Luminosity yield of one or many LED in series against power consumed...

[Externet]

How does it work ?
If only D1 and D5 were in the capacitive supply, there will be a given luminosity.

Several diodes in series as in the schematic will reduce the current a bit; but there would be more light emission. What would be a total luminosity against power consumption plotting ? In other words, the efficiency of a pair or many pairs.

D1 to D8 = 5000 Lm 20mA white LED
R1 = 1.5K 0.5 W
R2 = 250K 0.25 W
C1 = 0.33uF 250VAC NP
AC = 120V 60Hz

I made several of these for yard illumination and am very happy with them 2 years now, but wondered how does mathematics works with a single counterparalleled pair and with many counterparalleled pairs.

 

[ronsimpson]

I am not certain of the question.
With 2 LEDs there will be 3V across the LEDs.
With 8 LEDs there will be 12V across the parts.
With 8 LEDs, the current will be down slightly. Much like the line voltage dropping 9 volts. So the current is down 9% but the light emitters are up by 4X. If you make C1 larger by 10% the current is back up.

What current is in the LEDs? Because the duty cycle is 50% you could push the current up.

 

[ronv]

Lower current yields highest effecency.

**broken link removed**

 

[Externet]

Thanks, gentlemen.
RonS : prefer to keep current to ~20mA peak for longevity respecting suggested operating point instead of pushing it.
Expressing the original question in another point of view; for the shown circuit, how many LED emitters will yield more luminosity ?

RonV: seen that chart before, for a single emitter. Seems from it, adding a number of emitters in series to the point of decreasing each current peak to about 64% from the recommended working current of a single one would be more luminosity efficient.

 

[ronsimpson]

This depends on what you think EFFICIENCY is.
I think you are talking "current from the power line" to "light out". In that case increase the number of LEDs to the max.

example>:
1 pair of LEDs light=1
4 pair of LEDs light= 3.7 With no cap change. Light=4 with a cap change.

(white LEDs Vf=3.0v)
20 pair of LEDs, Vf=60V, Cap .33 changes to .68 (2x larger) Current is the same. Change 1.5k to 750 ohms.
Line current is the same. Light=20x

Now we get 20x the light for the same current.

 

[Externet]

Thanks.
A twisted form of 'efficiency' ---> more luminosity from the same circuit. I was after not altering the components values, as C1, R1, R2 kept the same.

Some removed from service as need to be bugnesting proof, some fine ----> https://i588.photobucket.com/albums/ss323/Innernet/P1010435_zps200dcafc.jpg

 

[ronsimpson]

Back to post #6.
Don't change the cap. 20 pairs of LEDs. Current is 1/2 but the emitters are 20x.
From post #3. If the current is 1/2 the light is 1.3 of 0.5 so 0.65

20 x 0.5 x 1.3=13
20 emitters, 1/2 current, 1.3x = 13 times the current.

You really should increase the cap and decrease its voltage.

 

[ronv]

Well, a little extreme, but you ask.
You could drop the current down to about 2.5 ma. Then use 51 of them in each string. You would gain not only from the higher efficiency but the voltage drop would go from about 3.6 volts each to around 2.8 volts so more leds with the same components. You would pick up a little more from the lower temperature as well.

 

[ronsimpson]

This depends on what you think EFFICIENCY is.
In my life efficiency is: Getting a useable product out for the least dollars.
Usefulness /dollar

I would drive a smaller number of LEDs at high current.