This might start an argument.
First, yes you can, and I have done it before with 24 vac
I played around for over 2 hours last night with this, and I now have something rigged up with test leads (and working), but the math (ohm's law) doesn't add up.
Half-wave rectified AC might be the problem as to why.

Generic T5 LED - first I tried a 4700 ohm series resistor only, the LED lit, but the 1/2 watt resistor started cooking after a few minutes. 120 / 4700 = 25.5 mA
I squared x R gives over 3 watts dissipated in this 1/2W resistor.

Next I tried higher resistance values trying to find a value that would not cook a 1/4W resistor. I need to fit this in a very small space, which is why I want a minimum of components. Plus I bought some 120v acdc panel mount LEDs and you can see that there is only 1 additional series component inside (I assume a resistor), but I can't read the value. Anyways, 100K ohm gives 1.2 mA with .144 watts dissipated in the resistor. For whatever reason, the LED was pulsating (60 HZ?) and then blew.

New LED with same 100K resistor plus a 1N4001 diode added in series. This worked and lit the LED to an acceptable brightness. I tried to lower the resistance value and this LED blew also. I found a 1/4W 120K resistor and this is how I have it set up right now (with diode).

How the heck do you analyze this circuit? Here is some measurements done with my Harbor Freight multimeter (obviously not true RMS):
Line: 121 vac
series resistor = 120K ohm
diode: 1N4001
1 mA calculated (120 / 120,000)
DC amps measured = 440 micro amps
118v AC across the resistor
53.3v DC across the resistor
AC volts across diode: 34v (one direction), .3v (the other direction)
DC volts across diode: 20.8v (both directions)

 

Never heard of a LED running off A/C...

 

Neither did I...………… until I did it...………… and it works.

I found this on another site: What's the simplest way to connect a 5v LED to a 120v AC circuit device?if by “simplest” you mean the least number of uncomplicated components… then a diode and resistor. The diode is needed to block the reverse biased current which will destroy the LED. The resistor will drop almost 115V and waste a lot of power:

for a 10ma LED current… about 10K, needs to dissipate over 1watt…

Best to use a 10Kohm 5watt power resistor, in series with the diode (like 1N4001) and the LED.

if your LED is bright enough at 1ma, then 100Kohm 0.5watt resistor could work.

 

So it's still DC. You basically constructed a rectifier power supply.

 

Well, sort of. It's just that it is impossible to calculate any proper values for any components because I am not dealing with straight DC. This is half-wave AC. That's why you just connect components together until you get what you want.


You can (sort of) get away with calling it DC only because it never goes negative.

 

What's the end goal? It seems like an exercise in futility, unless you just want to do it to figure it out

 

For a project. I only have space for an LED with minimal components to make it work on 120 vac.
BTW, they are now making LEDs for AC.

 

you can easily calculate the Vdc from the Peak or RMS of the AC voltage so you can analyze this... half wave rectifiers are very common, here is some quick reading for you that should help as an example

https://www.electronics-tutorials.ws/diode/diode_5.html

To summarize though

 

Thanks Alan. I read the link and here's what I came up with:
V dc would equal 38.178 volts. But a cheap HF meter will not read this correctly since this is pulsating DC.
So for simple analysis, the diode does not drop hardly any voltage, so we will ignore that.
38 volts across a 120K resistor (brown, red, yellow) gives a current of .316 mA. And the power dissipated by the resistor then would be (I squared x R) = 12 milli watts.

LEDs typically run with 20 milli amps through them. The calculations come up with 316 micro amps, which should not even illuminate the LED.
But it does, I already put the circuit together and installed it.