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Totally new to electronics here.

Without the red LED "charging indicator", the capacitor charges fully to 10V, but with the diode the capacitor maxes out at about 8.42V.

I don't understand why the LED causes the capacitor to not fully charge.

Running two LEDs in series drops the voltage across the capacitor down to 6.7V or so.

schematic

simulate this circuit – Schematic created using CircuitLab

Thanks

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  • \$\begingroup\$ Do you what a 20mA RED LED Vf does ? Drops Out at 1.6V and ESR of LED starts at 12 ohms so RC = 1200us then rises quickly as it dims . So that all you see is the LED drop voltage when extinguished \$\endgroup\$ – Sunnyskyguy EE75 Jun 26 '18 at 23:32
  • \$\begingroup\$ As the capacitor charges up, the voltage across the diode reduces and at some point the diode will stop conducting current, which in turn means your capacitor voltage will stop rising. \$\endgroup\$ – user159625 Jun 26 '18 at 23:45
  • \$\begingroup\$ You say red, why is the part number in the schematic for a 569 nm yellow-green led? \$\endgroup\$ – Misunderstood Jun 27 '18 at 1:43
  • \$\begingroup\$ Red yellow same voltage ok? \$\endgroup\$ – Sunnyskyguy EE75 Jun 27 '18 at 3:47
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A LED is a device with very non-linear internal impedance as a function of applied voltage. When the forward voltage is under 1-1.5 V, LEDs have very high impedance, few u-Amps at 1 V, which amounts to 300- 500 kOhm of effective impedance.

enter image description here

The LED current in this area is so low that it is not even characterized in many datasheets.

On the other hand, electrolytic capacitors have substantial leakage, as shown in this model picture:

enter image description here

Per this article, the leakage can be in 5 to 20 uA per 1 uF of capacitance. Therefore, a 100 uF capacitor is expected to conduct maybe 500 uA of DC current.

So what you see is a voltage divider between the low-voltage impedance of LED and the parasitic leakage impedance of electrolytic capacitor. At 500 uA your LED has forward voltage of about 1.6V, this is what your experiment reveals.

ADDITION: Here is the typical I-V curve for a LED in area of interest, at low forward voltages, all on SE EE site. enter image description here

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  • \$\begingroup\$ Why did you not use the IV curve of the LTL-307GE LED in the OP's schematic? media.digikey.com/pdf/Data%20Sheets/Lite-On%20PDFs/… \$\endgroup\$ – Misunderstood Jun 27 '18 at 1:41
  • \$\begingroup\$ @Misunderstood, Lite-ON doesn't have the low-current I-V region resolved to any useful degree. I found only OSRAM who has the data into microampere area. \$\endgroup\$ – Ale..chenski Jun 27 '18 at 2:07
  • \$\begingroup\$ Cool, I did not notice. Upvote for that. \$\endgroup\$ – Misunderstood Jun 27 '18 at 6:13
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Kirchhoff's voltage Law States that for a closed loop series path the algebraic sum of all the voltages around any closed loop in a circuit is equal to zero. This is because a circuit loop is a closed conducting path so no energy is lost.enter image description here

Photo credits go to this site: Electronics Tutorials

That means " in the same closed loop when adding more elements to the circuit, the total consumed voltage is equal to the total applied voltage " Vs = Vr1 + Vr2 "

Typically, the forward voltage of an LED is between 1.8 and 3.3 volts. It varies by the color of the LED. A red LED typically drops 1.8 volts, but voltage drop normally rises as the light frequency increases, so a blue LED may drop from 3 to 3.3 volts.

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  • \$\begingroup\$ Why doesn't the voltage drop across R1 or the value of R1 affect the final voltage of the capacitor? \$\endgroup\$ – Jelly User Jun 26 '18 at 23:42
  • \$\begingroup\$ They do, but you have to consider the current flowing through the resistor because V = I x R. So when current is very small the voltage drop across the resistor will be very small, and can be negligible. \$\endgroup\$ – Ahmed M.Zahran Jun 26 '18 at 23:45
  • \$\begingroup\$ @JellyUser So long as current flows, the voltage across C1 keeps increasing. The question is what voltage C1 reaches when the current stops flowing, because that's what C1 will charge to. Once current stops flowing, the voltage drop across R1 is zero by Ohm's law. \$\endgroup\$ – David Schwartz Jun 26 '18 at 23:46
  • \$\begingroup\$ Every Capacitor has a charging time and discharging time. it can be calculated from the capacitors value. so whenever you apply the capacitor required voltage within the required time to charge, it's considered to be completely charged. \$\endgroup\$ – Ahmed M.Zahran Jun 26 '18 at 23:50
  • \$\begingroup\$ But you should always consider to choose a capacitor working voltage greater than the voltage you intended to use to prevent the capacitor from explosion due to the unexpected rise of supplied voltage. i.e; you have to apply a factor of safety \$\endgroup\$ – Ahmed M.Zahran Jun 26 '18 at 23:52
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The V-F curve of a HE (High Efficiency) red LED looks like this (will vary with LED type and color, for example super-bright will be different):

enter image description here

As the current drops the voltage drop will decrease but even at very low current it will still be more than 1.5V. There will be a bit of current always because of your meter and because of the leakage of your 100uF capacitor.

If you want it to charge all the way up in a reasonable length of time you can put a resistor in parallel with the LED, but the LED will go out before it is fully charged.

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  • \$\begingroup\$ Not that it changes your answer but the LED in the OP's schematic is 1.8V at IV baseline. Why do you call this an HE Red? I think Cree XP-E red and OSRAM Oslon SSL red are much more efficient. \$\endgroup\$ – Misunderstood Jun 27 '18 at 1:53
  • \$\begingroup\$ @Misunderstood The manufacturer calls it "high efficiency" (see the datasheet link). This was developed before more modern super-bright red LEDs and after the original GaN deep red LEDs. So it's historical. You're 100% right, definitely not "highest efficiency". ;-) \$\endgroup\$ – Spehro Pefhany Jun 27 '18 at 2:03

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