But now, my secondary question is (in general). Are my current values correct?
Your current values seem correct.
According the datasheet of the PS2802, \$I_f\$ = 14 mA gives a \$V_f\$ = +/- 1.2V.
1.2V across 220 ohm yields a current of 5 mA.
14 mA + 5 mA = 19 mA = \$I_{total}\$.
19 mA through 1.2 kohm gives a 22.8V voltage drop.
And 22.8V + 1.2V makes 24V.
The other way around, when you don't know \$I_f\$ and \$V_f\$ yet (and you don't have LTspice or that specific component in LTspice) you want an expression of \$V_f\$ that gives a corresponding \$I_f\$ (or the other way around) to be used in the FORWARD CURRENT vs. FORWARD VOLTAGE figure given by the PS2802 datasheet.
So, let's start with
- \$I_{total} = I_f + I_{R1}\$
The voltage across the LED of the PS2802 is the same across R1:
- \$V_f = R1 \cdot I_{R1} = 220 \Omega \cdot I_{R1}\$
Rewriting gives
- \$ I_{R1} = \frac{ V_f }{ 220 \Omega } \$
The current through R2 is the input voltage minus the forward voltage, divided by R2
- \$I_{total} = \frac{ V_{in} - V_f}{R2} = \frac{24V}{1200 \Omega} - \frac{ V_f}{1200 \Omega} = 20 mA - \frac{ V_f}{1200 \Omega}\$
Substituting 4 in 1 gives
- \$ 20 mA - \frac{ V_f}{1200 \Omega} = I_f + I_{R1} \$
Substituting 3 in 5 gives
- \$ 20 mA - \frac{ V_f}{1200 \Omega} = I_f + \frac{ V_f }{ 220 \Omega }\$
Rewriting gives
\$ I_f = 20 mA - \frac{ V_f}{1200 \Omega} - \frac{ V_f }{ 220 \Omega }\$
\$ I_f = 20 mA - V_f \cdot \frac{71}{13200} \frac{ 1}{\Omega}\$
Now you can add (\$ I_f, V_f \$) points to the figure called FORWARD CURRENT vs.
FORWARD VOLTAGE. (You cannot draw a linear line through these points, because the FORWARD CURRENT axis is logarithmic.)
Where these points intersect the (e.g.) 25°C curve, that point gives you the exact \$ V_f\$ and \$ I_f \$.