I made a simple LED circuit using a battery, a variable resistor, and a LED. When the light intensity satisfied my demand (I do not need the full intensity) by changing the resistance, the condition was: R1 voltage: 0.5V LED1 voltage: 2.3V Current: 86 uA

Now, I want to add one more LED in a series to make a dual LED system. But I don't know how to calculate the appropriate supply voltage and resistor to keep the light intensity the same (= same current of 86 uA through 2 LEDs). For example, what should be the resistance if I use 4.5 or 6V supply? Any thoughts or suggestions?

LED: https://www.digikey.com/products/en?keywords=XPEBGR-L1-0000-00E02CT-ND

enter image description here

Okay, now I update a new circuit to supply 5.5V using a voltage regulator. It is battery powered (6V: 4 x 1.5V AA), so I want to keep the current (or voltage) constant ( ~ 86 uA) during the period of operation. Do you think this circuit would work as intended? (Let's ignore the product error between the same LEDs)


enter image description here

Or what about this design for the parallel arrangement?

I want to supply the constant voltage while the battery voltage goes down. That's why I need to use the voltage regulator.

Voltage regulator: https://www.digikey.com/product-detail/en/texas-instruments/LP3985IM5X-2.5-NOPB/LP3985IM5X-2.5-NOPBCT-ND/3527410

enter image description here

  • 1
    \$\begingroup\$ Two LEDs, even from the same batch, can have different brightnesses even when they both have exactly the same current flowing through them. Given your low level of current, I'm guessing these are being viewed by "dark-adapted" human eyes. Is that right? The reason I'm asking is because it is quite possible to get the exact same current through two LEDs, without much difficulty. But if that doesn't help you with the brightness, then it's barking up the wrong tree, so to speak. \$\endgroup\$
    – jonk
    May 10, 2019 at 21:41
  • 1
    \$\begingroup\$ Why not add the second LED with another resistor (!) in parallel to R1 and LED1? \$\endgroup\$
    – Huisman
    May 10, 2019 at 21:55
  • \$\begingroup\$ Jonk: That's a good point. I just ignore the difference of two LEDs next to each other in the LED reel. I measured the power in the dark room, so you're right. The LEDs would be used in the dark environment. \$\endgroup\$
    – H. Yu
    May 10, 2019 at 22:12
  • \$\begingroup\$ Huisman: Thank you for the comment. If I connect two LEDs in a series, at least (regardless of the difference between the LEDs) I can provide the same current to the LEDs. I am not sure whether I can provide the same current for the parallel connection though it is great suggestion theoretically. \$\endgroup\$
    – H. Yu
    May 10, 2019 at 22:25
  • \$\begingroup\$ @H.Yu I don't know what you expect to do with these. If you expect LEDs not to "flicker" for example, or wander around in apparent brightness over periods of hours, then you are mistaken (or, last time I checked a decade back.) We tried to create standard candles out of them and wound up throwing away 98% of them as unusable -- that's after a 48 hour "bake-in" period operating them with a 0.05% accurate (NIST traceable) current source and holding them thermally stable the entire time. Not good. \$\endgroup\$
    – jonk
    May 10, 2019 at 22:36

2 Answers 2


For a supply voltage of 4.5V, what you're asking is impossible. Apparently, if you run 86uA through your LED, it has a forward voltage of 2.3V. (Forward voltage increases as the current through the LED increases.) So for two LEDs in series you'd need at least 2*2.3V=4.6V

On a side note: Applying excactly 4.6V to your two LEDs in series without a resistor is not a good idea. LEDs need a series resistor, or something else to regulate the current through the LED. LEDs work on constant current, not on constant voltage.

So two LEDs in series need 2*2.3V=4.6V. At 6V supply, that leaves 6V-4.6V=1.4V across the resistor.

U=I*R so R=U/I > R=1.4V/86uA=16.279kOhm

You can't get a resistor with that excact value, so get 16kOhm. That will give you a current of 1.4V/16kOhm=87.5uA

Now looking at the datasheet of your LED, you should realise that you're not even close to using the LEDs at the intended current and brightness. The LED is intended for a current somewhere around 350mA, where their light intensity is quite high. They are called Xlamp for a reason... :-) You might get better results with an LED intended for lower current and brightness. When operated near the recommended or intended current, you will probably get more consistent light output.

Also, as others have commented already, brightness might be different for the two LEDs, even though the current is the same. Especially at these low brightness levels, a small difference in LEDs will be noticeable very quickly.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Thank you so much MartinF. Your logic helped me a lot. I updated a new circuit using a voltage regulator. Do you think it would work? \$\endgroup\$
    – H. Yu
    May 10, 2019 at 23:21
  • \$\begingroup\$ No, it will not work this way. The voltage regulator you drew is of the adjustable type. You need to set the output voltage using resistors. Look at page 20 of the datasheet. \$\endgroup\$
    – MartinF
    May 11, 2019 at 10:10
  • \$\begingroup\$ S-1133B55 type is internally set with 5.5V output. Though I can adjust by adding resistors, I don't need to, I think \$\endgroup\$
    – H. Yu
    May 13, 2019 at 6:15
  • \$\begingroup\$ What do you think about the 2nd design then? \$\endgroup\$
    – H. Yu
    May 13, 2019 at 6:36
  • \$\begingroup\$ If you would get the S-1133B55 you're good. But that one doesn't have the ADJ pin. Because the part you drew does have the ADJ pin, I assumed you would get the adjustable version, and in that case your schematic wouldn't work. So yes, getting the fixed 5.5V version with your second schematic would work. \$\endgroup\$
    – MartinF
    May 13, 2019 at 14:56

Just add another resistor and LED in parallel with the existing.
If the intensities do not match then you can change the value of the resistor(s).
If you put them in series and the intensities do not match you do not have a simple fix.

It does not matter what the supply voltage is, you just need two resistors and two LEDs. You should not need a regulator, it's a total waste. Any voltage above 2.3 V will be sufficient.

You can try two in series with a higher supply voltage and a higher resistance, that may work fine. Seems like additional and unnecessary work.

At this low current most of the recommendation for LED design go out the window.

At this low current it would be acceptable to power the LEDs without a resistor if you can supply the exact voltage required (e.g. 2.3V).

As far as the difference in intensity, Cree LEDs are fairly consistent and the human eye is not very good at perceiving the difference in intensity.

  • \$\begingroup\$ Thank you for the comments. But I need a voltage regulator to supply constant voltage over time because the battery voltage would go down. \$\endgroup\$
    – H. Yu
    May 13, 2019 at 6:18
  • \$\begingroup\$ @H.Yu Being battery powered you definitely do NOT want the LEDs in series. You should try a 2.3 V regulator and connect both LEDs directly to the 2.3 V supply with no resistor. You will only need resistors if you have to adjust the intensity where the values of the resistors would be different. The lower the output voltage of the regulator, the longer the battery will last. \$\endgroup\$ May 13, 2019 at 19:39
  • \$\begingroup\$ Thank you. I would try that one! \$\endgroup\$
    – H. Yu
    May 14, 2019 at 20:15

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