# Where's my mistake?

This is partially an electrical engineering problem, but the actual issue apparently lays in my maths.
I have the equation $\frac{V_T}{I_0} \cdot e^{-V_D\div V_T} = 100$
$V_T$ an $I_0$ are given and I am solving for $V_D$.
These are my steps:
Divide both sides by $\frac{V_T}{I_0}$: $$e^{-V_D\div V_T} = 100 \div \frac{V_T}{I_0}$$ Simplify: $$e^{-V_D\div V_T} = \frac{100\cdot I_0}{V_T}$$ Find the natural log of both sides: $$-V_D\div V_T = \ln(\frac{100\cdot I_0}{V_T})$$ Multiply by $V_T$: $$-V_D = V_T \cdot \ln(\frac{100\cdot I_0}{V_T})$$ Multiply by $-1$: $$V_D = -V_T \cdot \ln(\frac{100\cdot I_0}{V_T})$$ Now if I substitute in the numbers; $V_T \approx 0.026$, $I_0 \approx 8 \cdot 10^{-14}$ $$V_D = -0.026 \cdot \ln(\frac{8 \cdot 10^{-12}}{0.026})$$ Simplify a little more: $$V_D = \frac{26}{1000} \cdot -\ln(\frac{4}{13} \cdot 10^{-9})$$ and you finally end up with $V_D \approx 0.569449942$

There is an extra step to the problem as well: the question calls not for $V_D$, but for $V_I$ which is a source voltage and that can basically be determined by solving this: $$V_I \cdot \frac{1}{40} = V_D$$ I.e. $V_I = 40V_D$; which makes $V_I \approx 22.77799768$.
However, this is off by quite a bit (the answer is apparently $1.5742888791$).

### Official Solution:

We find $V_D$ to be $\approx$ 0.57.

(Woo! I did get that portion right.)

Since we know that $\frac{V_D}{i_D}$ is 100, we find $i_D$ to be 0.00026.

Background: $\frac{V_D}{i_D}$ is the resistance that I was originally solving. $V_D$ was the voltage across the element, and $i_D$ was the current through the element.

However, either I'm making some stupid mistake but if $i_D = \frac{V_D}{100}$ then how did they get 0.00026? Completing the rest of the solution's method (quite convoluted in comparison to mine, checked mine though another method; $V_I$ does in fact equal $40V_D$), with the correct value, 0.0057, I arrived at exactly the same final value as before.

Would it be fair to say that it is likely that my logic is correct?

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I got a different answer from either of the two you mentioned. $V_D \approx 0.56945$, and $V_I = 40V_D \approx 22.778$. Were there any conversions of units not accounted for along the way? – Shaun Ault Apr 12 '12 at 15:48
Also, why do you have $0.002$ in the denominator after the sentence "Now if I substitute the numbers...." ? Isn't that supposed to be $V_T = 0.026$? – Shaun Ault Apr 12 '12 at 15:49
Thank you! I'll fix these now – Kian Apr 12 '12 at 15:50
Are you sure $V_T \approx 0.026$? If you use $0.0026$ in both places (before you had it as 0.026 in one place and 0.002 in another) you get something close to the answer you think it should be. – user23784 Apr 12 '12 at 15:58
Which is to say there is something wrong with the numbers. The math is correct (you can tell that you have solved for $V_D$ correctly by just plugging all the values back into your first equation and seeing that you do, indeed, get $100$.) So, either the answer is not $1.57 \ldots$ or one of the numbers for the other parameters is off. – user23784 Apr 12 '12 at 16:03

You are confusing resistance with incremental resistance, I think. The incremental resistance only matters for small signal analysis. The problem is to set the operating point so that the incremental resistance will be the required value. This involves computing $V_D, I_D$. However, you cannot use the incremental resistance to compute $I_D$ in terms of $V_D$.

Also, you are forgetting to account for the $3.9k\Omega$ series resistance.

You correctly computed $V_D = 0.57 V$ required so the incremental resistance is $100 \Omega$.

However, the current through the diode at $V_D$ is given by the diode equation, which you haven't included above. The diode equation is $I_D = I_0 (e^{V_D/V_T}-1)$. Plugging in your numbers gives $I_D = 260 \mu A$ (basically $\frac{V_T}{100}$, since $\frac{V_D}{V_T}$ is large).

The question was to figure the input voltage $V_I$ that will set the diode operating point at $V_D =0.57 V, I_D=260 \mu A$. In the first instance, there is a series resistance of $R_S = 3.9 k\Omega$, so to figure the required $V_I$ you need $$V_I = R_S I_D+V_D$$ Plugging in the numbers gives $V_I = 1.58V$ or so.

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I find it convenient to distinguish small signal & operating point values using lower & upper case resp. In this case, I would use $I_D$ for the operating point current, and $i_D$ for the small signal current (which does not appear in this analysis). Serves as a reminder of what you are dealing with. – copper.hat Apr 12 '12 at 19:49
I wasn't forgetting the series resistance; I just miscalculated with it (that was the $40V_D$) – Kian Apr 12 '12 at 23:22
My professor uses the following notation: $i_D$ is total current, $I_D$ for large signal current, and $i_d$ :) – Kian Apr 12 '12 at 23:23
Also, you may be interested in electronics.stackexchange.com – Kian Apr 12 '12 at 23:29
@KianMayne: Thanks! – copper.hat Apr 13 '12 at 2:35