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Voltage drop for red LED is around 1.7V, but in my experiment, which consist of a voltage supply connected to a 330ohm resistor and a red LED all in series, the min voltage supply needed for LED to light up is 1.7V? by using Ohm's law the current is 0 ? but the LED light up ?

(I'm supposed to find the min voltage (and current) needed to light up the LED)

M.Katt
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  • Have you measured the voltage drop when PS is 1.7V? I bet you haven't... – Eugene Sh. Jun 22 '17 at 19:17
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    What makes you say the current is zero? – Chris M. Jun 22 '17 at 19:19
  • no i didn't :( could it be lower than 1.7V though ? – M.Katt Jun 22 '17 at 19:19
  • Of course. Why not? – Eugene Sh. Jun 22 '17 at 19:19
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    @ChrisM. The underlying logic as I see it: The voltage on the LED is 1.7V. The power supply is 1.7V. Therefore the voltage on the resistor is 0 V, therefore the current is 0. – Eugene Sh. Jun 22 '17 at 19:21
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    The LED lights up, for this to happen current **must** flow. This means that the current cannot be zero. Small, but not zero. Your assumption that at 1.7 V the LED does not allow current to flow is therefore false. – Bimpelrekkie Jun 22 '17 at 19:22
  • Is OP thinking 1.7v / 330 = .00515 (5mA) is too little current or something, if only applying 1.7v at the source? – raddevus Jun 22 '17 at 19:22
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    I believe the OP is thinking of an ideal diode, which doesn't start conducting until \$V_{th}\$ is reached. Real diodes don't work in such discontinuous fashion. – Chris M. Jun 22 '17 at 19:23
  • With dark-adjusted eyeball, have seen a little glimmer with 50 micro-amps flowing through a red LED...measure the voltage drop across your 330 ohm resistor with voltmeter on a sensitive scale. – glen_geek Jun 22 '17 at 20:53
  • measure the voltage across the LED _and_ the voltage across the resistor (separately). – Ed Randall Jun 23 '17 at 06:18

4 Answers4

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Your model of the LED is too simple for this situation. Here is the first LED I found a datasheet for:

enter image description here

As you can see, the voltage varies with the current. Different LED construction and materials will result in a bit different curve (and it will vary slightly from unit-to-unit even for components of the same model), but the important thing qualitatively is the shape of the curve, it takes less voltage across the LED to push a very small amount of current through it, and your eye is quite sensitive to low light levels so a tiny amount of current may be enough to produce visible light with a good LED.

Spehro Pefhany
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    Not to take away from your answer in any way, but this might be a helpful I-V curve for the OP: http://d2vlcm61l7u1fs.cloudfront.net/media%2F9ee%2F9ee45ab5-2cf1-43ed-9292-a1a7206ea86b%2FphpKb5S9Y.png – Chris M. Jun 22 '17 at 19:25
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    Great guess at what OP was looking for and interesting answer. I learned from it. – raddevus Jun 22 '17 at 19:26
  • @ChrisM. Yes, I didn't want to confuse the issue with multiple colors. Feel free to edit that nice set of curves in if you can link the source. – Spehro Pefhany Jun 22 '17 at 20:39
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Current in a diode (LED or otherwise) obeys Shockley's diode equation (please check the link, I won't bother copying it here).

Parameters vary depending on the actual diode or LED, and also some "diodes" (as in: a physical component) actually contain several diodes in series, but the idea is that I varies exponentially with Vf.

This means we don't have a sharp, yes-or-no transition. Even at low Vf, some current will flow. It will be very small, but sometimes enough to produce a noticeable effect...

Let's do a quick simulation:

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

I used a log axis for current, and a linear axis for voltage, which means the I-V characteristic is almost a straight line. At higher current, it is no longer straight, as the LED's internal resistance begins to matter (it is not modelled by Schockley's equation, but it is modeled by the simulator). The model used here was the default one in the simulator for a red LED, but it seems OK.

At 1V, we will still have 25µA current. With a modern high-brightness LED, and in low light, this will be visible to the naked eye.

bobflux
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  • The use of the semi-log plot here with the logarithmic current is really excellent for illustrating how small but non-zero currents will flow event below the nominal "forward voltage" of any diode or LED. I could imagine intersecting that with the plot of the resistor's I-V curve overlaid on the same plot. – compumike Jun 22 '17 at 23:20
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    "I used a log axis for voltage, and a linear axis for current" umm your graph has a log axis for current and a linear axis for voltage. – Peter Green Jun 23 '17 at 00:45
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With a DMM using V across series 1k with 3 digits you can measure lower currents than the meter in current mode which uses a much smaller internal shunt R.

Try putting DMM to get V across 330ohm (I=V/R) or use 1k resistor instead and get easier conversion on current .

Ultrabright Red and Yellow LED's in 5mm at 20mA are 2.0V +/-xx%

The difference from 10% to 100% of rated current is more like a linear bulk resistance than a diode curve and we call it ESR since it "saturated" over this range. The ESR for an average 85mW LED is around 1/0.085W = 12 Ohms. Better is lower ESR.

Working backwards from 2.0V you can estimate the breakpoint at 0 mA

Tony Stewart EE75
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The sparkfun is saying their basic 5mm Red LED has maximum 1.8V-2.2 forward voltage drop and maximum current through it will be 20mA. Since you get 1.7 V drop across LED. I believes you are using 330 ohm resistor(R) to protecting your LED. Then according to Ohm's law,

R=(Vmax-2.2V)/20mA;where Vmax is the maximum supply voltage.

Since you are using 330ohm resistor,

330=(Vmax-2.2V)/0.02A.

i.e., Vmax=8.8V

That is you can use upto 8.8V to turn on LED. If you want to protect your LED from 10V supply you should go for 390ohm resistance. I believes from this answer you can find out minimum voltage(Vmin) required to turn on LED by applying minimum forward voltage drop and minimum current on above equation. But most of the data sheets are totally silent about minimum requirements to light up the LED.

Arun
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