Diode behaviour in a simple circuit

Question

I have tried using the formula: \$I_d = I_s \cdot e^{V_d \over V_k}\$ where \$I_d\$ = diode current, \$V_d\$ = diode voltage and \$V_k\$ is a known and given voltage. I got nothing.

Edit: here's what I tried:

Answer 1

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.

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

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

Where \$\text{I}_\text{D}\$ is the diode current, \$\text{I}_\text{S}\$ is the reverse bias saturation current, \$\text{V}_\text{D}\$ is the voltage across the diode, \$\text{q}\$ is the electron charge, \$\text{k}\$ is the Boltzmann constant, \$\text{T}\$ is the temperature and \$\eta\$ is the ideality factor.

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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:

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

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

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

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

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

For the LED, let's use assume: \$\eta=1\$. And let's choose for \$\text{I}_\text{S}=3\cdot10^{-14}\space\text{A}\$ (as given in your question). Assume \$\text{V}_\text{T}:=\frac{\text{kT}}{\text{q}}=\frac{8094745087}{320435326800}\approx0.0252617\:\text{V}\$, of course (where we assumed the diode at room temperature). Then we find:

• If \$\text{R}=1000\space\Omega\$: $$\text{V}_\text{i}\approx2.02775\space\text{V}\tag5$$
• If \$\text{R}=400\space\Omega\$: $$\text{V}_\text{i}\approx2.03161\space\text{V}\tag6$$

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I solved for all the knowns using Mathematica. The code is given below.

In[1]:=Clear["Global`*"];
q = ((1602176634/(10^9)))*10^(-19);
k = ((1380649/(10^6)))*10^(-23);
T = 20 + ((5463)/20);
Is = 3*10^(-14);
\[Eta] = 1;
R = 1000;
V2 = 675/1000;
FullSimplify[
 Solve[{I1 == I2 + I3, I1 == Is*(Exp[(q*(Vi - V1))/(\[Eta]*k*T)] - 1),
    I1 == Is*(Exp[(q*(V1 - V2))/(\[Eta]*k*T)] - 1), 
   I2 == Is*(Exp[(q*V2)/(\[Eta]*k*T)] - 1), I3 == V2/R, 
   I1 > 0 && I2 > 0 && I3 > 0 && V1 > 0 && Vi > 0}, {I1, I2, I3, V1, 
   Vi}]]

Out[1]={{I1 -> (3 (22499999999 + E^(216293845590/8094745087)))/
   100000000000000, 
  I2 -> (3 (-1 + E^(216293845590/8094745087)))/100000000000000, 
  I3 -> 27/40000, 
  V1 -> 27/40 + (
    8094745087 Log[22500000000 + E^(216293845590/8094745087)])/
    320435326800, 
  Vi -> 27/40 + (
    8094745087 Log[22500000000 + E^(216293845590/8094745087)])/
    160217663400}}

In[2]:=N[%1]

Out[2]={{I1 -> 0.0127418, I2 -> 0.0120668, I3 -> 0.000675, V1 -> 1.35137, 
  Vi -> 2.02775}}

In[3]:=Clear["Global`*"];
q = ((1602176634/(10^9)))*10^(-19);
k = ((1380649/(10^6)))*10^(-23);
T = 20 + ((5463)/20);
Is = 3*10^(-14);
\[Eta] = 1;
R = 400;
V2 = 675/1000;
FullSimplify[
 Solve[{I1 == I2 + I3, I1 == Is*(Exp[(q*(Vi - V1))/(\[Eta]*k*T)] - 1),
    I1 == Is*(Exp[(q*(V1 - V2))/(\[Eta]*k*T)] - 1), 
   I2 == Is*(Exp[(q*V2)/(\[Eta]*k*T)] - 1), I3 == V2/R, 
   I1 > 0 && I2 > 0 && I3 > 0 && V1 > 0 && Vi > 0}, {I1, I2, I3, V1, 
   Vi}]]

Out[3]={{I1 -> (3 (56249999999 + E^(216293845590/8094745087)))/
   100000000000000, 
  I2 -> (3 (-1 + E^(216293845590/8094745087)))/100000000000000, 
  I3 -> 27/16000, 
  V1 -> 27/40 + (
    8094745087 Log[56250000000 + E^(216293845590/8094745087)])/
    320435326800, 
  Vi -> 27/40 + (
    8094745087 Log[56250000000 + E^(216293845590/8094745087)])/
    160217663400}}

In[4]:=N[%3]

Out[4]={{I1 -> 0.0137543, I2 -> 0.0120668, I3 -> 0.0016875, V1 -> 1.35331, 
  Vi -> 2.03161}}