Forget all the fancy charts. It's actually pretty simple. If you have a device that requires about \$3\:\textrm{V}\$ across it and are applying \$120\:\textrm{V}\$ across it through a resistor, then you have a very good current source. So the LED itself won't matter that much. It's this simple:

$$I = \frac{V_{120\:\textrm{V}}-V_{LED}}{R}$$

This works out to:

$$\textrm{d} I = \left[\frac{-V_{LED}}{R}\right]\textrm{d}V_{LED}$$

So with \$R\approx 50\:\textrm{k}\Omega\$, this means that if \$\textrm{d}V_{LED}=1\:\textrm{V}\$, then \$\textrm{d} I=60\:\mu\textrm{A}\$. That's not a lot of change in the current for a full volt change in the LED voltage. A current of \$2.4\:\textrm{mA}\$ would become \$2.34\:\textrm{mA}\$, or else \$2.46\:\textrm{mA}\$, depending on a full volt change in the LED's required value. That's not a lot of variation.

In terms of percent, you'd compute:

$$\frac{\frac{\textrm{d}I}{I}}{\frac{\textrm{d}V_{LED}}{V_{LED}}}=\frac{V_{LED}}{I}\cdot \frac{\textrm{d}I}{\textrm{d}V_{LED}}=\frac{3\:\textrm{V}}{2.4\:\textrm{mA}}\cdot\frac{-3\:\textrm{V}}{50\:\textrm{k}\Omega}=-0.075$$

In short a 33% change in the LED voltage would result in \$-0.075\cdot 33\%\approx -2.5 \:\%\$ change in the current in the LED.

So that's pretty good regulation. The reason is mainly because of the huge resistor value you are using here. If it were smaller, the regulation would be poorer.

Another reason is that LEDs vary in their voltage based upon the Shockley equation:

$$I_{LED}=I_{SAT}\cdot\left(e^{\frac{V_{LED}}{n V_T}}-1\right)$$

Or,

$$\textrm{d}V_{LED}\approx \frac{n V_T}{I_{LED}}\textrm{d} I_{LED}$$

Here, with \$V_T\approx 26\:\textrm{mV}\$ and \$n\approx 2\$, you would expect about \$50\:\textrm{mV}\$ change in the LED voltage for a doubling of the current through it. That's not much of a voltage change for a fairly significant change in the current through the LED. And to get that kind of change, you'd have to double the current through \$R\$, too. And you know that cannot happen. So the result is that there is only a very modest change in the voltage across the LED even with a significant change in the current through it.

That's another way of seeing how this regulates the voltage across the LED.

No curves required. It's just that you have a LOT of voltage headroom here and this accounts for good regulation.

If you need to see this another way, just imagine a huge voltage of \$1,000,000\:\textrm{V}\$ across a resistor of \$100\:\textrm{M}\Omega\$. The current will be \$10\:\textrm{mA}\$, right? Suppose you have an LED inserted there? How much would the current change if the LED needed \$10\:\textrm{V}\$ instead of just \$3\:\textrm{V}\$? Not much, right? Because the source voltage is so large, and because the dropping resistor must drop so much of the voltage, variations in the LED voltage have almost no effect at all on the current through the LED.

It's as simple as that.

Forget all the fancy charts. It's actually pretty simple. If you have a device that requires about \$3\:\textrm{V}\$ across it and are applying \$120\:\textrm{V}\$ across it through a resistor, then you have a very good current source. So the LED itself won't matter that much. It's this simple:

$$I = \frac{V_{120\:\textrm{V}}-V_{LED}}{R}$$

This works out to:

$$\textrm{d} I = \left[\frac{-V_{LED}}{R}\right]\textrm{d}V_{LED}$$

So with \$R\approx 50\:\textrm{k}\Omega\$, this means that if \$\textrm{d}V_{LED}=1\:\textrm{V}\$, then \$\textrm{d} I=60\:\mu\textrm{A}\$. That's not a lot of change in the current for a full volt change in the LED voltage. A current of \$2.4\:\textrm{mA}\$ would become \$2.34\:\textrm{mA}\$, or else \$2.46\:\textrm{mA}\$, depending on a full volt change in the LED's required value. That's not a lot of variation.

In terms of percent, you'd compute:

$$\frac{\frac{\textrm{d}I}{I}}{\frac{\textrm{d}V_{LED}}{V_{LED}}}=\frac{V_{LED}}{I}\cdot \frac{\textrm{d}I}{\textrm{d}V_{LED}}=\frac{3\:\textrm{V}}{2.4\:\textrm{mA}}\cdot\frac{-3\:\textrm{V}}{50\:\textrm{k}\Omega}=-0.075$$

In short a 33% change in the LED voltage would result in \$-0.075\cdot 33\%\approx -2.5 \:\%\$ change in the current in the LED.

So that's pretty good regulation. The reason is mainly because of the huge resistor value you are using here. If it were smaller, the regulation would be poorer.

Another, separate reason to consider is that LEDs vary in their voltage based upon the Shockley equation:

$$I_{LED}=I_{SAT}\cdot\left(e^{\frac{V_{LED}}{n V_T}}-1\right)$$

Or,

$$\textrm{d}V_{LED}\approx \frac{n V_T}{I_{LED}}\textrm{d} I_{LED}$$

Here, with \$V_T\approx 26\:\textrm{mV}\$ and \$n\approx 2\$, you would expect about \$50\:\textrm{mV}\$ change in the LED voltage for a doubling of the current through it. That's not much of a voltage change for a fairly significant change in the current through the LED. And to get that kind of change, you'd have to double the current through \$R\$, too. And you know that cannot happen. So the result is that there is only a very modest change in the voltage across the LED even with a significant change in the current through it.

Those fancy charts are showing you this effect. And that would take over and explain things, if you had a much smaller voltage to drop. But in your case, even if it were not the Shockley equation but some other behavior instead, you'd still have good regulation because of that huge voltage you are tossing away. You'd have good regulation with a low voltage light bulb, too, for example. So while this effect would account for good regulation with a low voltage source, it isn't the reason here.

No curves required. It's just that you have a LOT of voltage headroom here and this accounts for good regulation.

If you need to see this another way, just imagine a huge voltage of \$1,000,000\:\textrm{V}\$ across a resistor of \$100\:\textrm{M}\Omega\$. The current will be \$10\:\textrm{mA}\$, right? Suppose you have an LED inserted there? How much would the current change if the LED needed \$10\:\textrm{V}\$ instead of just \$3\:\textrm{V}\$? Not much, right? Because the source voltage is so large, and because the dropping resistor must drop so much of the voltage, variations in the LED voltage have almost no effect at all on the current through the LED.

It's as simple as that.

Forget all the fancy charts. It's actually pretty simple. If you have a device that requires about \$3\:\textrm{V}\$ across it and are applying \$120\:\textrm{V}\$ across it through a resistor, then you have a very good current source. So the LED itself won't matter that much. It's this simple:

$$I = \frac{V_{120\:\textrm{V}}-V_{LED}}{R}$$

This works out to:

$$\textrm{d} I = \left[\frac{-V_{LED}}{R}\right]\textrm{d}V_{LED}$$

So with \$R\approx 50\:\textrm{k}\Omega\$, this means that if \$\textrm{d}V_{LED}=1\:\textrm{V}\$, then \$\textrm{d} I=60\:\mu\textrm{A}\$. That's not a lot of change in the current for a full volt change in the LED voltage. A current of \$2.4\:\textrm{mA}\$ would become \$2.34\:\textrm{mA}\$, or else \$2.46\:\textrm{mA}\$, depending on a full volt change in the LED's required value. That's not a lot of variation.

In terms of percent, you'd compute:

$$\frac{\frac{\textrm{d}I}{I}}{\frac{\textrm{d}V_{LED}}{V_{LED}}}=\frac{V_{LED}}{I}\cdot \frac{\textrm{d}I}{\textrm{d}V_{LED}}=\frac{3\:\textrm{V}}{2.4\:\textrm{mA}}\cdot\frac{-3\:\textrm{V}}{50\:\textrm{k}\Omega}=-0.075$$

In short a 33% change in the LED voltage would result in \$-0.075\cdot 33\%\approx -2.5 \:\%\$ change in the current in the LED.

So that's pretty good regulation. The reason is mainly because of the huge resistor value you are using here. If it were smaller, the regulation would be poorer.

Another reason is that LEDs vary in their voltage based upon the Shockley equation:

$$I_{LED}=I_{SAT}\cdot\left(e^{\frac{V_{LED}}{n V_T}}-1\right)$$

Or,

$$\textrm{d}V_{LED}\approx \frac{n V_T}{I_{LED}}\textrm{d} I_{LED}$$

Here, with \$V_T\approx 26\:\textrm{mV}\$ and \$n\approx 2\$, you would expect about \$50\:\textrm{mV}\$ change in the LED voltage for a doubling of the current through it. That's not much of a voltage change for a fairly significant change in the current through the LED. And to get that kind of change, you'd have to double the current through \$R\$, too. And you know that cannot happen. So the result is that there is only a very modest change in the voltage across the LED even with a significant change in the current through it.

That's another way of seeing how this regulates the voltage across the LED.

No curves required. It's just that you have a LOT of voltage headroom here and this accounts for good regulation.

If you need to see this another way, just imagine a huge voltage of \$1,000,000\:\textrm{V}\$ across a resistor of \$100,000,000\:\textrm{M}\Omega\$. The current will be \$10\:\textrm{mA}\$, right? Suppose you have an LED inserted there? How much would the current change if the LED needed \$10\:\textrm{V}\$ instead of just \$3\:\textrm{V}\$? Not much, right? Because the source voltage is so large, and because the dropping resistor must drop so much of the voltage, variations in the LED voltage have almost no effect at all on the current through the LED.

It's as simple as that.

Forget all the fancy charts. It's actually pretty simple. If you have a device that requires about \$3\:\textrm{V}\$ across it and are applying \$120\:\textrm{V}\$ across it through a resistor, then you have a very good current source. So the LED itself won't matter that much. It's this simple:

$$I = \frac{V_{120\:\textrm{V}}-V_{LED}}{R}$$

This works out to:

$$\textrm{d} I = \left[\frac{-V_{LED}}{R}\right]\textrm{d}V_{LED}$$

So with \$R\approx 50\:\textrm{k}\Omega\$, this means that if \$\textrm{d}V_{LED}=1\:\textrm{V}\$, then \$\textrm{d} I=60\:\mu\textrm{A}\$. That's not a lot of change in the current for a full volt change in the LED voltage. A current of \$2.4\:\textrm{mA}\$ would become \$2.34\:\textrm{mA}\$, or else \$2.46\:\textrm{mA}\$, depending on a full volt change in the LED's required value. That's not a lot of variation.

In terms of percent, you'd compute:

$$\frac{\frac{\textrm{d}I}{I}}{\frac{\textrm{d}V_{LED}}{V_{LED}}}=\frac{V_{LED}}{I}\cdot \frac{\textrm{d}I}{\textrm{d}V_{LED}}=\frac{3\:\textrm{V}}{2.4\:\textrm{mA}}\cdot\frac{-3\:\textrm{V}}{50\:\textrm{k}\Omega}=-0.075$$

In short a 33% change in the LED voltage would result in \$-0.075\cdot 33\%\approx -2.5 \:\%\$ change in the current in the LED.

So that's pretty good regulation. The reason is mainly because of the huge resistor value you are using here. If it were smaller, the regulation would be poorer.

Another reason is that LEDs vary in their voltage based upon the Shockley equation:

$$I_{LED}=I_{SAT}\cdot\left(e^{\frac{V_{LED}}{n V_T}}-1\right)$$

Or,

$$\textrm{d}V_{LED}\approx \frac{n V_T}{I_{LED}}\textrm{d} I_{LED}$$

Here, with \$V_T\approx 26\:\textrm{mV}\$ and \$n\approx 2\$, you would expect about \$50\:\textrm{mV}\$ change in the LED voltage for a doubling of the current through it. That's not much of a voltage change for a fairly significant change in the current through the LED. And to get that kind of change, you'd have to double the current through \$R\$, too. And you know that cannot happen. So the result is that there is only a very modest change in the voltage across the LED even with a significant change in the current through it.

That's another way of seeing how this regulates the voltage across the LED.

No curves required. It's just that you have a LOT of voltage headroom here and this accounts for good regulation.

If you need to see this another way, just imagine a huge voltage of \$1,000,000\:\textrm{V}\$ across a resistor of \$100\:\textrm{M}\Omega\$. The current will be \$10\:\textrm{mA}\$, right? Suppose you have an LED inserted there? How much would the current change if the LED needed \$10\:\textrm{V}\$ instead of just \$3\:\textrm{V}\$? Not much, right? Because the source voltage is so large, and because the dropping resistor must drop so much of the voltage, variations in the LED voltage have almost no effect at all on the current through the LED.

It's as simple as that.