It is good practice to avoid using one resistor to limit the current through more than one parallel LEDs. Sharing a resistor among LEDs puts the same voltage across the parallel LEDs. If the V-I curve for the diodes is different, different amounts of current will flow. There is commonly a good deal of variation in the forward voltage of diodes, even of the same part number, so we use separate diodes.
Suppose we want to set two levels of brightness for a pair of LEDs. We know we need a separate resistor for each LED, but can we share transistors as in the diagram below?
At first glance this looks okay because each LED has its own resistor. I was working on a circuit like this last week. When I measured the resistance in the circuit, I found them to be lower than expected.
To see why this is, consider the case when D2 is removed. R1 still connects D1 to ground. But there is another path from D1 to ground: Through R2 to R3 to R4. So we actually have R1 || (R2+R3+R4) = 200 || (100+100+200) = 133.3ohms. This becomes clearer if you redraw this circuit (without changing the netlist) this way:
What happens if D2 isn’t missing but simply has a significantly different forward voltage? How do we analyze this circuit? We can draw the circuit with Q2 turned off and therefore omitted:
We can use the delta-wye transform to generate an equivalent “wye”. This a the transform I have not used since Circuits I. I’m thrilled finally to have a practical use for it!
(Note: I used alphabetic reference designators like R[a] in the wye topology and numeric designators like R[1] for the delta. This is opposite of how my text did it, but I did it this way to avoid confusion with the numeric designators in the original circuit.)
For the case where D2 is removed, R[c]+R[a] = 133.3 ohms, the same as we obtained for this case without the delta-wye transforms. In the wye topology, however, we could now work out the current variations at the maximum and minimum forward voltage of the diodes.
Conclusion:
When I look at the equivalent wye circuit, I see that some of the resistance is separated and some is common to both LEDs. My first thought is this is about half way between the ideal of each LED having its own resistor and the undesirable practice of two LEDs sharing a single resistor. I showed this to my colleague Bryan Piernot, however, and he showed me how even with 100s of mV of variation in V[F] among the diodes, disparity due to "wye" resistor sharing are minor compared to the effects V[F] variation with the current paths kept completely separate. The only significant effect of transistor sharing is if one of the LEDs is removed: Effective current-limiting resistance decreases from 200 to 134 ohms in our example.
N-CH FETs cost $0.15 at 1k quantities and use up 2 mm^2 of real estate apiece. If neither LED will be removed from the circuit, it is fine to share transistors among two LEDs. If, however, an LED may be removed or switched off, engineers must be aware removing the LED will affect the brightness level of the other LED.