1

This might have a really basic answer, but its been bugging me. In a lot of circuits where you are controlling a series of leds, circuits will be draw with a separate current-limiting resistor for each led, instead of one single on the common side.

Why would you do this?

schematic

simulate this circuit – Schematic created using CircuitLab

Instead of this?

schematic

simulate this circuit

SomeNorwegianGuy
  • 123
  • 6
  • for reasons of isolated current limits rather than shared current limits which means the brightness of one is affected by the state of all others – Tony Stewart EE75 Jul 24 '18 at 23:35
  • 1. What they said. 2. With a single shared resistor, if LEDs are well matched it may work well enough. But with typical manufacturing spreads the current drawn by various LEDs at the same Vf may vary very substantially. This may only cause unequal brightnesses, but may also overload the brightest LED enough to cause early failure. The remaining LEDs then share the same current and any imbalance may lead to even higher current in the lowest Vf LED and ... . This is a genuine problem in practice. – Russell McMahon Jul 25 '18 at 08:15

2 Answers2

0

Assuming the LEDs are multiplexed, you can use a single resistor if only one LED will be on at once (instantaneously on, that is).

If the ground symbol is replaced by a switch you can multiplex n x 4 LEDs with 4 resistors, where a practical value for n might be something like 4 or 8.

In general if you want to use persistence of vision to have the display have any combination of LEDs appear to be on simultaneously then you will use an n x m matrix and you will need either n or m resistors. Limitations on peak current for LEDs mean that you will not generally want to make an n x 1 matrix where n is larger than 8 or 10 (the peak to average ratio will be too large for a bright enough display with a given LED).

It is conceivable that you could use matched crummy drivers (essentially acting as ballast resistors) and change the duty cycle to equalize the apparent brightness with your 2nd circuit, but that would be complex and might not match all that well (and resistors are really, really cheap).

Spehro Pefhany
  • 376,485
  • 21
  • 320
  • 842
0

Because due to manufacturing variation, parallel diodes will not share current equally -- some will take more than others, so they will be unequally bright, and my understanding is that the effect can be dramatic because a very small change in forward voltage yields a very large change in current. (And this is for otherwise-identical diodes from the same source; if you're talking about mixing them, it's much worse.) Giving each one its own series resistor mitigates this problem.

I believe that in some cases you can get away with doing this, but I don't have a good understanding of when. Also, in particular, I believe in some cases there is a risk of thermal runaway -- one diode takes more current than others, which causes it to heat up, which causes it to take yet more current, until it is destroyed.

Glenn Willen
  • 2,250
  • 1
  • 11
  • 15