# Diode behaviour in a simple circuit

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:

• Hint 1: draw the schematic larger and write down all voltages and currents you know or can easily determine. Hint 2: The diode equation gives you the current through the diode for a certain voltage across the diode. $V_o$ is given and it not only the output voltage but also the voltage across a diode so that means you can calculate... Then use Ohm's to calculate the current through .... The sum of those currents flows through ... and .... Use the diode equation to calculate the resulting voltage. Add 3 voltages up and you know $V_i$. Commented Nov 20, 2021 at 22:15
• Current flows through the path of least resistance. At steady state, what is the equivalent resistor for the diode in parallel to the resistor? and why?
– Abel
Commented Nov 20, 2021 at 22:25
• @Bimpelrekkie thanks for your hints i think i got it using the equation and the voltage on the rightmost diode's branch. than the resistive current and diode current combine at ground and go to the main and the rest is easy. Commented Nov 20, 2021 at 22:31
• @Abel Current flows through the path of least resistance. So are you saying that if I connect a 1 Mohm resistor in parallel with a 1 Ohm resistor and apply a voltage across the parallel resistors, all the current will flow through the 1 Ohm resistor? It has the least resistance of the two resistors. I assure you that Ohm's law still applies which would contradict your statement. Commented Nov 21, 2021 at 12:27
• @Bimpelrekkie summary least resistance model: each infinetecimal unit of current flows through only the path of least resistance, increasing it (near infinitely until energy is dissipated). When combined with proper energy-resistance relations and movement of energy through materials, it can model many different components including resistors that are overheating. It's a model that logically explodes against the unreality of constant resistance without flow of energy and has some interesting implications about superconductors (and how they cannot create perpetual motion machines).
– Abel
Commented Nov 21, 2021 at 21:02

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.

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\$$: $$\text{V}_\text{i}\approx2.03161\space\text{V}\tag6$$

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}}

• pretty sure the textbook solution would set ideality factor to 1 unless otherwise specified...
– Abel
Commented Nov 21, 2021 at 22:53
• @Abel I edited my answer. Thanks for the comment, I think that you're right. Commented Nov 22, 2021 at 16:28
• You might want to ignore the '1' factor, which allows an exactly closed form solution to be written down with using a numerical solver. The difference is negligible at sensible forward currents. Using your value for Vt, I get: Input voltage: 2.027749963314493 for R = 1000 Input voltage: 2.031613126579359 for R = 400 Commented Nov 22, 2021 at 18:06