I'm using an 18650 (4.2v unknown mah) to power 14 leds (2v 20ma) in parallel with a 100ohm 1w resistor on each led. Everything has been running fine for 12 hours now with no problem.

My understanding is the resistors should drop voltage from 4.2 to 2v but my volt meter shows no change. Am I understanding resistors right? Any help in explanation would be greatly appreciated.enter image description here

  • \$\begingroup\$ Sometimes its easier to follow pictures. See dummies.com/programming/electronics/components/… \$\endgroup\$
    – Passerby
    Jan 9, 2020 at 9:38
  • 1
    \$\begingroup\$ @sharky1984 I don't see anything immediately wrong in your schematic. You are probably measuring across the wrong component. \$\endgroup\$
    – DKNguyen
    Jan 9, 2020 at 19:32

2 Answers 2



simulate this circuit – Schematic created using CircuitLab

Figure 1. How to measure battery and LED voltage.

When taking the measurement across an LED measure at the points shown on the left most voltmeter.

From the comments:

LEDs are flickering yellow 2.1 V - 2.6 V around 20 mA and operates as low as 1.8 V and as high as 3 V. ... VUPN527.


enter image description here

Flickering LEDs are not LEDs. They are an integrated package containing an LED and a blinker circuit. There is no datasheet available on the Flicker Flame Orange LED 5mm web page. The only specifications are:

Just supply 2.1 to 2.6 Volts DC and you are ready to go. They will NOT operate on any less than 1.8 Volts, and are quite dim at that voltage. You can run them at 3 Volts DC too, but you may shorten the life if you do that. Anything above 3 Volts is almost sure to kill the LED.

These use around or less than 20 mA each.

A couple of points to note:

  • The 20 mA current is not specified as peak or average. I suspect average.
  • The current will vary from close to zero while off to, I suspect, a lot more than 20 mA when on, depending on the voltage applied and the current limiting resistor.
  • The blinker circuit will cause some additional voltage drop so the lamp's on forward voltage will be higher than a standard orange LED.
  • Therefore the voltage drop you measure across your series resistor will vary between 0 V (LED off and no current through the resistor) and whatever the voltage drop is when the LED is on.
  • You (presumably digital) meter will give the average voltage. So, if the LED is on for 50% of the time and off for 50% of the time we would get the LED voltage switching between 4 V (off) and 2 V (on) at 50% duty cycle. The average is 3 V and that's what you saw on your voltmeter.
  • \$\begingroup\$ I've attached the leads of my volt meter to match the schematic and it only drops down to 3v instead of 2v. I've double cheched the resistor with my meter and it's 990hms. \$\endgroup\$
    – sharky1984
    Jan 10, 2020 at 18:29
  • \$\begingroup\$ What colour are the Leads? \$\endgroup\$
    – Transistor
    Jan 10, 2020 at 18:45
  • \$\begingroup\$ Red to the positive black to the negative \$\endgroup\$
    – sharky1984
    Jan 10, 2020 at 19:28
  • \$\begingroup\$ Sorry, I didn't spot my auto-correct typo. What colour are the LEDs? \$\endgroup\$
    – Transistor
    Jan 10, 2020 at 19:30
  • \$\begingroup\$ Leds are flickering yellow 2.1v-2.6v around 20ma and operats as low as 1.8v and as high as 3v. \$\endgroup\$
    – sharky1984
    Jan 10, 2020 at 19:43

My understanding is the resistors should drop voltage from 4.2 to 2v but my volt meter shows no change.

Allow me to clear up your misunderstanding.

The battery under a light load drops from the charge voltage of 4.2 to 3.8 fairly quickly. (minutes) The current in each LED is always defined by the voltage drop across each R. or I=V/R

I think you would like to measure your unknown battery capacity at a 1C = 20 hour rate. You have chosen 14 LEDs (2v 20ma) which are presumably Red or Yellow high brightness... These act like Zener diodes with a saturation threshold voltage , Vt and an internal resistance , Rs to create a forward voltage Vf=Vt+If*Rs. In addition to this you have your fixed resistor of 100 Ohms.

Due to the decline in battery voltage the light load current will reduce slowly from 3.8 V (100% SoC state of charge) to 3.0V = 0 to 10% SoC. Since battery has <0.1 Ohm Rs, it also rises sharply < 10% SoC which affects Vbat under heavy load more significantly. The Rs (or ESR) rises to 10x the initial level when dead, but you ought to avoid going this low for long life.


simulate this circuit – Schematic created using CircuitLab

The battery mA capacity can be measured best by recording the V drop across R every hour or every 0.1V drop from ~2V down to ~ >1.2V

The 1C capacity is the number of hours at I=V/R * h where the total h is 20 hours.

With an average load current of (245+153)/2 mA = 199 mA

Thus if 1C Rate is 2600 mAh I expect <10% Soc will occur in 2600mAh/200mA avg in 13 hours.

Or in other words if my assumptions and calculations are correct, mAh capacity = 200mA * hours to drop to 3.0V


Since someone assumed incorrectly that LEDs perform worse than Zeners, I assumed there is more than 1 person without the experience.

The variations of all LEDs, Diodes and Zeners is due to the power ratings & Mfg tolerance within a given chemistry or nominal voltage.

I am stating that basic Zeners are lower power parts and as a result have higher incremental series resistance and thus are less stable than LEDs.

The Rs value is generally inverse to the power rating.

A typical 5mm THT White LED has a Vth of 2.85V and an incremental resistance of about 15 Ohms +/-50% @ 20mA. Now with 5mA the LED will be more stable with a lower Vf ~ 2.9V You can verify on your own. e.g. @ 20mA Vf=2.85+0.15~0.45V. Older LEDs had much higher tolerances on high side, but quality has improved in the last 20 years.

Now compare with an ON Semi 3.3V Zener

enter image description here

  • \$\begingroup\$ The 1.8V + 10 Ohms is an estimate based on my experience of HB Red LEDs but datasheet will have much higher tolerance and tells you how to expect Vf to drop with If current to about 10% but becomes logarithmic at low currents.. \$\endgroup\$ Jan 9, 2020 at 3:23
  • \$\begingroup\$ @ tony-stewart-sunnyskyguy-ee75, the statement “These act like Zener diodes...” isn’t technically valid. LED voltage will vary depending on LED current. Zener diodes hold relatively fixed voltage with varying current (within limits). \$\endgroup\$
    – Leoman12
    Jun 17, 2020 at 0:26
  • \$\begingroup\$ @Leoman12 Disagree. Go look at ONSemi Zener specs for Zth Ztk which are the incremental resistances at rated threshold and knee and if you compute as I have that Rs or Zt varies exactly alike all diodes and LEDs with inverse variations for zener voltages and power levels. Your statements are incorrect in this view.. If you use a string of LEDs @ 5mA, they will be just as accurate as Zeners. \$\endgroup\$ Jun 17, 2020 at 3:37
  • \$\begingroup\$ @ tony-stewart-sunnyskyguy-ee75 if you take a look at this datasheet (vishay.com/docs/83010/tlhe510.pdf ) for a LED, you’ll see in either figure 4 or 8 that for a change of 1mA to 20mA, the forward voltage changes from 1.75 to 2V. So under your statement, this is essentially acting like a Zener diode? This can’t be. What I’m saying is incorrect is that a stating an LED acts like a Zener diode.While they may have similarities, the LED isn’t operating in reverse bias as a Zener diode would and it operates under zener effect whereas LED is in forward bias and emits light. \$\endgroup\$
    – Leoman12
    Jun 17, 2020 at 19:54
  • 1
    \$\begingroup\$ Interesting. Seems like zeners aren’t what I thought they were. Poor Zener diodes. Thanks for the info. \$\endgroup\$
    – Leoman12
    Jun 17, 2020 at 20:34

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