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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:

enter image description here

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:

enter image description here

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?

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  • \$\begingroup\$ Try replacing the diode with a voltage source with the value of the forward voltage: \$V_t\$ = 0.7 V. Can you then solve this? The advantage then is that the voltage at "A" is known (0.7 V) so you can treat the 2 V, 2 kohm and the 5 V, 5 kohm circuits as separate circuits making things easier. \$\endgroup\$ Aug 7, 2021 at 21:29
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    \$\begingroup\$ Your calculation of the Thevenin voltage is not correct. You need to find the voltage from (a) to (b) if the diode is removed, and it is not just a matter of subtracting the two voltages. \$\endgroup\$ Aug 7, 2021 at 21:31
  • \$\begingroup\$ Thanks a lot! Yes, adding 0,7 v at A seems easier, since I can just treat A as a separate voltage in counting the currents into I (diode) \$\endgroup\$ Aug 8, 2021 at 8:04

3 Answers 3

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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.

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  • \$\begingroup\$ Thank you so much for your answer! When you calculate the current between the two resistors, do you see them in one big mesh and then take the resistors in series to only calculate the largest mesh-current going from the 5 v source and then you add it with the same current times the 2 v source so you add two currents going inside the diode and then at the end take the Vth - Vt to get the voltage over the diode and divided by Rth (Resistance over the diode)? Did I understand you correctly? \$\endgroup\$ Aug 8, 2021 at 7:31
  • \$\begingroup\$ Also why not take 3/7 mA* 5kOhm? Since the current is going from 5 v source? \$\endgroup\$ Aug 8, 2021 at 7:40
  • \$\begingroup\$ @DiodeLight97 No, I avoided using the mesh method. The other answer has a very nice and easy approach. So look at it. What I did is just compute the Thevenin equivalent of the two voltage sources and the two resistors and used that to work out the current. \$\endgroup\$
    – jonk
    Aug 8, 2021 at 7:42
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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

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  • \$\begingroup\$ Thanks a lot for your answer! I understand very well, thanks! \$\endgroup\$ Aug 8, 2021 at 7:37
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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:

schematic

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:

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

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    \$\begingroup\$ Hi Jan! Thank you very much for your guidance and thorough explanation. As you guessed, I did not know about the Shockley diode equation. Now I do! :) I gather that the Shockley equation provides a more exact answer to I-V relationship in diodes with specified properties and also in specified temperatures such as room temperature as you mentioned. Useful also in practice when I work with real life amplifiers. \$\endgroup\$ Aug 11, 2021 at 20:07
  • \$\begingroup\$ @DiodeLight97 yes indeed completely right! I am glad that I could participate in your understanding of this equation ;). \$\endgroup\$ Aug 11, 2021 at 20:12

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