LED driver without IC and Transformer

[Muhammed Arshid]

I think to make LED lamps commercially. So I got circuit for it. But resisters are heating.
What is the power conception in this circuit ?
Please help me..

 

[alec_t]

That circuit is an accident waiting to happen. Each cap C1-C3 passes ~70mA, so the total current is ~210mA. The voltage drop across R2 or R3 is therefore ~21V, so R2 and R3 will each dissipate ~4W. They will get very hot. But R2 is only rated 1W, so is likely to catch fire! R3 doesn't even have a rating but is equally likely to catch fire!

 

[ronsimpson]

The original circuit was made by yet another person that thinks LEDs need voltage. He regulates down to 4.7 volts then used a resistor and LED.

LEDs are current devices not voltage devices.
Addressing what Alec_t said:
R2 is 18 ohms to get the heat down below 1 watt. You could use two 18 ohm 1W resistors.
Remove the other resistors and Zener.
The voltage on C4 (and C5 if you use it) will be 2.8 to 3.3 volts so you you can use smaller caps.
I have made these with out any capacitors.
C1, C2, C3 make 200mA so use it to directly drive the LED. Forget the 4.7V idea.
Many people use the 1N4007 diodes to with stand the 230VAC. The diodes in this design will not see more than 5 volts so use the 1N4001.

What about this. Comments please.

 

[Externet]

I do not get why too many people use bridge rectifiers for leds from mains, raising the already high voltage even more.
Besides, if you want to to use rectifiers, why use rectifier diodes ??? Use led diodes instead of rectifying diodes ! They shine themselves while doing its job !

One way I do it:
AC phase---------------||-------------leds counterparalleled-------------/\/\/\/--------------neutral

It is convenient to add to the above, a ~270K resistor in parallel to the capacitor. Leds can be a counterparalleled series string.
I have a bunch in operation for years --->https://s588.photobucket.com/user/Innernet/media/P1010435_zps200dcafc.jpg.html?sort=6&o=73

 

[ronsimpson]

Something like this.

If you use the 4 diodes and one LED and NO capacitor there will be 100hz or 120hz flashes of light. I had one design rejected because these flashes can be seen if you walk by fast.
Using the two LEDs design with no (4 diodes) you will get 50/60hz flashes. The two LEDs do not look like one. Under some conditions it looks just fine. Used as a night light while moving you eyes from side to side it is not good. (quote marketing)

Externet, I agree with your idea. Very simple. I could not get it sold.

 

[Muhammed Arshid]

Thanks for all..

I think, life of LEDs will reduce in this circuit. 21V will reach in LED. LED rated 3.6v only.
I have one more doubt, I want add 5+ LEDs in series.
What is the power conception in bellow circuit (if I add 5 LEDs)?

 

[alec_t]

21V will reach in LED

How do you get that figure?
You can simply put 5 LEDs in series in that circuit. Just make sure the voltage rating of the cap is greater than 5 x LED forward voltage.

 

[Muhammed Arshid]

But I think , it will reduce life of LEDs. So we want use zener diodes .

 

[ronsimpson]

I think, life of LEDs will reduce in this circuit. 21V will reach in LED. LED rated 3.6v only.

There is no 21V. Stop thinking voltage it is current. The capacitor divider is like a constant current source. It will put 200mA into the LED and the LED will limit the voltage to what it needs. 1 LED will have about 3V. 2 LEDs will have about 6V total. 5 LEDs will have about 15V. The current will be about the same in all cases. At 6 to 7 LEDs the current will be down about 10%.

Alec_t is right. 5-LEDs The voltage will be 5X so the capacitor and diodes will see 15V. (about 15, it will change with temperature and from one batch of LEDs to another)

A LED acts like a Zener diode. A 3V LED is like a 3V +/- 20% Zener.

The circuit in post #1 was never built by the author. The resistors will be too hot. It is just an idea. He uses a 1W Zener to limit the voltage to a (3V 1W LED). You don't need to limit the voltage to a LED with a Zener. The LED is a Zener diode. You need to limit the current.

 

[Muhammed Arshid]

Now I will connect 1 LED as bellow (for testing)

 

[Muhammed Arshid]

Can I use 100uf, 100V capacitor (C4) ?

 

[alec_t]

You can; but the greater the capacitance the less noticeable the flicker.

 

[Muhammed Arshid]

I used 50v capacitor ( C4 ) heating .. Not working now ..

 

[ronsimpson]

C4 is hot?
In backwards?
Too high a voltage?
Leakey?
C4 should not be hot!

 

[Muhammed Arshid]

I add new capacitor (c4) . But it is heating. I connect -ve (black line) to -ve terminals.

 

[alec_t]

Are you sure all the rectifier bridge diodes are correctly orientated and in good working order?
Have you checked for wrong wiring connections anywhere in your circuit?

 

[Muhammed Arshid]

yes .. I have 100% sure ..

 

[alec_t]

If you are using the circuit in post #10 and the LED becomes disconnected or fails open-circuit then the cap voltage will rise to the mains voltage peak level and the cap will overheat and die (perhaps explode).

 

[Muhammed Arshid]

Sorry .. I was disconnected LED first. But my doubt is what is the efficiency and using power of this circuit ?

 

[alec_t]

Efficiency is high (in terms of power for which you get billed). The 1uF caps drop most of the mains voltage but dissipate very little power because the current is ~90 degrees phase shifted relative to the voltage across them. The power used is mainly due to the LED (1W), plus ~200mW due to the bridge.