circuit-conditions, the some of the voltages around the close loop must be zero. Hence the current would be zero.
If you are not convinced enough till now, then, let us assume the current to be non zero. Then metal wire gets heated and it dissipates energy in the form of heat. Due to conservation of energy and non-availability of any external sources in the circuit, and since the wire dissipates energy, the semiconductor diode should be cooled since it has to lose equal amount of energy. This violates the principle of thermal equilibrium as simultaneous heating of one part (metal wire) and simultaneous cooling of the other part (semiconductor bar) of the same physical system without the presence of any external energy generating source is not possible. So there should not be any current flowing in the short circuit.
Hope its help you
=======☺☺☺
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