Capacitor-Based Power Factor Correction

I might be missing something fundamental here but why are you powering the load (LEDs) through the parallel combination of R1 & C1?

So, this C1 in paralel circuit R1C1 is a "resistance" for AC, where the capacitor has reactance Xc of the suitable value calculated for the current LED (20mA).Such circuit practically does not heat.

The formula for the "resistance" Xc is here: https://en.wikipedia.org/wiki/Electrical_reactance

Calculation of this value should be able to even elementary school student.

None of the elementary students I asked about that equation could do it.

They didn't have that app on their cell phones.

I can't read the schematic, but what is the power source coming in on the left side? Why are you using the bridge rectifier? What are the power supply exact specifications - volts, amps?

Wait, you are using line voltage and using the 2 resistors as a voltage divider? This is a good way to get injured or killed. Count me out.

Apparently, his is = Unsave at Any _^Speed Use.

What are RC-time constants of the TWO -- series & parallel -- RC networks and do they match or buck the 220VAC 50HZ input voltage?

If I understand correctly, you are paralleling 3 to 4 of these circuits on the output of an inverter. I think the inverter is breaking into a parasitic oscillation. Get rid of the parallel capacitor for the resistor at the left of the schematic. Better yet, run the lights directly from the battery, and skip the inverter.

C1 is vital for this particular circuit topology (an ac voltage divider) to work correctly.

;)

Not at 50 or 60 Hz.

Yes at 50 or 60Hz!

http://cr4.globalspec.com/comment/133640

It works well if properly designed and the current is fairly constant.

I'm impressed, you used yourself as a reference!

In a compensated voltage divider, the resistors are adjusted at low frequencies, and the capacitors at high frequencies (usually above 1 MHz). There is no need for a capacitor here. The resistor just wastes the extra power. This whole approach is completely unnecessary. He should just use battery that is supplying the inverter.

I'm impressed, you used yourself as a reference!

Why shouldn't I, I have hundreds of products in the market using this style of power supply.

The circuit posted isn't that efficient anyway, R1 should be in series and R4 would be better to be a Zener (as mentioned in the posted reference).

This whole approach is completely unnecessary

Which is what I already pointed out in post #9 and #12.

Let me guess, the inverter you are using is a standard square wave inverter not a true sine wave type inverter right? You are lucky the inverter hasn't blown up, they don't like highly capacitive loads!

Why don't you just power the lights of whatever DC source is powering the inverter? Yes if it is a battery supply the battery voltage will vary a bit as they discharge, but not really enough for it to be noticeable, and cheaper than having to buy a better quality true ac sine wave inverter!

Dear Sir, thanks for your post, you understand my problem exactly, actually here in india 70% UPS user have only square wave inverter (UPS), and thats why instead of changing hues amount of square wave inverter, i want to make my device suitable for square as well as true sine wave inverter, as you wrote in your post, square wave inverter not run large capacitive load, so if in each circuit we add some inductor value like 10mH Line filter at Main AC line, so can i reduce this problem please guide... thanks a lot.

It may be possible to modify your circuit to allow it to run on a more square than sine wave form but it is not very simple and will likely involve a lot of trial and error and possibly damage to your square wave inverter (which is going into current limit as the current increases as you have found). You would have to match the circuit characteristics to the inverter characteristics.

I cannot recommend this approach using your existing circuit. It would be far better to use a small transformer to step down and isolate the inverter voltage from the LED circuit instead of this voltage divider circuit. Not only would this fix the problem but it would also make the circuit safer, the price of a transformer is not that great especially if weighed against having to replace a broken inverter.