Question: [![enter image description here](https://i.stack.imgur.com/UBFqR.png)](https://i.stack.imgur.com/UBFqR.png)
The supply voltage is 15 V and RL is 100 kΩ. Find the voltage across each diode and RL.

Is = 0.1 μA, η = 2, and VT = 25 mV.

How can I calculate the net current through RL? I have three unknowns: the net current and the voltages across the diodes.

Answer: 
You'd use the Shockley equation,
$$ I(V\_D)=I\_S \left( e^\frac{V\_D}{\eta V\_T} - 1 \right), $$
to figure it out.

You're given \$I\_S\$, \$V\_T\$ and \$\eta\$.

We can invert this equation: given \$I\$, solve for \$V\_D\$. This will be more useful, since the current \$I\$ is the same for both diodes:

$$
V\_D(I) = \ln \left( \frac{I}{I\_S}+1 \right) {\eta V\_T}
$$

Now, the diode D2 is in reverse saturation, since the voltage across it is on par with \$-\frac{1}{\eta V\_T}\$. The reverse saturation current is then \$-I\_S\$, and that diode acts as a current limiter, and sets the current in the entire circuit.

Then, the voltage drop on the resistor is \$I\_S \cdot R\_L\$, the voltage drop on D1 is, per Schockley equation, \$V\_{D1}=\ln(2)\eta V\_T\$, and the voltage drop across D2 is whatever voltage is left in the circuit.

We can simulate it and see if it agrees with our figuring:

![schematic](https://i.stack.imgur.com/6a9uB.png)

[simulate this circuit](/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2f6a9uB.png) – Schematic created using [CircuitLab](https://www.circuitlab.com/)

The diode models in the simulation have had their \$I\_S\$ and \$n\equiv\eta\$ set according to the values given in the assignment. The thermal voltage is slightly different since the value used by CircuitLab is not exactly 25mV, but we see that the simulation and the computed results agree to within 1mV or so.
Assuming the circuit is properly drawn and it is assembled with good parts you will not get any current flow through the circuit except for a small leakage current through the D2. D2 will have the supply dropping across it - the Vf of the first diode. There should be no voltage drop across RL other then what is caused by the leakage of D2.