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Representing the relationship between depletion layer charge and bias voltage graphically

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Heidi

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I was trying to figure out how to graphically represent the relationship between the charge stored in immobile ions in the depletion region of a pn junction and the reverse bias voltage across the junction.

What I know is that when the reverse bias voltage is greater, more bound charge is stored in the depletion region of a pn junction, with positive ions in the n side and equal amount of negative ions in the p side. What I can't understand is that when the relation between the stored charge and the bias voltage is represented graphically as in Figure 3.62(a) in the attached files 1 and 2, why the depletion layer charge, Qdep, is plotted as negative when reverse biased? And when forward biased, the depletion layer charge should be reduced, but why is Qdep positive and increasing with vD?

Do you think the depletion charge Qdep is referring to the positive bound charge or the negative one? Would you please show me how I can understand Figure 3.62(a)?
 

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Last edited:
I'm not 100% sure of your question but you can think of depletion charge as the relative strength of the electric field (electric force per charge) between two plates (P-dope/N-doped) and a dielectric (depletion region). As the space between plates is decreased as the depletion region width narrows with forward bias the field strength/capacitance increases in that depletion region like in a small compression trim capacitor when the screw is tightened.

**broken link removed** **broken link removed**
 
Last edited:
I'm not 100% sure of your question
Please help me interpret the graph of Figure 3.62(a) in 2.pdf attached above. If Qdep represented the stored charge of positive/negative ions in the depletion layer of a pn junction diode, vD represented the voltage applied across the diode, what would the negative value of Qdep mean while reverse-biasing the diode? 1.pdf is the related material in my textbook.
 
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