IV Curve help!
Hi everybody.
I'm really struggling on this question. The last time I did this I only got 3/5, which confuses me as I thought I answered properly. Can anyone help me?
I'm really struggling on this question. The last time I did this I only got 3/5, which confuses me as I thought I answered properly. Can anyone help me?
Original post by TobyReichelt
Hi everybody.
I'm really struggling on this question. The last time I did this I only got 3/5, which confuses me as I thought I answered properly. Can anyone help me?
I'm really struggling on this question. The last time I did this I only got 3/5, which confuses me as I thought I answered properly. Can anyone help me?
My suspicion is you have not stated how the diode junction works and understood how the energy barrier across the junction operates.
Original post by uberteknik
It would help to see where you are losing marks if you gave us your answer and also posted the paper and mark scheme.
My suspicion is you have not stated how the diode junction works and understood how the energy barrier across the junction operates.
My suspicion is you have not stated how the diode junction works and understood how the energy barrier across the junction operates.
good point, why does a diode graph characteristic look like it is? i have no idea, but i do know it does look like the pic OP posted
Original post by thefatone
good point, why does a diode graph characteristic look like it is? i have no idea, but i do know it does look like the pic OP posted
In the forward-biased state, as long as the applied voltage is less than the knee voltage, the current is roughly exponential in the voltage. (This is the Shockley diode equation). This exponential relationship can be derived from statistical mechanics, which says that in the unbiased state, the number of charge carriers that can cross the energy barrier of the depletion layer follows a Boltzmann distribution (i.e. the same distribution that energy follows in an ideal gas). You can see some more details here:
http://www.feynmanlectures.caltech.edu/III_14.html
Once the applied voltage is greater than the knee voltage, the I-V curve tends asymptotically towards an ohmic relationship as V increases. This is because the current is limited by the ohmic properties of the diode material, rather than by the ability of charge carriers to cross the energy barrier of the depletion layer. I don't know enough about the details of this region to say much more than that, though.
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