#### lordeos

Joined Jun 23, 2015
33
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

#### dannyf

Joined Sep 13, 2015
2,197
- 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
1ma through at least 1MOhm -> what voltage? Experimenting at that kind of voltage levels can be deadly.

Is my resistance not high enough so i can't get the voltage on the LED
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.

#### Wendy

Joined Mar 24, 2008
22,428

#### Alec_t

Joined Sep 17, 2013
12,060
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.

#### mcgyvr

Joined Oct 15, 2009
5,394
An LED is a diode..
Like a diode it has a (fairly stable) forward voltage drop stated in its datasheet..

#### blocco a spirale

Joined Jun 18, 2008
1,546
An LED is not a light bulb it is a diode and therefore it behaves like a diode; the two are not comparable.

#### dannyf

Joined Sep 13, 2015
2,197
the two are not comparable.
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.

#### Wendy

Joined Mar 24, 2008
22,428
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

#### dannyf

Joined Sep 13, 2015
2,197

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.

#### Attachments

• 31.4 KB Views: 9

#### dl324

Joined Mar 30, 2015
12,871
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) ?
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.

#### dannyf

Joined Sep 13, 2015
2,197
Here is the same simulation for a typical red led, from 1na to 10ma. Identical behavior.

#### Attachments

• 30 KB Views: 11

#### Wendy

Joined Mar 24, 2008
22,428
Simulators are not real, they are computer analogs, or analogies. Real circuits tend to have their own rules.

In theory there is no difference between theory and practice, in practice there is.

#### dannyf

Joined Sep 13, 2015
2,197
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.

#### Wendy

Joined Mar 24, 2008
22,428
And have you ever built this? I pretty well have. Simulators are not real.

LEDs are current devices, not voltage.

#### dannyf

Joined Sep 13, 2015
2,197
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.

#### Attachments

• 65.3 KB Views: 4

#### Wendy

Joined Mar 24, 2008
22,428
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.

#### dl324

Joined Mar 30, 2015
12,871
I simulate between my ears. Not all of my stuff works, but when it does, I remember and learn.
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...

#### dannyf

Joined Sep 13, 2015
2,197
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.

#### blocco a spirale

Joined Jun 18, 2008
1,546
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.
I disagree, but yes they are similar in that they both emit photons.

#### dannyf

Joined Sep 13, 2015
2,197
Yeah: any two things can be similar or different, depending on the purposes of that comparison.

Nothing absolute here.