RS-232 Tx / Rx monitoring LEDs?

[Brent Hilpert]

On 2015-Aug-22, at 11:55 PM, drlegendre . wrote:
> On Sun, Aug 23, 2015 at 1:17 AM, Chuck Guzis <cclist at sydex.com> wrote:
>> On 08/22/2015 10:23 PM, dwight wrote:
>> 
>> I would think the reverse voltage sum of the diodes is enough.
>>> Different diodes also can handle different voltages. Since the sum
>>> of the forward voltages is enough to handle AC, I'd suspect the
>>> reverse voltages each would handle is quite small as well.
>>> The problem is when the current limiting is done with a resistor
>>> that in the forward direction drops a lot of voltage.
>>> The diode has to handle the voltage until breakdown when reversed.
>>> If the resistor was handling 1 Watts, with the right break down,
>>> the LED could be taking .5 Watts. This is more than most are designed
>>> for.
>> 
>> ...and that's just the nub of it.  The success of this depends largely on
>> the consistent characteristics of every LED in the string.  Since LEDs tend
>> to fail short if submitted to overvoltage, I've often wondered if a spike
>> in the AC supply would precipitate a cascade failure in the string.  I've
>> looked hard and there are no rectifier diodes in the string--just the LEDs
>> themselves.  Probably saves about 5 cents or so of manufacturing cost.
>> 
>> I've also seen LED "night lights" from China that employ nothing more than
>> a safety capacitor (usually about 104) in series with a resistor connected
>> to two back-to-back LEDs, all across the AC line.
>> 
>> I've wondered what the lifetime of such a setup is.
>> 
>> --Chuck.
>> 
> I've also seen C-R series voltage dropping circuits, here & there.
> 
> Correct me if I'm wrong, but doesn't the series cap dissipate power just as
> it would, were it a series resistor? I mean, if the LED is passing 20mA,
> the cap is also doing 20mA - and at whatever the Vdrop is.
> 
> Right? If not, why?

I doubt if any brief explanation here is going to the topic justice. Look up power factor or reactive power.
FWIW:

The impedance (capacitive reactance, Z=Xc=1/(2*pi*f*C) of the C does produce the desired voltage drop but the C also shifts the phase of the current relative to that of the V. To apply the power equation P=VI properly, you can't just multiply the RMS values together, you multiply the instantaneous values of the V & I sine waves together through a cycle. You get a third sine wave, that for power. If V & I are in phase, the power sine wave will all be in the positive region and is real power consumption. When they are out of phase, some portion of the power sine wave will be negative: a portion of the energy the C sucked down the line is being returned during each cycle.

Yes, it does reduce energy consumption relative to a purely R solution.

On a large scale, the power company doesn't like it because it unnecessarily adds to the currents circulating in the system, but then, this is from C which shifts the current in one direction, so it's doing some compensation for the inductive wall warts you have plugged in around the house, which as L shift the current in the other direction.