Which circuit is better ?

[Externet]

As in reliability, safer, better design, and the rationale ? All diodes are LED.

 

[ronsimpson]

I do not have a good reason. I have thought about this long and hard years ago.
I chose the right side. The current was high and the LEDs ran hot so I wanted the large amounts of copper to help with heat. The point where 4 LEDs come together is a large square of copper. The PCB was about 95% copper.

I think the answer is more to do with the layout of the PCB. If you wanted the LEDs in a circle the left picture is best.

 

[Externet]

Thanks. Same here. Hard to discern which is more convenient.
But see no relation to layout nor heat. The voltage peaks and valleys produce LED current peaks and valleys at the LEDs recommended operation point.

About Vr for each LED, the right schematic ensures each LED independently will stay below maximum; and the left schematic handles Vr as a sum of the string.
If leds are (-say green, at Vf = 2V, Vr = 5V, I = 20ma-) both circuits work just fine with AC supply being 8V peak.

The right schematic applies Vr = 2V to each led, the left schematic applies Vr = 8V to each string of four.

 

[ronsimpson]

I think we agree we are right. (that a dangerous situation)

I have thought about failure modes.
If a diode fails shorted, the current goes up about 2%. That's nice. Current regulated not voltage regulated.

 

[Ratchit]

Externet,
If you are going to string LEDs in series like that, the the configuration on the left has my vote. Let's say that D13 shorts. That will allow the full voltage to be applied across the remaining three bottom diodes of the left branch, AND the three bottom diodes of the right branch. Only one string is affected for the left configuration if a diode shorts.

Ratch

 

[Externet]

The other failure mode is an open LED

 

[ronsimpson]

What is the signal source. If it is a voltage source then Ratchit is right.
I think it must be a current source and noting bad happens if a diode shorts.

LEDs should never be driven form a voltage. All wise use a current source.

 

[Ratchit]

Externet,

An open LED will not damage other LEDs in either branch. A shorted LED will. Specifying a current source is equivalent to putting a high value resistor in series with the LEDs and increasing the voltage source until they light up satisfactorily. Good practice and easy to calculate.

Ratch

 

[Externet]

An open LED, say D2 will cause full supply (divided) Vr to D5, D6, D7, D8. If supply voltage is beyond a point, will there be a chance for smoke, as in a mains/capacitive current limited circuit ? Or the capacitor will just charge to peak and current won't flow ?

 

[ronsimpson]

Or the capacitor will just charge to peak and current won't flow ?

 

[kubeek]

An open LED, say D2 will cause full supply (divided) Vr to D5, D6, D7, D8.

This really depends on what your power supply is. Normal voltage source will not cause what you descirbed. A current source surely will.

 

[Externet]

OK. Let's stay then to this specific supply circuit from mains :

C1 = 0.33uF 250V NP
R1 = 1.5K
R2 = 250K
AC = 120VAC rms 60 Hz
D1 to D8 = all white leds, I = 20mA, Vf = 3.4V, Vr = 5V, 5000Lm

 

[Ratchit]

An open LED, say D2 will cause full supply (divided) Vr to D5, D6, D7, D8. If supply voltage is beyond a point, will there be a chance for smoke, as in a mains/capacitive current limited circuit ? Or the capacitor will just charge to peak and current won't flow ?

Just because the Vr of a diode is exceeded does not mean the diode will fail. The reverse current has to be exceeded, which won't happen if the current is limited by resistance or a current source. Now you are introducing capacitors and other elements?

Ratch

 

[eTech]

As in reliability, safer, better design, and the rationale ? All diodes are LED.
View attachment 86613

Hi

I gotta ask, when you state "safe", what do you really mean?
You mean safe as in "failsafe"? As in preventing a failure from harming someone?

What are these Lights used for? Are they status indicators meaning Safe or Unsafe?

eT