1
\$\begingroup\$

I am trying to show that from the Shockley equation Vin >> nVt.

Given a simple diode circuit like this

enter image description here

So far, I have equated the current through the resistor \$I = \frac {V_{in} - V_{D1}} {R_1}\$ to the Shockley equation \$I_{D1} = I_s(e^\frac {V_{D1}}{nV_t} -1)\$.

Thus

\$\frac {V_{in} - V_{D1}} {R_1} = I_s(e^\frac {V_{D1}}{nV_t} -1)\$

\$V_{in} - V_{D1} = I_s R_1 (e^\frac {V_{D1}}{nV_t} -1)\$

Expanding right-hand-side:

\$V_{in} - V_{D1} = I_s R_1 e^\frac {V_{D1}}{nV_t} - I_s R_1 \$

Assuming that \$I_s\$ is small \$I_s R_1\$ can be ignored, giving

\$V_{in} - V_{D1} \approx I_s R_1 e^\frac {V_{D1}}{nV_t} \$

Taking logs on both sides

\$ln(V_{in} - V_{D1}) \approx ln(I_s R_1 e^\frac {V_{D1}}{nV_t}) \$

\$ln(V_{in} - V_{D1}) \approx ln(I_s R_1) + \frac {V_{D1}}{nV_t} \$

From this I would like to interpret the equation such that I can show that when \$V_{in}\$ is positive, \$V_{in} >> nV_t\$, for the diode to be conducting current.

I know that \$n = [1,2]\$ and that \$V_t = \frac{kT}{q}\$ which is around 25mV for room temperature. I know that the diode will start conducting at around \$0.6V = 600mV.\$

\$600mV >> 2 \times 25mv = 50mv\$
So these values all show that what I am trying to show is correct.

However, can this be extracted from the equation derived alone, without knowing any of the values?

\$\endgroup\$
11
  • 2
    \$\begingroup\$ Welcome to EE.SE. \$V_{in}\$ can have any positive or negative value providing a valid solution to all equations in this circuit. That should be obvious. Hence I think you intend to know something different from what you are asking. \$\endgroup\$
    – Ariser
    Commented Oct 20, 2019 at 11:22
  • 1
    \$\begingroup\$ Since no measurements are shown, not much can be said. It is possible, for example, that Vin is zero volts. You'll need to add something to the question, I think. \$\endgroup\$
    – jonk
    Commented Oct 20, 2019 at 11:23
  • \$\begingroup\$ @Ariser I have been trying to modify the Shockley equation, along with adding assumptions, to be able to come up with a form of the Shockley equation where I can show that Vin >> nVt. Up till now, I have equated the current through the resistor as I = (Vin-Vd1)/R to the Schokley equation. This way I can have Vin and nVt in the same equation. However I have not been able to show the result that Vin >> nVt yet. It may be that I might not be taking the right assumptions? \$\endgroup\$
    – user234465
    Commented Oct 20, 2019 at 12:28
  • 1
    \$\begingroup\$ Add all your calculations and equations you have done so far to your question. The proven wrong ones, too. And all you are in doubt. And I recommend editing your question rather than adding comments. Then you can use proper formatting. \$\endgroup\$
    – Ariser
    Commented Oct 20, 2019 at 12:37
  • \$\begingroup\$ Post has been updated with equations and my intuition with real life values for what I'm trying to show. \$\endgroup\$
    – user234465
    Commented Oct 20, 2019 at 13:52

1 Answer 1

0
\$\begingroup\$

The equation to determine the current in your circuit is transcendental. There is no closed form solution. To solve it you need to use an approximate iterative method like Newton-Raphson, or use a computer to do a numerical solution (which will likely also use an iterative algorithm).

But you already know the approximate solution. For a high enough voltage source value, the diode forward voltage is 0.6 or 0.7 V. And you know that 0.6 or 0.7 V is much higher than 25 mV. So you already have your intuitive solution.

Or you can change the circuit to use a current source instead of a voltage source. Then you know the current a priori. Say the current is 10 mA. Then you have

$$ I = 10\ {\rm mA} = I_s \exp\left(\frac{V}{nV_T}-1\right)$$

If you take a typical value for \$I_s\$ as \$10^{-15}\ {\rm A}\$ or so (and neglect the -1 term inside the exponential), then you have

$$ \exp\left(\frac{V}{nV_T}\right) = 10^{13}$$

Then you can immediately get

$$ \frac{V}{nV_T} \approx 30 $$

which should satisfy any requirement for \$V \gg nV_T\$.

\$\endgroup\$
2
  • \$\begingroup\$ Yes, this makes sense when trying to relate the voltage across the diode \$V_{D1}\$with \$nV_t\$, however I was trying to relate the input voltage \$V_{in}\$ with \$nV_t\$. \$\endgroup\$
    – user234465
    Commented Oct 20, 2019 at 16:24
  • 1
    \$\begingroup\$ @Gurtyo, your circuit shows an ideal independent voltage source as \$V_{in}\$. It's voltage can be any value you want it to be. Only if the voltage is substantially larger than \$nV_t\$ will large currents flow through the circuit. How much current do you want to flow? Use the analysis I showed above to figure out what will be the diode voltage for that current. Now add the resistor voltage based on Ohm's law. Add those up and you get what the source voltage has to be in your original circuit to get that current. \$\endgroup\$
    – The Photon
    Commented Oct 20, 2019 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.