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I noticed this LED setup in a datasheet for the Kingbright SA40-19EWA. Instead of the leds being in a simple series or parallel setup, each segment of the display has four sets of two leds in parallel, but each set is in series.

enter image description here

I know the lights work, but don't have one of the displays to cut up to confirm that they are actually connected like this. I'm not asking for how this specific part works, but generally, can this wiring setup, as presented, actually work? How does the current flow, how would the voltage drop be calculated? I know some of the users here have simulation software, maybe they can confirm if a simulation of this would work?

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  • \$\begingroup\$ What's your concern here? You wouldn't have trouble if they were all resistors? Is it the non-linear elements that concerns you? \$\endgroup\$ – DrFriedParts Apr 29 '13 at 3:31
  • \$\begingroup\$ @DrFriedParts I am completely at a loss by the setup. I don't understand how it should work at all. Given that the usual series/parallel setup doesn't have connections between each string, and usually, the warning is to give each parallel led string it's own resistor. With resistors in this setup, it's easier to understand, since two resistors in series are additive, in parallel are essentially dividing, you can treat this as one giant/single resistor, but I'm not even sure of the theory behind it. \$\endgroup\$ – Passerby Apr 29 '13 at 3:55
  • \$\begingroup\$ Gotcha... ok, I'll expand answer... \$\endgroup\$ – DrFriedParts Apr 29 '13 at 4:44
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    \$\begingroup\$ The manufacturer probably has a smart way to ensure the LED's are reasonably closely matched to work in this setup. \$\endgroup\$ – jippie Apr 29 '13 at 8:33
  • \$\begingroup\$ Parallel LEDs have their own resistor because otherwise they are constrained to the same voltage, and any difference in the V-I curve will cause uneven distribution of current. However, if you have two exactly matched diodes, they can be connected in parallel. Things like that are possible on integrated circuits. The cross connections help to keep paired LEDs at the same voltage. Without them, the whole stacks of four LEDs are being paralled, which invites cumulative error. If one pair is matched to 1%, a parallel pair of four matches to just 4%. \$\endgroup\$ – Kaz May 1 '13 at 3:44
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Functional Reasons

Placing LED's in a parallel-series combination like this is a technique to average out the brightness given constant voltage. It helps present a more even looking illumination cheaply (as opposed to active constant-current circuitry).

The Q-point of the two LED's in parallel will be determined by the voltage assigned to the pair by the current allowed through the 4 series steps.

Diodes in Parallel

  1. All diodes have a forward voltage drop that is non-linearly related to current.
  2. This curve is not identical even for diodes made in the same batch

The final Q-point will be V1 = V2 (diode voltages equal), I1 ≠ I2 (currents unequal).

However, in your case the parallel combination is itself in series.

That means that I1 + I2 = Ip is the same for each parallel pair of diodes in the same string.

Light Emitting Diodes

In the case of LED's in the forward region, brightness is almost linearly related to current. So holding Ip constant gives you approximately the same amount of light from the pair of diodes even though it is difficult to know how the light output will be distributed between them.

Extending this to the string of pairs, you cannot find a closed form solution (I might be wrong, but I don't see one) so you would need to solve it iteratively (I like Newton-Raphson) like SPICE does.

In general, however, the more diodes in each parallel group and the more parallel groups in each string, the closer the light output of the parallel groups approach each other given a batch of LEDs, normally distributed, by light output.

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