LED drive from AC main

[yassin.kraouch]

Hi,
Can Anyone please explain to me this schematics ?
i was not able to understand how they calculate the voltage drop ? to drive LED with capacitor ?
https://320volt.com/el-yapimi-enerji-tasarruflu-led-lamba-220v-50-led/

 

[betwixt]

The capacitor is used as though it was a resistor with equivalent value to it's reactance. In other words, it behaves like a resistor with value of 1/(2*pi*f*C) where f is in Hz and C is in Farads.

Beware that this provides NO safety isolation so no part of the circuit should be touchable and that it is susceptible to damage from high voltage spikes on the AC input. The capacitor MUST be non-polarized and rated for the AC voltage. Ideally it should also be a safety rated capacitor. Do NOT use a normal resistor as the fuse if you use the top schematic, the type it uses is guaranteed to go open circuit if it overheats, a normal resistor will remain conductive and fail to cut the current.

Brian.

 

[yassin.kraouch]

thaks for the amswer, but how do i calculate the output voltage ?

 

[betwixt]

You don't. The output voltage is whatever the LEDs need. That comes from the manufacturers data sheet. If you use more than one LED in series, add their Vf together to find the total. So for example if your LEDs had Vf of 2V and you use 10 of them, the output voltage has to be 20V. From there you can calculate how much voltage has to be dropped from the AC and knowing the LED current you use Ohms law to work out the required resistance. Then find the capacitor value that gives the same reactance as that resistance using the formula I gave.

Brian.

 

[yassin.kraouch]

can you please give me a real example with calculation ? i am a bit confused ?
thanks

 

[betwixt]

I'll try but I have to make some assumptions: Vf of each LED is 1.6V, you want them to pass 20mA, your AC voltage is 230V and your AC frequency is 50Hz. If yours is different, just change the figures and recalculate.

Say you have 10 LEDs in series, the total voltage across them will be 10 x 1.6 = 16V.
The voltage you have to drop is 230 - 16 = 214V.
The current through the LEDs is 0.02A so the resistance needed to drop 214V at 0.02A is (R = V/I) 214/0.02 = 10,700 Ohms.

So the total resistance in series has to be 10,700 Ohms but that is partly from a resistor and partly from the reactance of a capacitor. The resistor will generate heat so we want to keep that part of the total relatively small. Capacitors come in fixed values so the best strategy is to find the nearest standard capacitor value that gives a reactance less than needed then make up the difference with a real resistor.

The frequency of the current through the capacitor is twice the AC frequency because the bridge rectifier is conducting on both polarities of each cycle so the calculation is based on 50 x 2 = 100Hz.

The ideal capacitor value with a reactance of 10,700 Ohms at 100Hz is 1/(2*pi*100*10,700) = 148.74nF.

The nearest standard value with a lower reactance (higher capcitance) is 150nF (0.15uF) which has a reactance of 10,610 Ohms so you make up the difference with say 100 Ohms of resistor.

Brian.

 

[yassin.kraouch]

I'll try but I have to make some assumptions: Vf of each LED is 1.6V, you want them to pass 20mA, your AC voltage is 230V and your AC frequency is 50Hz. If yours is different, just change the figures and recalculate.

Say you have 10 LEDs in series, the total voltage across them will be 10 x 1.6 = 16V.
The voltage you have to drop is 230 - 16 = 214V.
The current through the LEDs is 0.02A so the resistance needed to drop 214V at 0.02A is (R = V/I) 214/0.02 = 10,700 Ohms.

So the total resistance in series has to be 10,700 Ohms but that is partly from a resistor and partly from the reactance of a capacitor. The resistor will generate heat so we want to keep that part of the total relatively small. Capacitors come in fixed values so the best strategy is to find the nearest standard capacitor value that gives a reactance less than needed then make up the difference with a real resistor.

The frequency of the current through the capacitor is twice the AC frequency because the bridge rectifier is conducting on both polarities of each cycle so the calculation is based on 50 x 2 = 100Hz.

The ideal capacitor value with a reactance of 10,700 Ohms at 100Hz is 1/(2*pi*100*10,700) = 148.74nF.

The nearest standard value with a lower reactance (higher capcitance) is 150nF (0.15uF) which has a reactance of 10,610 Ohms so you make up the difference with say 100 Ohms of resistor.

Brian.

thanks for the explanation, but are you sure that the frequency is @x%) hz ?
because i found in this qrticle that they put only %) hz https://www.electroschematics.com/5678/capacitor-power-supply/

- - - Updated - - -

sorry i mean 2*50 hz

 

[chuckey]

Both the diagrams waste money!!! If you put the LEDs in back to back pairs then each combination will run of AC, as one conducts on alternate half cycles, so protecting the other against excessive reverse voltage. As each LED is now driven by 1/2 wave AC, the current during the conducting phase should be twice that of the running current. This arrangement should not be used to illuminate machinery as there will an enormous amount of 100 HZ "flicker", which will not effect the eye but might cause rotating parts to strobe so appear stationary.
Frank

 

[betwixt]

You 'hit the nail on the head' Frank, the original circuit used a bridge rectifier AND a smoothing capacitor to prevent that happening. The effects of strobing, especially on synchronous motors should always be considered, especially in places where health and safety are not considered as important than here. If a chain of anti-parallel LEDs are used, they should be wired as two leds in parallel at each step of a serial chain rather than two parallel chains of serial LEDs. The reason being that in long chains the differences in Vf can accumulate and exceed the PIV of the other chain, it also gives a degree of protection should any single LED die.

I apologize for my calculation error yassin.kraouch, the same formula applies, just use the native line frequency. I must be working on too many projects simultaneously and I'm muddling them up! (or it's old age, which is more likely :-( )

Brian.