Re: AC Led circuits

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Chris Colorado

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Dear all,
Needed a circuit to power up few LED from 240v AC. so designed the attached circuit1 ( refer picture ). as per calculations voltage across B ( refer picture) should be ~50vdc, but when chkd with a digital voltmeter it is ~240vdc, and across A is ~140vac.
But in circuit2 it is ~330vdc ( voltage across 2.2µfd capacitor ).
when I replaced the 2.2µfd cap with a 220v/5w bulb, voltage across it is ~50vdc, and before bridge ( across A ) is ~25vac.
Why it is not always ~50vdc across B?


Many Thanks.
Chris
 

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The output capacitor charges to the peak voltage of the rectified waveform.
Without the output capacitor the output is the average DC that changes quickly from 0V to the peak voltage over and over.
Then the average pulsing voltage is much less than the smoothed peak voltage.
 
You don't need a bridge. Just use this circuit:


**broken link removed**
 
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Thanks,
So how do I adjust voltage @ B to connect 18 white LED in serial ( each 3.0v , TTL ~50v). Thanks again.


Hi there,


If you are using 18 white LEDs in series then you dont have to change anything except maybe the 110 ohm resistor. You do have to connect the LEDs before applying power however, do not connect them after power has already been applied.
You should probably change the 110 ohm resistor to 330 ohms however to limit fast input change currents to around 1 amp. 110 ohm is probably too low to limit transients.
After you connect the 18 LEDs and turn the power on, the 18 LEDs load the output DOWN to a voltage more appropriate for them. This happens because of the reactance of the capacitor.

Just to note, using a bridge rectifier does in fact allow for better utilization of parts. That is, the white LEDs get used to their full potential rather than just about half. What this boils down to is that when compared to a half wave rectifier, the bridge rectifier allows the white LEDs to be used to full power without a larger peak current and that means longer life.
 
The current is limited by the value of the capacitor. We are talking about 7mA per 100u for full wave and 3.5mA for half wave. You can drive a white LED to 17mA for 50,000 hours.
Problems with reduction in life of the LED only comes in when the actual WATTAGE is exceeded. Pulses of high current do not affect the life appreciably.
That's why we can overdrive our fluorescent tube substitutes when they are mounted in a freezer.
A LED on halfwave with a slightly higher current produces more perceived illumination as the eye sees peaks. So this method is more efficient.
 
You can get a much higher efficiency from a LED by pulsing it with a high-current-spike for a very short period of time. Providing wattage limitations are met, you can increase the current to 300% and obvioulsy the ON duration will be a lot less than 30% - from 1% to 15%.
 
Hello again,


An LED pulsed at twice the current as an LED that has a constant current of the same average value produces less light output.
For example, a 20ma LED run at 20ma constant puts out more light than that very same LED pulsed at 40ma 50 percent duty cycle. The reason is because the efficiency goes down with current (look at the curve for a white 20ma LED). You would have to pulse the LED at *more* than 40ma to get the same light output as a 20ma LED running constant current.
 
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You can get a much higher efficiency from a LED by pulsing it with a high-current-spike for a very short period of time.
No.
An LED is efficient at a low current. At high currents its forward voltage increases which causes a lot of wasted heat (poor efficiency).
Also, your vision sees the average brightness, not the peak. Which is why PWM is used for dimmers.

Here is the efficiency curve of an ordinary red LED and the relative brightness curve vs current for a Luxeon Super-Flux LED:
 

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You should probably change the 110 ohm resistor to 330 ohms however to limit fast input change currents to around 1 amp. 110 ohm is probably too low to limit transients.





Thanks

So with a 330Ω what will be the output current @ B? I guess I need the 2.2µf cap to smooth out the waveform as well. If output @ B is 20mA what is the value of smoothing cap? ( it is 10000µf per 1A @ 1% ripple ....?).
 
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Thanks everyone,

This circuit allows a blinking LED to be powered directly frm AC. If I want to add more LED ( 20 or so ) in parallel & to keep the blinking timing same ( 1/2 sec on, 1/2 sec off ) how should I change the values ( diac is DB3 & firing @ ~30v )?

Thanks again.
 

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Pulsing can make use of eye response and persistence to make LED appear brighter. There is a lot of factors to consider with regard to initial brightness, color, and angle of entry at the eye.

**broken link removed**
 
Pulsing can make use of eye response and persistence to make LED appear brighter. There is a lot of factors to consider with regard to initial brightness, color, and angle of entry at the eye.

**broken link removed**


I believe that is correct, pulsing *can* make use of eye response and persistence.

The operative word in there however is, "can". The problem is, most circuits that are used for lighting can not take advantage of this because the period between pulses would have to be too long for a human being to be comfortable with. In other words, nobody wants to use a flashlight like they did when flashlights historically first came about and batteries would run down quick, where they had to literally "flash" the light on (then the battery voltage would die down and the light would dim out) then turn it off and wait a few seconds, then turn it back on to get another flash.

Most people these days want a light that at least appears to be of constant intensity like modern flashlights or home lighting. They dont care if it is pulsing, as long as those pulses are close enough together to make it appear to the eye that the light is of constant intensity. The problem with this is that when the LED is pulsed fast enough to appear constant, the eye actually responds as if it were really constant and so it integrates as usual. This of course means that the LED curve is the main limiting factor yet again, even though the LED is flashing.

There is one exception to this rule that i can think of, and that is for someone lost in a cave and their flashlight battery is running down. By pulsing the LED once every five seconds they can probably see where they are going and conserve battery power and actually get some benefit from the light that might appear to be brighter than usual and thus get more overall light from the same battery power which would seem like an increase in efficiency. As noted however the pulses have to be quite far apart in time, on the order of several seconds.

This isnt to say that there are some applications that can't benefit from fast pulsing (dimmers for example) but the benefits will never include higher efficiency as the overall efficiency will either be exactly the same for some apps or much lower for other apps. Thus, pulsing never increases efficiency.
This isnt too hard to prove looking at the manufacturers data sheet for say the Nichia 20ma white LED. The light output vs input current goes DOWN as the current goes up, meaning that for example double the current produces LESS THAN double the light output.
This isnt too hard to prove experimentally either...

Experiment To Prove It:
Take two LEDs, check their light output with a constant current source to make sure they are about the same, pulse one at 15ma average at 25 percent duty cycle at 1kHz and drive the other at a constant 15ma. Since the 15ma 25 percent duty cycle will require a 60ma pulse at 25 percent duty cycle, we will be driving the pulsed LED at four times the current level for one quarter of the time period. Now both LEDs have the same average current flow, yet one is pulsed and one is not. Point both LEDs at a white wall (the target) keeping them the same distance from the wall and note that the pulsed LED light output is quite a bit lower than the constant current driven LED. It's not as low as half as bright, but will look around 25 to 35 percent lower than the other LED, which is equivalent to the constant current driven LED being farther away from the target. This also means that we can illuminate the same target to the same level of brightness from a farther distance with the constant current driven LED. The other thing to note is that there will be a color shift, toward the blue or purple, at the higher current level thus the pulsed LED color will change somewhat also.
Lastly, to prove that both LEDs are really the same under pulsed conditions, swap them and note the same results.

Finally, if we look at the voltage vs current for the LED we find that the voltage goes up as the current goes up, and since we are driving the LED with pulses of 60ma that means the voltage goes up. It doesnt go up by that much, but it does in fact go up, which means we loose efficiency. The loss isnt that much, on the order of 10 percent, but it's still a loss. That's for the LED itself though. If we dont use a switching circuit the loss will stay the same, but we have to remember that the pulsed LED will look around 30 percent dimmer which still means a loss in efficiency. If we want to keep the light output the same we'd have to pulse at 85ma and then we would be exceeding the manufacturers recommendations for pulsing vs current as well as consuming a lot more power.
 
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