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?

  • 1
    $\begingroup$ 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? $\endgroup$ – Shaun Ault Apr 12 '12 at 15:48
  • 1
    $\begingroup$ 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 $V_T = 0.026$? $\endgroup$ – Shaun Ault Apr 12 '12 at 15:49
  • $\begingroup$ Thank you! I'll fix these now $\endgroup$ – Kian Apr 12 '12 at 15:50
  • 1
    $\begingroup$ 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. $\endgroup$ – user23784 Apr 12 '12 at 15:58
  • 1
    $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – copper.hat Apr 12 '12 at 19:49
  • $\begingroup$ I wasn't forgetting the series resistance; I just miscalculated with it (that was the $40V_D$) $\endgroup$ – Kian Apr 12 '12 at 23:22
  • $\begingroup$ My professor uses the following notation: $i_D$ is total current, $I_D$ for large signal current, and $i_d$ :) $\endgroup$ – Kian Apr 12 '12 at 23:23
  • $\begingroup$ Also, you may be interested in electronics.stackexchange.com $\endgroup$ – Kian Apr 12 '12 at 23:29
  • $\begingroup$ @KianMayne: Thanks! $\endgroup$ – copper.hat Apr 13 '12 at 2:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.