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enter image description here

I just want to know that my logic is sound for the above question. So, since the diode is connected to a resistor in parallel, for the voltage values where V is less than the breakdown voltage, the resistor is the only contributor to the current I and thus the graph for the first part follows Ohm's law.

Once the V reached the breakdown voltage, however, the PN diode also starts contributing to the current, but since the resistor is still in play, there isn't the steep rise in current, characteristic of the diode, but it is enough to indicate a non-ohmic presence. Thus, the answer would be A.

Is that correct?

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  • $\begingroup$ At each voltage $I_{\rm A}=I_{\rm C}+I_{\rm D}$ You are adding the currents which pass through each of the two components $\endgroup$ – Farcher Oct 2 '17 at 7:12
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Correct, except one slight comment about reasoning post turn on voltage. Just by elimination, C and D are wrong, as they are the pure element IVs. If one considered the diode as a "perfect" element, it really would be a step function, limited only by the real series resistances elsewhere. The fact that diodes cannot physically do this means they are still a (very small) resistor post turn on, and open reverse and pre turn on. So in that case, you can approximate it as just the one resistor before turn on, then a tiny resistor in parallel with a large one post. Clearly most current would flow throw the smaller one, so that rejects B.

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Well, we know that:

$$ \begin{cases} \text{V}_{\space\text{in}}\left(t\right)=\text{V}_{\space\text{D}}\left(t\right)=\text{V}_{\space\text{R}}\left(t\right)\\ \\ \text{I}_{\space\text{in}}\left(t\right)=\text{I}_{\space\text{D}}\left(t\right)+\text{I}_{\space\text{R}}\left(t\right)\\ \\ \text{I}_{\space\text{D}}\left(t\right)=\text{I}_{\space\text{S}}\cdot\left\{\exp\left(\frac{\text{V}_{\space\text{D}}\left(t\right)}{\text{n}\cdot\frac{\text{k}\cdot\text{T}}{\text{q}}}\right)-1\right\}\\ \\ \text{V}_{\space\text{R}}\left(t\right)=\text{I}_{\space\text{R}}\left(t\right)\cdot\text{R}\\ \\ 1<\text{n}<2\\ \\ \text{k}\approx1.3806485279\times10^{-23}\\ \\ \text{q}\approx1.602176620898\times10^{-19} \end{cases}\tag1 $$

So, we get:

$$\text{I}_{\space\text{in}}\left(t\right)=\text{I}_{\space\text{S}}\cdot\left\{\exp\left(\frac{\text{V}_{\space\text{in}}\left(t\right)}{\text{n}\cdot\frac{\text{k}\cdot\text{T}}{\text{q}}}\right)-1\right\}+\text{V}_{\space\text{in}}\left(t\right)\cdot\frac{1}{\text{R}}\tag2$$

Where:

$$0.000086<\text{n}\cdot\frac{\text{k}}{\text{q}}<0.00017\tag3$$

And when $\text{T}$ is room temperature we get:

$$19.7928<\frac{1}{\text{n}\cdot\frac{\text{k}\cdot\text{T}}{\text{q}}}<39.5856\tag4$$

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  • $\begingroup$ Uh, I didn't understand :/ What exactly is the conclusion you've come to? $\endgroup$ – Antara Kulkarni Oct 3 '17 at 12:01

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