Discussion in 'General Electronics Chat' started by lordeos, Sep 18, 2015.

1. ### lordeos Thread Starter Member

Jun 23, 2015
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Hi all,

I'm just experimenting with LED's to better understand the behaviour. The test i"m doing is putting as many high resistors in series with a LED to drop the voltage over the LED to 0V. The reason why is that in every circuit i made i'm always noticing that an LED (red led in my case) always has a minimum of 1.4 V as with other components (lightbulbs) i can reduce the voltage to 0 V

- When i put an LED in series with a Potmeter and a very high fixed resistor (for example 1 megaohm) the voltage on the LED never drops below 1.4 V (it's balancing on the edge between on - off, my lab voltage shows current = 1MA
- When i put a 6 V lightbulb in series with a Potmeter and a very high fixed resistor (for example 1 megaohm) i can reduce the voltage on lightbulb to +- 0V

Is my resistance not high enough so i can't get the voltage on the LED to 0 or is it normal that an LED can't go lower then his lowest voltage (depending on the color) ?

When i look at voltage current graphs on the internet i see that some graphs indeed show that the lowest voltage on the x-axis is 1.4V and some other graphs do show 0v as the lowest voltage.

Thx for the help

2. ### dannyf Well-Known Member

Sep 13, 2015
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1ma through at least 1MOhm -> what voltage? Experimenting at that kind of voltage levels can be deadly.

Resistance doesn't matter -> as long as you get the current sufficiently low (however you do that), you will see that the voltage drop across the led approaches 0v.

Because there just isn't any other way around it.

Mar 24, 2008
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4. ### Alec_t AAC Fanatic!

Sep 17, 2013
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There's something wrong with your measurement technique if you can get 1mA (that's 1 milliamp, certainly not 1MA ) through a 1 megohm resistor. Either that or your supply must be at least 1000V !
A lightbulb is a resistive device whereas a LED is a semiconductor device which will drop its Vf (forward voltage) which increases only slightly with current above a very low value.

5. ### mcgyvr AAC Fanatic!

Oct 15, 2009
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An LED is a diode..
Like a diode it has a (fairly stable) forward voltage drop stated in its datasheet..

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6. ### blocco a spirale AAC Fanatic!

Jun 18, 2008
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An LED is not a light bulb it is a diode and therefore it behaves like a diode; the two are not comparable.

7. ### dannyf Well-Known Member

Sep 13, 2015
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they are quite comparable: both generate light; and both have variable resistance (vs. current).

You can view a diode (led or otherwise) as a resistor whose "resistance" is a function of the current going through it.

8. ### Wendy Moderator

Mar 24, 2008
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Fraid not, you really do not understand LEDs very well. They are diodes, and have the characteristics of a diode, with a very high Vf. I wrote the tutorial I posted because many people who think they understand them don't. Like the old Mark Twain quote, "It ain't what folks know that get them into trouble, but what they don't know that ain't so.".

First you have to achieve the Vf (forward dropping voltage, which is different for each color and batch), then the resistance is negligible. Their PIV (peak inverse voltage, the voltage they short out if reversed biased) is also very low, between 4-10V, I usually treat it as 5V max.

LEDs, 555s, Flashers, and Light Chasers

9. ### dannyf Well-Known Member

Sep 13, 2015
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It shows the voltage drop over a diode (in this case a 1n4148 - the same principle applies to other types of diodes as well). The diode is driven by a current source from 1na to 1ua, since we are only interested in the near cut-off behaviors of the diode.

The green trace shows the voltage drop over the diode: it is logrithmic, as expected and way below silicon diode's 0.7v nominal voltage drop.

The blue trace shows the equivalent resistance of the diode: it varies with the current levels and as the current goes up, the equivalent resistance goes from 15MOhm down to less than 1Mohm.

Another way to put it, if you could produce a resistor whose resistance vs. current characteristics track the blue trace, this "resistor" would be indistinguishable from the diode in this example.

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10. ### dl324 Distinguished Member

Mar 30, 2015
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Look at the IV curve for your diode (or a similar one) and take note of the current required for the diode to drop 0V and use an appropriate series resistance to achieve it. Lowering your supply voltage is another option.

A simple test is to use an infinite resistance (i.e. an open circuit) and observe that the current in the diode is indeed zero.

11. ### dannyf Well-Known Member

Sep 13, 2015
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Here is the same simulation for a typical red led, from 1na to 10ma. Identical behavior.

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12. ### Wendy Moderator

Mar 24, 2008
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Simulators are not real, they are computer analogs, or analogies. Real circuits tend to have their own rules.

13. ### dannyf Well-Known Member

Sep 13, 2015
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As you can see, the voltage drop over the led declines with current, and reaches 0.2v @ 1na for the red led.

So yes, if you can lower the current going through the diode, its voltage drop will decline, to as low of a level as your current can go.

14. ### Wendy Moderator

Mar 24, 2008
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And have you ever built this? I pretty well have. Simulators are not real.

LEDs are current devices, not voltage.

15. ### dannyf Well-Known Member

Sep 13, 2015
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Here is Fairchild's characterization of their 1N4148. It goes from 1ua to 10ma, in two graphs. As you will notice that the voltage drop the diode is <300mv @ 1ua and 500mv @ 0.1ma, in both cases substantially less than the nominal 700mv voltage drop for a silicon diode.

You will probably also notice that the curves in the datasheet follow our simulation remarkably well.

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16. ### Wendy Moderator

Mar 24, 2008
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Do you ever built anything, I simulate between my ears. Not all of my stuff works, but when it does, I remember and learn.

You will note the OP ran a real experiment, and his results do not agree with your theory.

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17. ### dl324 Distinguished Member

Mar 30, 2015
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Hear! Hear!

Call me old fashioned, but my preferred simulator is a breadboard. I knew a guy who breadboarded the 80286. They had simulators back then, but they were so slow that breadboarding gave quicker results...

18. ### dannyf Well-Known Member

Sep 13, 2015
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Another way to look at it is through the Schokley diode equation: as the voltage drop over the diode approaches 0v, the exponential term approaches 1 and the value within that bracket goes to 0 so the current goes to zero.

Alternatively, a situation where there is a voltage across the diode without current going through it (assuming no photo diode effect here) would be pretty weird.

The difficulty here is that without some specialized equipment, it would be very difficult for you to drop the current in a controlled fashion and to measure it reasonably accurately.

But math, datasheets, simulation and reasoning hopefully suffice.

19. ### blocco a spirale AAC Fanatic!

Jun 18, 2008
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I disagree, but yes they are similar in that they both emit photons.

20. ### dannyf Well-Known Member

Sep 13, 2015
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Yeah: any two things can be similar or different, depending on the purposes of that comparison.

Nothing absolute here.