A Simple AC powered LED Lamp

[Apple-iPad]

**broken link removed** All your replies can help me too. Thank you.

 

[zmint]

The method I used was to subtract the total voltage drop of all of the LED's in series from the RMS voltage of the lines. Then use that value to calculate the needed impedance of the capacitor at the normal line frequency and voltage

This, is to calculate the value of C1 and I guess this must be a non-polarized (without + -) Cap., Right ? What about the Voltage rating of this Cap ?

For the R1 and C2 values I just use any spare capacitor I have laying around thats between 20 - 250 uf and has a higher voltage rating than the voltage drop the LEDS produce.

And this is a polarized Cap. Am I Right ?

 

[tcmtech]

Yes its a non polar capacitor like a poly or mylar type.
As far as the voltage rating I try to use at least a 2X over the RMS to LED voltage difference.

C2 is a electrolytic capacitor.

 

[zmint]

Yes its a non polar capacitor like a poly or mylar type.
As far as the voltage rating I try to use at least a 2X over the RMS to LED voltage difference.
C2 is a electrolytic capacitor.

In relation to this method, here is a relevant Extract' from a circuit, which one of the respectable members, suggested, in one of his post
............... “The capacitor is around 4.7µF 440Volts. The schematic looks something like this...

I'm shooting from the hip on some of these values, but they are in the ballpark. I was figuring on 10µF 250V polarized capacitors, but if you can find 4.7µF non-polarized 500VAC caps they will work (but are harder to find)”

Keeping in view the last few words, please put some lite on using polarized caps (C1, C2) instead of AC caps. for limiting Voltage & current. What are the points to be kept in mind while working with, this scenario. How do it function ? Say for e.g I need a .22 uF 400 or 600V AC non pol. cap., but I want to use polz. cap. instead, what will be the values of C1 & C2 pol caps.

 

[zmint]

It seems everybody is following the quote "Issue a general safety warning. Then look the other way and allow stupidity the chance to eliminate itself."

But for other enthusiastic DIY hobbyist, I managed to dig out some information which may me helpful to someone we reads this post later.

The article states :
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"A polarized capacitor forms the dielectric between the conducting plates electrochemically. Voltage must be applied in one direction only across the conducting plates to form the dielectric insulating film. The applied voltage across the capacitor must always be in a particular direction, and the capacitor is externally marked to indicate the proper voltage polarity. Reversing the applied voltage can cause the insulating film to rapidly fail, and can cause the capacitor to actually explode from rapid production of gas within the shell of the capacitor.

A non-polarized capacitor uses a dielectric material with uniform insulating properties (such as mylar, mica, glass, etc.), and can be installed in a circuit without regard for the polarity of the applied voltage.

Also polarized capacitors have big capacity in small size (above 1 microf), non-polarized capacitors can have big capacity but sizes are very big.(So they are unuseful in most projects) "

Question : "Can I paralell 2 polar cap to form a non-polar cap ? "

Answer : NO YOU CAN'T.
Dielectric of polarized capacitor is Al2O3 for Aluminium and TaO2. for tantalum capacitors. Polarized cap. must be under minimal 2V DC otherwise is dielectric resolved.
It is possible to form this cap. on a higher voltage.

You must connect them antiserial. That means + polarity connected together and - and - used externally. Or viceversa. Voltage rating is as for single capacitor.

In some applications you can use serial pol. cap (+--+ or -++-) with parallel diodes (CAAC or ACCA)

You can use caps back to back for non-critical applications, it is better to use them in conjunction with the diodes, but their characteristics will never be exactly the same as a true non-polarized cap.
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Although, that technical lingo did explain something to me, yet I am still on, square one.
How to find the value of Pol. cap. to be used for substitution for non-Pol. caps.

 

[tcmtech]

Electrolytic capacitors dont work in AC applications for continuous duty usage.
AC circuits need non polar capacitors. There is no substitution equivalent.

 

[Hero999]

I don't see why you can't use a non-polarised electrolytic in an AC application as long as the voltage and ripple current ratings aren't exceeded. I've seen plenty of non-polarised electrolytic capacitors connected to single phase induction motors.

 

[zmint]

Electrolytic capacitors dont work in AC applications for continuous duty usage.
AC circuits need non polar capacitors. There is no substitution equivalent.

Thanks, for putting the final lid on that one. It was bothering me for long.


Now that, I have starting to understand the involved concepts and also have Electronics Assistant's calculators to help me with my electronic calculations, still have some confusions.

Already knowing about role of rectification, Cap for eliminate flicker, LED just bother for the current scenario, Total Vf of LED connected series is added Vf of All etc. etc., For better understanding of the concept of Capacitive Reactance what Voltage should I expect to measure at points A and B in all the attached 4 circuits (1, 2, 3 & 4) , given the Capacitive Reactance of .47uF is 6.773 Kilohms at 50 Hrz, LED Vf=3.5 & If = 20 mA ?



What I figured out is, that the Cap. will give constant current but the Volt will vary according to the load, but what is the LED load in each of first three cases ? The forth one is obviously (3.5 x 4 = 14 V). Guess, I am right ?

 

[tcmtech]

Your voltages across the A and B points will be the toatal number of LEDs in series times their forward voltage drop. If you have 10 white LEDs with a forward drop of 3.5 volts each then your A to B voltage would be around 35 volts.

In the ballast capacitor equation you would take the line voltage RMS and subtract the 35 volts from it and then calculate for what current you need to supply. if ifs a 240 volt 50 Hz line and you had 10 LEDs running at 20 ma you would calculate for a impedance of (240 - 35)/.02 = 10250 ohms.

Then using that you can calculate your capacitance for your line frequency. On a 240 volts 50 Hz line you would need around .31 Uf.

 

[Ubergeek63]

Already knowing about role of rectification, Cap for eliminate flicker, LED just bother for the current scenario, Total Vf of LED connected series is added Vf of All etc. etc., For better understanding of the concept of Capacitive Reactance what Voltage should I expect to measure at points A and B in all the attached 4 circuits (1, 2, 3 & 4) , given the Capacitive Reactance of .47uF is 6.773 Kilohms at 50 Hrz, LED Vf=3.5 & If = 20 mA ?

there is an easy fix for it:

put a rectifier bridge before the LEDs
put your electrolytic in for the flickers
remove half the LEDs

dan

 

[rocky01]

Hi all,

I am posting one circuit i found over the net. please comment on it if any changes are to be made. View attachment 64229