Equations for voltage/current through a transistor
Coming into this circuit felt a bit weird for me. It's not something I'd consider doing and I'm not sure what you expect it to do. But that doesn't mean it can't be analyzed. And maybe something will fall out of that process.

Using Thevenin, we can easily proceed from the left side to the right side, below:







simulate this circuit &ndash; Schematic created using CircuitLab

All I did here was to take the resistor divider, much like you already attempted to do, and figure out the Thevenin voltage for it and its Thevenin resistance. I then replaced the resistor divider with this Thevenin equivalent. I think you should already know how to compute that voltage and that resistance, so I won't trouble you with the details. If you do need more, I can always add it later on.

Another thing I decided not to do is to include your parasitic resistances for the BJT. By the end of this, I think you can work out how to add those. So let's keep it simpler and see where that goes.

I'm going to discuss some a priori guesses about the LED and the BJT, below. So let's have at it:


We can just assume \$V_\text{LED}\$ is a fixed value no matter what, if we want. In this case, we might say \$V_\text{LED}=2\:\text{V}\$. But since you've had a look at the Shockley diode equation, I think you know better than that, now. One lucky thing about diodes is that if you don't change the current much, then the voltage also doesn't change much. So a more sophisticated "model" might be \$V_\text{LED}=1.6\:\text{V}+I_\text{LED}\cdot 20\:\Omega\$. Here, at \$I_\text{LED}=20\:\text{mA}\$ you would get \$V_\text{LED}=2\:\text{V}\$. But now, if the current is lower the computed voltage will be lower; etc. So this model now has two parameters; one for the basic voltage below which the LED doesn't work and the other for an intrinsic resistance that we assume exists and causes a voltage change when the current changes. Finally, you can go all-out and use the Shockley diode equation plus other parasitics to get closer to what Spice does. But so long as you don't deviate too far from the "calibrated" current, the two-parameter model for the LED is usually "good enough."
When supplied a base current via a forward-biased base-emitter junction, a BJT is either in "active mode" or else in varying degrees of "saturation." In simple terms, the BJT is active if the magnitude of the voltage difference between its collector and its emitter is greater than the magnitude of the voltage difference between its base and its emitter. Otherwise, it's in saturation. And when in saturation, the closer the collector voltage gets to the emitter, the deeper that saturation is. Usually, you start out assuming "active mode" and then see where that takes you. If it appears to cause the collector to be very close to the emitter, you switch gears and change your approach and instead analyze it as a saturated BJT. One way or the other, you usually get close enough.


So let's dive in on the assumption of "active mode" for the BJT. Using KVL provides:

$$\begin{align*}V_\text{TH}-R_\text{TH}\cdot I_\text{B}-V_\text{BE}-V_\text{LED}&amp;=0\:\text{V}\\\\&amp;\therefore\\\\I_\text{B}&amp;=\frac{V_\text{TH}-V_\text{BE}-V_\text{LED}}{R_\text{TH}}\end{align*}$$

The trouble with the above is that we don't actually know the LED voltage (except that you added it for a given current, which we also don't yet know) and we don't know \$V_\text{BE}\$ for the BJT because we don't know it's collector current. So, there's a few details to worry over.

Let's assume \$V_\text{LED}=2\:\text{V}\$ (simplest case) for now. Also, let's assume \$V_\text{BE}=700\:\text{mV}\$. (I'll get back to the saturation current you gave, later.) Then we'd find that \$I_\text{B}=\frac{2.439\:\text{V}-700\:\text{mV}-2\:\text{V}}{729\:\Omega}\approx -358\:\mu\text{A}\$. That's a problem because it means current is going the wrong way. So we already know we are in trouble with our assumptions. We know that if this circuit actually has any currents, and it probably does, we'll have to adjust our numbers.

One thing we know for certain is the Thevenin equivalent for our resistor divider. So we cannot change that part. This leaves just two values to adjust and we know one or both of them must be smaller than we assumed.

If we keep our approximate idea about the voltage for the LED, but allow ourselves a little bit of adjustment there, then we know that the collector cannot be lower than the emitter and therefore cannot be lower than the LED voltage, itself. Roughly, this means that the largest possible collector current will be about \$15\:\text{mA}\$. (Saturated BJT.) Let's try \$V_\text{LED}=1.6\:\text{V}+I_\text{LED}\cdot 20\:\Omega\$ with \$I_\text{LED}=15\:\text{mA}\$ to get \$V_\text{LED}=1.9\:\text{V}\$. That's lower, so nice. Also, given \$\beta=100\$ (your specs), we'd guess that \$I_\text{B}\approx 150\:\mu\text{A}\$. This means that the base voltage will be \$2.439-150\:\mu\text{A}\cdot 729\:\Omega\approx 2.33\:\text{V}\$ and that means \$V_\text{BE}\approx 430\:\text{mV}\$. Well, at least that's in the right direction now.

But now we can use your saturation current to see if that could even be possible. In general, for small signal BJT devices, you'll find (via that Shockley equation) that \$V_\text{BE}=\left(V_T\approx 26\:\text{mV}\right)\cdot\operatorname{ln}\left(\frac{I_\text{C}\approx 15\:\text{mA}}{I_\text{SAT}=1\:\text{fA}}\right)\approx 789\:\text{mV}\$. So that doesn't match up: \$430\:\text{mV}\ne 789\:\text{mV}\$.

If we lowered the collector current to \$1\:\text{mA}\$, then \$V_\text{BE}\approx 718 \:\text{mV}\$ (Shockley) and \$V_\text{LED}\approx 1.62\:\text{V}\$ (from our two-param model.) The base voltage would be \$2.432\:\text{V}\$ (Thevenin calculation), so now we are comparing a Shockley \$718 \:\text{mV}\$ to \$2.432\:\text{V}-1.62\:\text{V}=812\:\text{mV}\$. And that is a much closer match-up. But now we can see that the collector current can be just a little bit larger.

We have about \$100\:\text{mV}\$ of mismatch now. I know that there is about \$60\:\text{mV}\$ of Shockley equation change to \$V_\text{BE}\$ for each \$10\times\$ change in the collector current. (I remember this.) I also know that I only got an added LED voltage drop of \$20\:\text{mV}\$ at \$1\:\text{mA}\$. Going to \$10\:\text{mA}\$ would add another \$180\:\text{mV}\$, which is too much. But only \$3\:\text{mA}\$ would only add another \$40\:\text{mV}\$. But then the Shockley equation wouldn't be \$60\:\text{mV}\$ difference anymore, now instead only \$30\:\text{mV}\$.

But I think that \$I_\text{C}=4\:\text{mA}\$ might be about right. This should yield \$V_\text{BE}\approx 754 \:\text{mV}\$ (Shockley) and \$V_\text{LED}\approx 1.68\:\text{V}\$ (from our two-param model.) The base voltage would be \$2.41\:\text{V}\$ (Thevenin calculation.) Now, \$2.41\:\text{V}-1.68\:\text{V}=730\:\text{mV}\$. And that is very close.

The exact equation looks like this:

$$I_\text{C}=\left[\frac{\beta\cdot V_T}{\left(\beta+1\right)\cdot R_\text{ON}+R_\text{TH}}\right]\cdot\operatorname{LambertW}\left[\frac{I_\text{SAT}}{\left[\frac{\beta\cdot V_T}{\left(\beta+1\right)\cdot R_\text{ON}+R_\text{TH}}\right]}\cdot e^{^\frac{V_\text{TH}-V_\text{FWD}}{V_T}}\right]\approx 3.267\:\text{mA}$$

(Here, \$R_\text{ON}=20\:\Omega\$ and \$V_\text{FWD}=1.6\:\text{V}\$.)

As you can see, the math is esoteric. Also, I made a "model" for the LED and I've no idea at all if it is close, or not. It's just what old red LEDs were close to. Today? Could be different. Probably is. And, in any case, LEDs really obey something much closer to a modified Shockley equation. So if you think the above equation is bad, you should see it with the LED modeled more accurately with the modified Shockley.

The end result here is that we can probably say that the BJT is not saturated and is, instead, operating in its active mode. (The voltage drop across \$R_3\$ is not large enough to push the BJT into saturation.)

So... What exactly were you wanting from your schematic? I honestly don't know.

And none of the above is anywhere near as complicated as what Spice does.