Thevenin equivalent or mesh-current in a diode circuit

Question

I've been stuck on a problem for a couple days and many hours now and see no other option than to ask here and see if you can help me or give me a lead.

What I have been stuck on is this:

My initial solution for finding out the current through the diode has been to apply Thévenin equivalent where I first solve the Thévenin resistance according to:

The two resistors are in parallel where the voltage sources are shorted, the diode is open, and the Thévenin resistance is 10/7 kilo Ohm.

Then I got Thévenin voltage by taking 5V - 2V = 3V.

(Vth - Vt) / Rth = I (diode) -----> (3 - 0,7)/ (10/7) = 1.61 mA

The solution is supposed to be 1.5 mA. I am not sure what I did wrong here. I also tried the mesh-current method but I got strange numbers. What do you think?

Answer 1

The Thevenin voltage isn't \$3\:\text{V}\$, though it's not terribly far from it. There are a variety of ways to compute it but none of them look like what you did.

One way to think about solving for it is to just look at the current magnitude that would exist between the \$2\:\text{V}\$ and \$5\:\text{V}\$ sources, via the two series resistors. This current would be \$\frac{5\:\text{V}-2\:\text{V}}{2\:\text{k}\Omega+5\:\text{k}\Omega}=\frac37\:\text{mA}\$. That current, times \$2\:\text{k}\Omega\$, is the voltage that must be added to \$2\:\text{V}\$ to get the Thevenin voltage where the two resistors meet. This is \$2\:\text{V}+\frac37\:\text{mA}\cdot 2\:\text{k}\Omega=2\frac67\:\text{V}\$.

The Thevenin resistance is the two resistors taken in parallel, or \$1\frac37\:\text{k}\Omega\$, just as you said it is.

All you now need to do is subtract \$700\:\text{mV}\$ from your Thevenin voltage and then divide that result by the resistance to work out the current.

P.S. I use a different method to compute the Thevenin voltage in a divider like this: \$V_{_\text{TH}}=\frac{V_1\cdot R_2+V_2\cdot R_1}{R_1+R_2}\$. In your case, that would look like this: \$V_{_\text{TH}}=\frac{2\:\text{V}\,\cdot\, 5\:\text{k}\Omega+5\:\text{V}\,\cdot\, 2\:\text{k}\Omega}{2\:\text{k}\Omega+5\:\text{k}\Omega}=2\frac67\:\text{V}\$. It's just short-hand for the same process used, earlier.

Answer 2

The current through the diode is the sum of the currents through the two resistors.

ID = (2-0.7)/2k + (5-0.7)/5k = 1.51mA

Answer 3

Well, let's make a mathematical closed solution. I know that this is maybe above the OP's knowledge, but I think it is important to show it in combination with the other answers given.

The Shockley diode equation, gives the relation between the voltage across and the current trough a diode:

$$\text{I}_\text{D}=\text{I}_\text{S}\left(\exp\left(\frac{\text{q}\text{V}_\text{D}}{\eta\text{k}\text{T}}\right)-1\right)\tag1$$

We are trying to analyze the following circuit:

^{simulate this circuit – Schematic created using CircuitLab}

When we use and apply KCL, we can write the following set of equations:

$$\text{I}_3=\text{I}_1+\text{I}_2\tag2$$

When we use and apply Ohm's law, we can write the following set of equations:

$$ \begin{cases} \text{I}_1=\frac{\text{V}_1-\text{V}_3}{\text{R}_1}\\ \\ \text{I}_2=\frac{\text{V}_2-\text{V}_3}{\text{R}_2}\\ \\ \text{I}_3=\text{I}_\text{S}\left(\exp\left(\frac{\text{q}\text{V}_3}{\eta\text{k}\text{T}}\right)-1\right) \end{cases}\tag3 $$

Substitute \$(3)\$ into \$(2)\$, in order to get:

$$\text{I}_\text{S}\left(\exp\left(\frac{\text{q}\text{V}_3}{\eta\text{k}\text{T}}\right)-1\right)=\frac{\text{V}_1-\text{V}_3}{\text{R}_1}+\frac{\text{V}_2-\text{V}_3}{\text{R}_2}\tag4$$

For the LED, let's use parameters taken from a Luminus PT-121-B LED: \$\eta=8.37\$, and \$\text{I}_\text{S}=435.2\:\text{nA}\$. (Assume \$\text{V}_\text{T}:=\frac{\text{kT}}{\text{q}}=\frac{8094745087}{320435326800}\approx0.0252617\:\text{V}\$, of course.)

Using the known values, we find:

• $$\text{V}_3\approx1.60818\space\text{V}\tag5$$
• $$\text{I}_3\approx0.000874272\space\text{A}\tag6$$

I used Mathematica to find it, with the following code:

In[1]:=Clear["Global`*"];
q = ((1602176634/(10^9)))*10^(-19);
k = ((1380649/(10^6)))*10^(-23);
T = 20 + ((5463)/20);
Is = (4352/10)*10^(-9);
\[Eta] = 837/100;
V1 = 2;
V2 = 5;
R1 = 2*1000;
R2 = 5*1000;
FullSimplify[
 Solve[{I3 == I1 + I2, I1 == (V1 - V3)/R1, I2 == (V2 - V3)/R2, 
   I3 == Is*(Exp[(q*V3)/(\[Eta]*k*T)] - 1), 
   I1 > 0 && I2 > 0 && I3 > 0 && V3 > 0}, {I1, I2, I3, V3}]]

Out[1]={{I1 -> (-46909 + (
    131741976290925 ProductLog[(
      387370706176 E^(1780583630706176/131741976290925))/
      131741976290925])/11393256064)/109375000, 
  I2 -> 234307/546875000 + (
    752811293091 ProductLog[(
      387370706176 E^(1780583630706176/131741976290925))/
      131741976290925])/17801962600000000, 
  I3 -> (17 (-1 + (
      131741976290925 ProductLog[(
        387370706176 E^(1780583630706176/131741976290925))/
        131741976290925])/387370706176))/39062500, 
  V3 -> 312568/109375 - (
    752811293091 ProductLog[(
      387370706176 E^(1780583630706176/131741976290925))/
      131741976290925])/3560392520000}}

In[2]:=N[%1]

Out[2]={{I1 -> 0.000195909, I2 -> 0.000678363, I3 -> 0.000874272, 
  V3 -> 1.60818}}

---

Note: when we use the voltage drop of \$0.7\space\text{V}\$ and the characteristics of the Luminus PT-121-B LED at room temperature on the Shockley diode equation, we get the following current flow trough the diode:

$$\text{I}_3=\text{I}_\text{S}\left(\exp\left(\frac{\text{q}\cdot\frac{7}{10}}{\eta\text{k}\text{T}}\right)-1\right)=$$ $$\frac{17}{39062500}\cdot\left(\exp\left(\frac{2492274764000}{752811293091}\right)-1\right)\approx0.0000114902\space\text{A}\tag7$$