Solid State Physics - p-n junctions

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Homework Help Overview

The discussion revolves around the behavior of a diode in relation to its resistance under forward and reverse bias conditions, specifically using an ohmmeter to measure these resistances at a temperature of 300K.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the potential and emf, and how this affects the calculation of resistance. There is confusion regarding the signs of the potential in reverse and forward bias conditions, as well as the implications for resistance calculations.

Discussion Status

Some participants have offered guidance on using Ohm's Law to relate voltage and current for resistance calculations. There is an acknowledgment of potential errors in calculations, and participants are exploring the implications of negative values in the context of reverse bias.

Contextual Notes

Participants express uncertainty about previous knowledge and its application to the problem, indicating a need for clarification on fundamental concepts related to diodes and resistance measurements.

GrantB
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Homework Statement



An ohmmeter is sometimes used to determine the "direction" of a diode by connecting the ohmmeter to the diode one way and then reversing the ohmmeter leads. If the ohmmeter applies an emf of .5V to the diode in order to determine resistance, what would be the ratio of reverse resistance to forward resistance at 300K?


Homework Equations



I=I0(ee[itex]\varphi[/itex]/kBT-1) ?



The Attempt at a Solution



I know that for reverse bias, the potential [itex]\varphi[/itex] is negative, and for forward bias, it is positive. But that's basically as far as I've gotten.

I am mainly confused as to what the relationship between the potential and emf is, and how that leads you to getting the resistance.

I think I've forgotten too much of my previous physics class...

Thanks for the help.
 
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Use Ohm's Law to write [itex]R=\frac{V}{I}=\frac{V}{I_0 e^{qV/kT}-1}[/itex]. I used V instead of [itex]\phi[/itex] and q instead of e
 
I'm running into a problem.

When I take e-qV/kT-1 for the reverse bias it gives me a negative number.
 
Last edited:
You also have to take V in the numerator to be negative, then you will get a positive number
 
cbetanco said:
You also have to take V in the numerator to be negative, then you will get a positive number

Ahh, thank you.

I am getting an answer of 2.5x108 when the answer is 2.4x108. Although, I think its a rounding (error) somewhere since what I'm doing seems correct.

Thanks!
 

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