Ok, here is what I got: I used the formula R = (Vs - Vf)/If to come up with a 410 ohm resistor.
Great!
In the real world, resistors aren't available in all possible values. There are several series, called E6, E12, E24, E48, E96 and E192, that define what values are available. The E-number is the number of different values in each decade; E12 uses twelve values: 10, 12, 15, 18, 22, 27, 33, 39, 47, 56, 68, 82, then repeats for the next decade; E24 adds another twelve values in between those ones: 10, 11, 12, 13, 15, 16, 18, 20, 22, 24, 27, 30, 33, 36, 39, 43, 47, 51, 56, 62, 68, 75, 82, 91. And so on for the higher E-numbers, except that starting at E48, numbers resolve to three digits, not two, and are recalculated, so many E24 values don't have exact matches. Generally, more accurate resistors are available in higher E-number series. Capacitors, and off-the-shelf inductors, also use E-series numbering.
So in the E24 series, your closest options (closest to 410 ohms) would be 390 and 430 ohms. Or you could go to an E48 resistor and get a nominal value of 412 ohms. But the nearest E24 (even E12) value is close enough for this application.
At one point I had an average v.pot. reading of 10.4 and I used 2.2 as my v. drop across the LED divided by the 20mA continuous current required for the device. For some reason when I refreshed the schematic, my avg. volt reading is now 11.9, which yields a 490 ohm resistor, which gives me about 17mA avg current. Strange...
The average capacitor voltage will vary slightly depending on how much current you draw from it. Have a look at the capacitor waveform with different resistor values and you'll see that lower resistances (higher currents) cause deeper ripple, and therefore a lower average voltage. That's because the peaks are almost entirely determined by the AC input, which is constant, but when you draw more current from the capacitor, it discharges quicker, so it discharges further in between the peaks, making the mean voltage lower.
You can use the .step command in LTspice to do it for you - change the value field of R1 from "410" to "{R}" (without the quotes) and add a directive (not a comment; it must appear in black) that says ".step param R list 330 470 560" (without the quotes). Use whatever values you want in the list.
When you run the simulation, LTspice will run the simulation once for each value in the list, and when you click a node to graph its voltage, or Alt-click a wire to graph the current through it, LTspice will show the graphs for all of the simulations, overlaid on top of each other, so you can see how varying the resistance affects any aspect of the circuit you're interested in.
The design looks successful according to how much current is being driven through the LED, however, isn't the v.potential across the LED too high at 12v
No; the voltage across the LED isn't 12V. LTspice will tell you what it is; it will be around 2V for a red LED. The 12V is applied to the series combination of the resistor and the LED, and the resistor acts as a spring, and "takes up the slack" in terms of voltage, while limiting the current.
I also found interesting that if you put the resistor before the capacitor, the circuit displayed different behavior in the waveform - the variance in the current was greater.
Right. With the AC going straight into the bridge rectifier, feeding straight into the capacitor, the peak voltage on the capacitor is almost entirely determined by the AC voltage, so changing the load current only affects the troughs in the capacitor voltage waveform. If you insert a resistor between the AC and the bridge, or between the bridge and the capacitor, the peak voltage on the capacitor will also be dragged down by the load current.
I came up with an analogy for voltage, current, resistance and capacitance that I call the "DTS model". One day I want to make a nice animated demo for it, but in the meantime you can check out
https://www.electronicspoint.com/threads/.248766/#post-1470672 for a crude explanation. I hope it helps; please let me know what you think of it.
Thanks again for all your help and instruction
No problem! It's nice to work with someone who listens, learns, and takes some initiative. It's not as common as you'd think.