Semiconductor physics: Voltage below forward voltage

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CaptainMarvel1899
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Assume we have a diode closed circuit.We connect the p type region of the diode to the positive terminal of the battery(cathode).We connect the n type region of the diode to the negative terminal of the battery(anode).The voltage of the battery is 0.3V .The diode%s intristic se miconductor is silicon.Here is my issue . The net force moves the electrons from the n type to the p type.But because there is not enough voltage the valence electrons of silicon have a slight propability to overcome the band gap and become free charge carriers.But if they don't become become free charge carriers they will form bonds with B-atoms because charges do move and fill Boron holes.And this would result to a bigger potential energy of the system.What am I missing?
 
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Can you not just look at it as narrowing the depletion layer - in the same way that reverse bias will widen it? Work will have been done because the unbiased condition will be the lowest Energy situation.
 
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sophiecentaur said:
Can you not just look at it as narrowing the depletion layer - in the same way that reverse bias will widen it? Work will have been done because the unbiased condition will be the lowest Energy situation.
Widening of the depletion region in reverse bias does happen.Narrowing doesn't happen as you think it is happening.As voltage is increased the drift velocity of the free electrons is increased so they switch on and off the depletion region more frequently if the had less voltage , so less resistance.Also the depletion region is decreased by the applied voltage because electrons gain energy due to the the electric potential and have a bigger propability of becoming free charge carriers.So back to my question how do you explain this thing maybe the depletion region would be increased(not very much)by the movement of those carriers?
 
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Yes it gives me a clue.I will do some experiments on my own and tell you the results.