2 diodes in series: why ? - Page 1
Forum section: electronics / beginners

author: Nounours18200
author_rank: Contributor
timestamp: September 30, 2019, 09:12:39 am
content: HI,I have read in this topic:https://www.eevblog.com/forum/chat/vd1212-diodes-replacement/that a good replacement for a VD1212 diode is to use 2x1N4148 in series.I apologize for my lack of knowledge, but why a single 1N4148 is not enough ? I do not understand why to connect 2 in serie, because a diode pass or does not pass the current ?Thank you for your explanation...

author: tautech
author_rank: Super Contributor
timestamp: September 30, 2019, 09:22:37 am
content: The Vf required is the clue to why two in series are required.

author: bd139
author_rank: Super Contributor
timestamp: September 30, 2019, 09:25:24 am
content: Exactly that.That's the only characteristic required most most cases for these. They are usually used to give two semiconductor junction drops in amp biasing circuits which is conveniently the same as the B-E drops on the PA transistors added together (or near enough).You can also replace them with a BJT and a couple of resistors (or a trimmer) for a precise drop but it's hardly worth the effort in a reasonably well designed circuit. Look up "Vbe multiplier".

author: ledtester
author_rank: Super Contributor
timestamp: September 30, 2019, 09:59:43 am
content: Quote from: Nounours18200 on September 30, 2019, 09:12:39 amDiodes can have different V-I curves. THe forward voltage (the point in the V-I curve where the diode starts conducting) for the 1N4148 is around 0.65 V whereas the forward voltage for the VD1212 is around 1.3 V. Putting two 1N4148 diodes in series will approximate the V-I curve of a VD1212.The same thing is going on with LEDs -- red, yellow, green, blue LEDs all have different voltage drops ranging from around 2 V to 3.5 V.Update: Here's are some V-I curves for LEDs. Note that the curves start bending at different voltages.

author: 001
author_rank: Super Contributor
timestamp: September 30, 2019, 06:27:58 pm
content: Quote from: ledtester on September 30, 2019, 09:59:43 amIs it common constant voltages for LEDs? Can I use LED as a zener diode with low capasitance?

author: bd139
author_rank: Super Contributor
timestamp: September 30, 2019, 06:31:07 pm
content: Yeah you can do that. It's quite handy. You can also use LEDs as varactors.

author: Someone
author_rank: Super Contributor
timestamp: September 30, 2019, 11:29:48 pm
content: Quote from: bd139 on September 30, 2019, 06:31:07 pmNoting they have a "bonus" photodiode effect which can make those uses a little complicated.

author: AG6QR
author_rank: Frequent Contributor
timestamp: September 30, 2019, 11:46:42 pm
content: Quote from: 001 on September 30, 2019, 06:27:58 pmTo a reasonable approximation, the frequency of light emitted by a LED is related to its Vf in the same way that the frequency of a photon is related to its energy expressed in electron-Volts.The Planck constant allows you to convert from one to the other.Physics is pretty cool.

author: ledtester
author_rank: Super Contributor
timestamp: October 01, 2019, 02:01:32 am
content: Quote from: AG6QR on September 30, 2019, 11:46:42 pmIn researching my answer I came across this SE thread which suggests that Vf is determined more by the resistivity of the materials used in the manufacturing of the LED:https://electronics.stackexchange.com/a/396384/95488Lots of good comments and food for thought there.

author: AG6QR
author_rank: Frequent Contributor
timestamp: October 01, 2019, 05:24:38 am
content: Quote from: ledtester on October 01, 2019, 02:01:32 amThere are lots of good comments, and a lot of them talk about how Vf is related to the wavelength of the light produced. It's a common undergrad (or even high school) physics experiment to use colored LEDs to measure the approximate value of Planck's constant.https://www.google.com/search?q=led+planck+constant+experimentBut to get an accurate value, you need to determine the threshold voltage that just barely begins to produce light output, as explained in one of the links from the Google search. Here's one of the experimental procedures:https://www.scienceinschool.org/2014/issue28/planck. Others suggest a more crude approach:http://pages.physics.ua.edu/lab10x/ph102/PDF/Plank%27s_Constant_LAB.pdfThe Vf at a nominal current of 20mA or so may be significantly above the threshold voltage, and may be influenced by a number of other factors, including resistivity of materials. Still, even at reasonable working currents, you'll find that the Vf is still related to the wavelength of light produced. Everyone agrees that, at similar light outputs and currents, red LEDs have a lower Vf than green ones, which have a lower Vf than blue ones. Everyone agrees that Vf is related to the bandgap voltage of the doped semiconductors, and that the color produced is also related to the bandgap voltage.

author: Nounours18200
author_rank: Contributor
timestamp: October 01, 2019, 02:53:15 pm
content: usefull info !I currently only have 1N4149 : can I use them (in series) instead of 1N4148 ?The datasheet shows that they have the same Vfmax = 1.0A, but i am not totally sure...Thanks for your replies,

author: rfeecs
author_rank: Frequent Contributor
timestamp: October 01, 2019, 06:25:41 pm
content: The frequency of light emitted depends on the bandgap. Vf also depends on the bandgap, but also depends on current. As you lower the current, Vf goes down, all the way to zero. So the number for Vf you get is arbitrary, depending on the current you choose to measure it.In these experiments, they choose a current that works out to give approximately the right answer.Here is yet another proposed school experiment that uses a potentially more accurate way to measure bandgap by varying the temperature:https://hal-amu.archives-ouvertes.fr/hal-01266699/document