I like to drive 5 LED (white) directly from 240V AC , the application is to replace normal lighting bulb..
The LEDs are connect in series, the Vf for each LED is 3.2V, current can be 100mA, so total voltage require is about 16Vdc 100mA.
Beside using transformer to step down the voltage, what is other possible way to drive the LEDs (100mA) directly from AC240V?
Is there any dedicated IC for this application? :roll:
I like to drive 5 LED (white) directly from 240V AC , the application is to replace normal lighting bulb..
The LEDs are connect in series, the Vf for each LED is 3.2V, current can be 100mA, so total voltage require is about 16Vdc 100mA.
Beside using transformer to step down the voltage, what is other possible way to drive the LEDs (100mA) directly from AC240V?
Is there any dedicated IC for this application? :roll:
A transformer is the only sensible way, either conventional, or a switch-mode design. Any crude 'dropper' technique is going to waste most of the electricity used - in your case 93% wasted! - which rather makes a mockery of trying to use low energy LED's.
connect a series 2k2 resistor and a diode in the other way the leds are. if the leds are from "+" to gnd then diode must be from gnd to "+".
(gnd is one hole in AC and "+" is the otherone)
the diode must take the voltage so use 1n400X series. It is there so that the reverse voltage wouldn't kill your diodes.
well, this schematic has always worked for me. nothing had birned down (jet). Only thing different is that in setead of this "protection diode" i used anothed LED with a resistor.
If you are so STUPID to try this circuit out with a 1\8W resistor you shouldnt be operating with AC at all. it HAS to be 2W or more. (mine is 5W).
well, this schematic has always worked for me. nothing had birned down (jet). Only thing different is that in setead of this "protection diode" i used anothed LED with a resistor.
If you are so STUPID to try this circuit out with a 1\8W resistor you shouldnt be operating with AC at all. it HAS to be 2W or more. (mine is 5W).
Why has no one suggested the usual method of using a capacitor to limit the current? I have usually seen it with a low-valued resistor also in series, for protection from transients. Of course, you still need the anti-parallel diode (or another LED) across the LED.
Why has no one suggested the usual method of using a capacitor to limit the current? I have usually seen it with a low-valued resistor also in series, for protection from transients. Of course, you still need the anti-parallel diode (or another LED) across the LED.
Because although standard diodes can have very high blocking voltages (little BAT16 can block 50V) the proporties of an LED comes at the price of some of a diode characteristics. Namely blocking voltages
It is not uncommon to have an LED that can only block 5V.
You connect an LED to the mains (in the correct configuration) with no anti-parallel diode, the LED will have to block all the volts and it wont be albe to
Why has no one suggested the usual method of using a capacitor to limit the current? I have usually seen it with a low-valued resistor also in series, for protection from transients. Of course, you still need the anti-parallel diode (or another LED) across the LED.
The capacitor won't dissipate any power. The resistor will, but it's much lower in value than it would be if it were doing all the current limiting. I'm thinking of maybe 220 ohms or so.
A 680nF cap will give an average LED current of about 20mA.
Without the anti-parallel diode, even if the LED were able to stand off the reverse voltage, no average DC could flow through the LED, so it wouldn't light up.
Thanks for reply.
1. I have seen the LED bulb that direct plug to 240VAC, they are compact , light weight, very little heat dissipate,
Anybody know what is the circuit inside?
2. If I want to use switch mode power supply that can deliver 16Vdc 100mA, or constant current mode 100mA,
what IC you suggest?
3. Is it possible to use SCR/Triac technique to perform above task?
I re-read the posts and now know I am referring to the case of resistor dropper and you guys referring to the capacitor dropper. I agree that in case of a capacitor dropper, an anti-parallel diode is needed across the LED for it to work.
The capacitor won't dissipate any power. The resistor will, but it's much lower in value than it would be if it were doing all the current limiting. I'm thinking of maybe 220 ohms or so.
A 680nF cap will give an average LED current of about 20mA.
If you're feeding the LED at the same current, then EXACTLY the same amount of energy HAS to be dissipated somewhere! - as most of the voltage is dropped across the capacitor, the capacitor will dissipate it. The resistor is mainly intended as a 'fuse', if the capacitor goes S/C.
Do the maths! - "dissipation = voltage across capacitor x current through capacitor", capacitors aren't some 'magic' component where standard rules don't apply :lol:
Capacitors used to pass high currents DO get hot, and their failure is a common problem.
Do the maths! - "dissipation = voltage across capacitor x current through capacitor", capacitors aren't some 'magic' component where standard rules don't apply :lol:
Sorry Nigel but you are way off on this one.
Dont forget that in a capacitor the voltage and current are at 90 degrees to each other, a phase angle of 90deg.
In an AC circuit Power = V x I x Cos(phi)
Here phi is 90 deg so Cos(phi) = 0
So no power dissipated in the capacitor.
Nigel Goodwin said:
Capacitors used to pass high currents DO get hot, and their failure is a common problem.
Ok, above I was referring to perfect capacitors, in practice there is a VERY SMALL Equivalent Series Resistance (ESR) to account for losses in the capacitor but for most capacitors it is very small.
However, with capacitors like big electrolytics the ESR is a bit higher, if they are passing large currents they warm up and start to age, the ESR will increase so they get hotter and eventually just give up or explode.
I have only seen the capacitor current limiting scheme discussed in forums - I have never tried it. I don't think it is a good idea unless the line (mains) is really clean (i.e., no big transients). Otherwise, peak currents could be really high, and might destroy the LED.
I stand by my power dissipation comments, which were explained well by JimB.