I'm kind of new here. I've been roaming around the forum for a while trying to figure out how to finish up this circuit I'm building and really need some help.

I'm trying to build a blinking LED circuit using 39 5mm yellow 2Vf 20mA LEDs, pn2222A transistors, and a 9V battery. I have the LEDs wired in 2 parallel strings(one string containing 12 LEDs and one containing 27 LEDs). From what I've learned so far, It would've been best for me to wire them in strings of 3-4 with each string having its own resistor, but I have them wired, soldered, and held in place already so I prefer not to dismantle everything and start over.

This is what I have so far:


simulate this circuit – Schematic created using CircuitLab

Sorry, part of the post is missing for some reason. Allow me to start over.

I'm working on an art piece that contains a spinning object which represents planet Earth. I'd like to add lights to the spinning earth and making the lights battery powered is best. I can't use a slip-ring to run power to everything because a slip-ring won't fit. The spinning Earth as a 8" diameter. I should be able to install all the components (battery, lights, transistors, capacitors, resistors, etc.) to the spinning object, as long as the overall circuitry is light-weight and simple.

I have resistors ranging from 10-10k ohms, pn2222A transistors, 10 and 100uF capacitors, and AAA & 9V batteries. So is there a better way to make 39 lights blink in a light-weight, remote, and efficient fashion?

Did I mention I'm way out of my comfort zone but really really really need this to work?

  • 1
    \$\begingroup\$ Go here: davidbridgen.com/astable.htm and learn. Start with only two LEDs. You also need some resistors. Once you have that blinking with two LEDs figure out how to blink more than two. \$\endgroup\$ Feb 19, 2019 at 16:11
  • \$\begingroup\$ Re-arrange your LEDs into series strings of 3 each, with a resistor in each string. The way you have it will result in tons of power wasted and (at best) uneven output from your LEDs. \$\endgroup\$
    – brhans
    Feb 19, 2019 at 16:16
  • 1
    \$\begingroup\$ If you want to run LEDs on batteries for any appreciable amount of time, you do not want to use resistors, but rather a switch mode current regulator. This will be non-trivial. Even for a simple project you should probably not use a 9v battery unless you only need a very short duration infrequently enough to keep throwing them away. Overall this question does not really fit here - it is too broad, you have started in the wrong direction, and it is not really clear what your actual goal is which makes it hard to offer redirection advice. \$\endgroup\$ Feb 19, 2019 at 16:18
  • \$\begingroup\$ What exactly is your question? \$\endgroup\$ Feb 19, 2019 at 16:21
  • 5
    \$\begingroup\$ Not wanting to dismantle a bad circuit idea and fix it is your first real problem. \$\endgroup\$
    – Andy aka
    Feb 19, 2019 at 16:31

2 Answers 2


You need to understand the power capacity of various battery technology in Ah, Wh and how this degrades due to ESR self-heating and when you exceed rated current.

Using a 9V battery rated for 550mAh in 20 hours when drops well below 8V.

Ideally under low current, a 9V battery can provide 550mAh * 9V <5Wh or 27.5mA for 20 h.

You have 39x 2V LEDs @ 20mA = 1.56 Amps or more than 50x times ~~ its 20h rated capacity. Then by dropping 7V in resistors you are wasting 7/9= 77% of the battery power.

Rules to remember

  • Choose a primary battery with a 20h capacity
  • choose a battery to match your string voltage at dim voltage when the battery is discharged ( e.g. 1.75V dim 2.1V +/-25% bright per YEL 5mm LED )
  • Choose a series R and Pd to current limit when the battery is new. I=ΔV/R current limit
  • Most LEDs have 25% tolerance on Vf at I rated.
    • Better suppliers are 5% from the same batch or bin e.g. 2.05+/-0.1V
    • or even better 2.0~2.1V for Red , Yellow
    • This is inherently due to MFG tolerances on bulk resistance of the diode and sorting.
  • e.g. 20mA @ 2.05V ~ 41 mW with an Rs of ~12 (+/-25%) Ohms
    • so @ 10mA the voltage drops 120mV +/-25% from 2.05 to 1.93V per LED * 10 mA ~ 19mW then drops to maybe 1.75V at for a dim LED. Consult with Datasheet for nominal VI curve or similar part

So your battery choice is N.G.

More Rules to Remember

  • Your transistor switch must handle much more current than you plan to use to allow cool operation as they are rated with infinite heatsinks or pulse current @ 25'C
  • Current switch choice must have low Vce(sat) at Imax or low voltage drop
  • 2N2222 cannot handle 1.5A.
  • If you cannot tolerate a 0.5 to 1V drop , use a low ROn FET that can.


simulate this circuit – Schematic created using CircuitLab


Nobody makes 2V LEDs. They have a range of forward voltage that might be from 1.8V to 2.2V. In parallel, all the 1.8V LEDs will light then probably burn out, then the 2V LEDs will light then burn out, then the 2.2V LEDs will light then burn out. That is why a current limiting resistor is used with one LED or a bunch of LEDs in series. I have an LED flashlight with 24 matched LEDs in parallel. The manufacturer probably bought thousands of LEDs and has somebody test the forward voltage of each one and make groups with the same forward voltages.

Will you blink the LEDs or simply alternate the groups of LEDs? Alternating uses a lot of current all the time so the 9V battery life will be short.

  • \$\begingroup\$ No, alternating does not "use a lot of current". It is actually fairly central to efficient designs especially within the current regulating mechanism itself. \$\endgroup\$ Feb 19, 2019 at 17:49
  • \$\begingroup\$ It doesn't really matter whether they blink or alternate but it sounds like alternating them will be the most efficient way. So it sounds like re-wiring the array in series strings of 3 each with the same Vf measurements in each group and each group having its own resistor is the right way to go. What would be most efficient power source in that case? \$\endgroup\$
    – Creslos2
    Feb 19, 2019 at 18:13

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