Calculate Current - Page 1
Forum section: electronics / beginners

author: etnel
author_rank: Contributor
timestamp: October 10, 2021, 02:05:06 pm
content: Hello, how would I calculate the current at the different points in this circuit?Vcc = 4vLED: Vf = 2v, 20mAR1 = 100 ohms

author: Capernicus
author_rank: Frequent Contributor
timestamp: October 10, 2021, 02:13:18 pm
content: You dont have to think in exact numbers when you do this... I think that people do too much maths and not enough real thinking when they do electronics.This circuit is so basic, why bother calculating any number at all.So you can see we have 1 led on the left with a resistor, and 3 leds on the right, 2 in parallel 2 in series, with no resistance, so the right is definitely alot brighter than the left, and if you used a small enough battery then you wouldnt be able to even see anything on the led with a resistor, regardless of the resistance.Thats worth more any amount of pages of confusing mathematics, and gets it over and done quicker, and you dont have to worry about pesky details that dont matter.There is more to conclude here and know than just that, like the led voltage drop, you need to be able to handle that as well... but that probably can be "read" as some approximate laws and the exact numbers of things isnt so important, the general concepts are way more important.

author: not1xor1
author_rank: Frequent Contributor
timestamp: October 10, 2021, 02:55:38 pm
content: Quote from: etnel on October 10, 2021, 02:05:06 pmSupposing ideal (identical) LEDs you wold get 2V/100Ω = 20mA through the left branch and something more than 20mA through the right one (since there are 2 LEDs in parallel and that means less current and less voltage drop).With real components you might get 15-25mA on the left side and larger variations or even burnt LEDs on the right sideLEDs should always be connected in series to a constant current source/sink or resistor.

author: bdunham7
author_rank: Super Contributor
timestamp: October 10, 2021, 03:07:15 pm
content: Quote from: Capernicus on October 10, 2021, 02:13:18 pmActually, in physics and EE the rule is that if you can't do the math, then you don't really understand it. But...QuoteProbably because it is a homework problem or something like that. But it is a very bad example of a problem because although perhaps you can 'calculate' currents, your calculations won't have any meaning in the real world the circuit design fails to accommodate the non-ideality of the components. And they give Vfof the LEDS only for 20mA, so it can be argued that there is not enough information to solve the problem. So, you're right in this instance--the circuit is so flawed that there's no point in calculating anything.

author: mariush
author_rank: Super Contributor
timestamp: October 10, 2021, 03:42:45 pm
content: You can split that circuit in the middle.On the left side you have the led and you have the resistor.Voltage = current x resistance according to ohm's lawSo input voltage - (number of leds x forward voltage) = current x resistance4v (vcc) - (1 x 2v ) = current x 100 = > current = (4-2)/100 = 2/100 = 0.02Aon the right branch, you have two leds in parallel, then one in series with the two in parallel.So your current would be input voltage - (1 group of leds in parallel x forward voltage) - (1 led in series x forward voltage) = current x resistanceIf you had a resistor to limit current on this branch, the section of circuit with 2 leds in parallel will have same current as the single led experiences, so the two leds in parallel will have half the current.As it is ... you have equivalent of 2 leds in series, so around 4v voltage drop and you don't have any resistance, unless you start to account for the wire resistance as current limiter.So if we assume ideal circuit where the leds are fully open at 2v and the wire has no resistance, then those leds would simply blow up because there's nothing to limit current going through them.

author: Terry Bites
author_rank: Super Contributor
timestamp: October 10, 2021, 04:28:59 pm
content: If you're not making circuits before you know the maths you're going to end up a tedious accademic. True.

author: Capernicus
author_rank: Frequent Contributor
timestamp: October 10, 2021, 04:38:57 pm
content: If you know whats going to happen, the exact mathematical intricacies didnt stop you from guessing it. But on the other hand its easy to get it wrong so I do still believe in alot of experimentation.

author: xrunner
author_rank: Super Contributor
timestamp: October 10, 2021, 04:47:43 pm
content: Someone should breadboard it and check whatreallyhappens ...Me right?

author: jmelson
author_rank: Super Contributor
timestamp: October 10, 2021, 04:56:18 pm
content: The current in the right branch is uncontrolled. VERY SLIGHT variations in Vcc or LED forward drop will cause HUGE changes in LED current. That's why all practical LED circuits use a series resistor or a current source. Anyway, as a homework exercise, the current in the right branch cannot be calculated, unless you assume the LEDs draw exactly 20 mA at 2.000 Volts. This is a pathological circuit, and should not even be given as a homework problem.Jon

author: Capernicus
author_rank: Frequent Contributor
timestamp: October 10, 2021, 04:58:59 pm
content: anything above 2 will light up the leds, below that they cut out because of the voltage drop. its 4 volts for 2 in a row. Xrunner - test it out if we are right!!!!!If its true, all you need is 120 leds and you can have a full 240v connection and the leds are fine with a 100 ohm resistor.

author: mariush
author_rank: Super Contributor
timestamp: October 10, 2021, 04:59:21 pm
content: Quote from: xrunner on October 10, 2021, 04:47:43 pmSadly they will probably use a couple CR2032 batteries or a 9v battery (and a linear regulator) to power the circuit and you wouldn't get the expected results... because they won't consider the internal resistance of the battery as a current limiter.So, it may not be quite the desired lesson.

author: xrunner
author_rank: Super Contributor
timestamp: October 10, 2021, 05:02:11 pm
content: Quote from: mariush on October 10, 2021, 04:59:21 pmI'll do it later I think I have time. The LED Vf won't be 2.0 V I'll have to use red, more like 2.1 or 2.2.That's a textbook problem someone made up as we all realize. I'll be back later ...

author: bdunham7
author_rank: Super Contributor
timestamp: October 10, 2021, 05:04:50 pm
content: Quote from: jmelson on October 10, 2021, 04:56:18 pmEven with that assumption it doesn't work unless they assume Vfis absolutely constant from 10mA to 20mA.

author: AG6QR
author_rank: Frequent Contributor
timestamp: October 10, 2021, 06:13:58 pm
content: Quote from: jmelson on October 10, 2021, 04:56:18 pmAgreed, except that the right branch is ill-defined even if you DO assume the LEDs draw exactly 20mA at 2.000 Volts.The two LEDs in parallel in the top half of the right side each must each be getting a portion of the current that goes into the bottom LED on that side. So they can't possibly all three be conducting 20mA at 2.000V!In a simplified classroom model of an LED, you might be able to assume that vF is 2.0000V, completely independent of current, but that's going to be a false assumption if you ever try to test it in the real world. If you make that false assumption anyway, you'll find that the bottom LED conducts 20mA at 2.000V, so each of the top LEDs would conduct 10mA, if you assume they're completely identical. The assumption that LEDs in parallel split the current symmetrically is another one that won't hold up to real-world experimentation, but might be an assumption that's used in a theoretical classroom exercise.So you get an answer where the voltage across each of the top two LEDs on the right side is equal to the voltage across the bottom LED, despite the fact that the bottom LED is carrying twice the current. That's preposterous, but it follows from the preposterous assumptions made.The circuit won't work in practice.

author: xrunner
author_rank: Super Contributor
timestamp: October 10, 2021, 06:33:15 pm
content: Quote from: AG6QR on October 10, 2021, 06:13:58 pmI got it all breadboarded up. Yep, the right side is not acting predictable as suspected because the two LEDs are not identical. However the left side pretty much matches calculations. Will be back with more results. It's a textbook problem that can't really be solved without more information.

author: CaptDon
author_rank: Super Contributor
timestamp: October 10, 2021, 06:57:52 pm
content: Xrunner, that theoretical problem goes against every practical designof a proper LED circuit and sets such a poor example of a real worldpractical circuit. If your teacher laid that out as a learning examplehe/she is an idiot and I would hate to even think that circuit was publishedin a training textbook. If that was in a published book no wonder allof my recent engineering applicants leave me wondering what theyactually taught in these engineering classes. The answers I got tobasic questions and examples on our entrance exam tests make meonly want to hire engineers over the age of 60. If this was a textbookexample circuit it would even be hard to use the old cliche "Book smartbut real world stupid". Anyway, best luck in your studies and hope youhave a rewarding career path! I dearly miss my old mentors many ofwhich learned electronics during WWII and then worked at places likeRaytheon, Westinghouse, GE, Martin Marietta and others. I was verylucky to have been mentored by them and did very well in my careerpath from knowledge gained from the old school guys!!!

author: xrunner
author_rank: Super Contributor
timestamp: October 10, 2021, 07:08:40 pm
content: Quote from: CaptDon on October 10, 2021, 06:57:52 pmI've already had a career.It's not my teacher or problem it's the OP's (etnel)

author: xrunner
author_rank: Super Contributor
timestamp: October 10, 2021, 09:25:56 pm
content: Refer to the diagram.I set the input voltage to be 4.00 V at the breadboard. The measured resistance of R1 was 98.7 ohms.Path A measured current was 17.1 mA and Vf of the LED was 2.14 V. Calculated current would be 18.5 mA. Reason for a little lower measured current is probably resistance in the breadboard clip connections.Path B current was 10.5 mA and Vf of LED B was 2.1 VPath C Vf (across both LEDs) was 1.93 V. The current measured through each of the LEDs in parallel was very touchy. I got from 2 to 4 mA depending on the stability of the connection I made with the probes or tapping on the desk. In fact, any slight push on either LED resulted in both LEDs variation of light output (due to slight resistance changes in the breadboard connection). In other words it's a very unstable situation as was suspected. Best way would have been to solder in all the components and leads for that kind of measurement. So no, you cannot calculate the solution for current through path C from the information given.Equipment usedhp 34401Ahp E3611ARigol DM3058E

author: BeBuLamar
author_rank: Super Contributor
timestamp: October 10, 2021, 10:41:53 pm
content: What you don't know is the bulk resistance of the diode. This is the determining factor of how much current will go thru the right branch.

author: mariush
author_rank: Super Contributor
timestamp: October 10, 2021, 10:46:39 pm
content: Also, measure the resistance of each wire from the power supply to the prototyping board, and the resistances of the thin wires connecting the proto board "bus bars" to the leds.Your circuit is basically[ + psu ] ---[ R - red wire from psu ] --- [ R - thin wire ] --- [ led 1 and led 2 in parallel] - [ led 3 ] - [ R - thin wire ] --- [ R - black wire to psu ] - [ - psu ]You will have around 0.1-0.2 ohm of resistance in the wires.

author: xrunner
author_rank: Super Contributor
timestamp: October 10, 2021, 11:41:26 pm
content: Yep guys all true. Where's etnel? Seems like he'd want to say something about all this by now.

author: Tomorokoshi
author_rank: Super Contributor
timestamp: October 11, 2021, 12:18:26 am
content: 1. Given the "pathological" nature of the question, along with the limited amount of data provided, we can approach it with certain ideal conditions:1.1. All diodes are exactly the same temperature.1.2. All diodes are identical.1.3. The ideal diode equation applies: i = ISe^(v/nVT)1.4. I leave it to the student to look up the various terms, derivation, etc.2. Calculations of the "A" circuit2.1. We are given that the forward drop is 2.0V at 20mA.2.2. The resistor is 100 ohms and the power supply is 4V.2.3. The current is conveniently (4V - 2V) / 100 ohms = 20mA, which matches the diode.2.4. The current through point A is 20mA.3. Calculations of the "B" and "C" circuits3.1. With some math we can derive that for the same diode we have this relation: (v1-v2) = nVT*ln(i1/i2)4. Relating the "B" circuit to the "C" circuit4.1. iB= 2iC4.2.vC= 4 -vB4.3.vB-vC= nVT*ln(iB/iC), or:4.4.vB- (4 -vB) = nVT*ln(2)4.5. 2*vB- 4 = nVT*ln(2)5. Relating the "B" circuit to the "A" circuit5.1. The voltage on diode B is greater than the voltage on diode A.5.2.vB-vA= nVT*ln(iB/iA), or:5.3.vB- 2 = nVT*ln(iB/ 0.02)6. Combining the results6.1. Multiply equation 5.3 by 2:6.2. 2*vB- 4 = nVT*2*ln(iB/ 0.02)6.3. Note that equations 4.5 and 6.2 have the same left side. Make an equality from the right sides:6.4. nVT*ln(2) = nVT*2*ln(iB/ 0.02)6.5. ln(2) = 2*ln(iB/ 0.02)6.6. ln(2) / 2 = ln(iB/ 0.02)6.7.e^(ln(2) / 2) =iB/ 0.026.8.iB=e^(ln(2) / 2) * 0.02, or:6.9.iB= 28.28 mA6.10.iC= 14.14 mA.6.11.iA= 20 mA.

author: rstofer
author_rank: Super Contributor
timestamp: October 11, 2021, 01:44:13 am
content: I think everybody is trying to add reality to a homework problem. No issues with the left LED, it gets 20 mA, no doubt.The two LEDs on the right feed into a series LED that will drop 2 volts at 20 mA and each of the parallel LEDs drop 2V at 10 mA and everything balances. Just as long as you don't think about real components.Current A = 20 mACurrent B = 20 mACurrent C = 10 mAGiven that no more information was presented, I think the 'ideal' component is all you can use. How you justify the 10 mA through the parallel LEDs, I don't know. The 10 mA is correct, whether that would actually produce 2V drop isn't given in the problem statement.Not a particularly good example.

author: EEVblog
author_rank: Unknown
timestamp: October 11, 2021, 01:52:33 am
content: Quote from: etnel on October 10, 2021, 02:05:06 pmIt's one of those questions that actually makes no sense in practice. You wouldn't put a supply (presumably low impedance) of 4V across non-linear LED's in series like that.But even theoretically the question is still stupid.If vf is 2V @20mA, that makes 2V/20mA throguh the lower LED on the right, but then makes no sense for the upper two LED's which must be 10mA each (assuming even distribution), so the Vf will no longer be 2V.

author: EEVblog
author_rank: Unknown
timestamp: October 11, 2021, 02:46:31 am
content: Could help myself, video coming shortly!

author: xrunner
author_rank: Super Contributor
timestamp: October 11, 2021, 02:48:23 am
content: Quote from: EEVblog on October 11, 2021, 02:46:31 amOh neat, I will be interested in your results so I can compare ...

author: EEVblog
author_rank: Unknown
timestamp: October 11, 2021, 06:05:59 am
content: 00:00 - The Forum question02:12 - Would this make a good job interview question?03:36 - Dave's Solution07:04 - The Solution08:08 - Other forum responses11:40 - Kirchoff's Current Law MUST Hold!12:51 - REBEL against The System!15:10 - Xrunner's practical test21:38 - Practical Test

author: strawberry
author_rank: Super Contributor
timestamp: October 11, 2021, 06:54:01 am
content: thermal dependant circuit that reach equilibrium at some specific pointcan use LED string as photo cell to lit up one LED

author: gbaddeley
author_rank: Regular Contributor
timestamp: October 11, 2021, 08:31:12 am
content: It’s a trick question.

author: Nusa
author_rank: Super Contributor
timestamp: October 11, 2021, 10:17:47 am
content: My guess is poster asked a question about the problem, but did not give us the actual question asked about the circuit. So we're all spinning our wheels for no particular reason.

author: not1xor1
author_rank: Frequent Contributor
timestamp: October 11, 2021, 11:04:50 am
content: My previous post was quite confuse and unclear, so I'm attaching a simulation screenshot to better describe what I meant.I adjusted the voltage supply and resistor to get about 20mA through the left LED.I think transient simulation renders better the non linear behaviour of diodes.

author: Capernicus
author_rank: Frequent Contributor
timestamp: October 11, 2021, 12:25:38 pm
content: How you power it makes a difference, If I charged it with 2 volts 20ma peak output, but with just a battery, I still have a theory that all the leds would be off.

author: xrunner
author_rank: Super Contributor
timestamp: October 11, 2021, 12:31:39 pm
content: Quote from: Capernicus on October 11, 2021, 12:25:38 pmWell why don't you test your theory, wire it up, and tell us what it actually did ...

author: mikeselectricstuff
author_rank: Super Contributor
timestamp: October 11, 2021, 12:46:24 pm
content: A completely literal answer would be that B and C pass no current, as there is no difference between total Vf and the supply.But as there is also zero resistance it could be infinte current.

author: TrickyNekro
author_rank: Frequent Contributor
timestamp: October 11, 2021, 01:11:36 pm
content: Of course, theoretically it can work... And you have to calculate it iteratively based on the I = Is * ( exp ( Vd / n * Vt ) - 1 ). You know theoretically the n and the Vt so you get the Is. Because you got two LEDs over the single one, it isn´t really sure that you are going to get 20mA through B, probably you are going to get less, going off my head here.Practically, you wouldn´t even put diodes in parallel because of the differences in manufacturing and temperature ( as in Vt ), which can lead to a runaway effect.Practically, you also have your average LED lamp committing that crime out there, only take seriously the ones that have many many LEDs in series and hope the lead resistance is enough to balance out.BTW, just for the extra mind boggling, you also have to make sure no light is shining on the LEDs, or equal amount of light for all that it matters, go have fun now :-P

author: Capernicus
author_rank: Frequent Contributor
timestamp: October 11, 2021, 01:24:59 pm
content: Quote from: xrunner on October 11, 2021, 12:31:39 pmI know with 2 leds in series, you need 4 volts to power it. I swear the initial post was 2 volts in, and he changed it!!!

author: Capernicus
author_rank: Frequent Contributor
timestamp: October 11, 2021, 01:53:40 pm
content: Think about it what you will, but I bet if I took a uni course on electronics I would get in the lowest 1% percentile of the class.Thanks Dave for the vid, If I had 120 leds Id make a video of me putting a 500 ohm 1/4 resistor on them and just plug it straight into the wall as a thnkyou vid, but I dont have the leds.

author: hadibaria
author_rank: Newbie
timestamp: October 11, 2021, 02:32:07 pm
content: Everycircuit simulation results. default LED settings Vf = 2v under 20mA

author: Wytnucls
author_rank: Super Contributor
timestamp: October 11, 2021, 03:00:28 pm
content: Your calculations exactly match the results given by the EveryCircuit simulation.Tomorokoshi (reply 21)

author: Tomorokoshi
author_rank: Super Contributor
timestamp: October 11, 2021, 03:15:19 pm
content: The range of answers and results just goes to show:You never can tell with D's.

author: Wytnucls
author_rank: Super Contributor
timestamp: October 11, 2021, 03:38:25 pm
content: I'm not sure the person that formulated that question expected this level of accuracy, but you are obviously on the right track.(Note to self : I must eat more corn!)https://eng.libretexts.org/Bookshelves/Materials_Science/Supplemental_Modules_(Materials_Science)/Solar_Basics/D._P-N_Junction_Diodes/3%3A_Ideal_Diode_Equation

author: lacek
author_rank: Regular Contributor
timestamp: October 11, 2021, 04:30:58 pm
content: Dave, sorry, but the solution you have shown in the video is wrong. I agree that this question is somehow wrong and it lacks the U(I) curve for the LED diode. But you can't assume that one diode drops 2V @ 10mA and other drops 20V @ 20 mA.I would say that Tomorokoshi's solution is as good as one can do. Just assume exponential dependence of U on the current.This is somehow what you do in your video. If you assume Ib = 20mA then you have Ic = 10 mA and the voltage drop Vc is < 2V. if you assume IC=20 mA, then the IV=40 mA and VB >2Vin first case you have total voltage < Vcc in the latter you have > Vcc. So the answer would be thatIB is "between 10mA and 20 mA"IC is "between 20mA and 40 mA"Tomorokoshi's wrote system of equations that give solutions that satisfy the aboveI agree that it is not a proper exam questions as it is not possible to solve without U(i) characteristis. If this is "open answer" (so not ABCD) then there is a place to write the above.I disagree with most of the criticism about idealizations. Yes the circuit is "academical", but the idealizations simplify the problem and allow to consider just the core of the problem, not to dwell on the detail: are the diodes matched or not. This is important, but for learning it is important to start from simple problem, not to dwell on such digressions.

author: lacek
author_rank: Regular Contributor
timestamp: October 11, 2021, 04:40:20 pm
content: Quote from: TrickyNekro on October 11, 2021, 01:11:36 pmI am sure this is problem from some sort of Circuits I class or some "electricity for physicists" course. They were taught during the lecture what is the dependence of diode voltage on the current and they are supposed to solve a problem of determining how to solve a simple circuit with a non-linear U(I) dependence.The formula that Tomorokoshi has used most likely was covered in the lecture and students are assumed to know about it and to use it for a solution. The problem stated as is is not "self-consistent" but it is written for a student group that all attend the lecture. It is not expected to remind all necessary formulas in problem text. The students are supposed to find them.I see no point in complaining about "unpracticality" of said situation. Its like taking any physics problem that deals with physics of the material point in outer space and complaining that thing almost always have non-zero diameter and are affected by light pressure from sun or relativistic effects that have to do with gravitation that modify the 1/r^2 force. Yes they do, and from the context of class it is obvious does whether they should be included or not.In Phyisics/Electronics one can always consider "more accurate model", but both "rough models" and "very accurate models" have their uses.

author: lacek
author_rank: Regular Contributor
timestamp: October 11, 2021, 04:42:51 pm
content: @Dave: assuming 2V at middle junction is simply wrong. You run in contradictions as you correctly state in the video.

author: bdunham7
author_rank: Super Contributor
timestamp: October 11, 2021, 05:17:00 pm
content: Quote from: lacek on October 11, 2021, 04:40:20 pmUnfortunately, Tomorokoshi's solution can't account for temperature because the thermal characteristics are unknown--so he appears to just assume the two temperatures are the same. Even if we had a datasheet, we might not know if the I/V curve was done quickly at a constant temperature or perhaps reflects an equilibrium at a standard ambient temperature. Any analysis of 'ideal diodes' or even real ones without considering temperature is not going to be very helpful.I understand the idea of progressing from basic to advanced, but I think it is not impossible to put a little thought into questions and examples to make sure that they aren't nonsensical under more advanced analysis. I can't think of any basic principle or theory that cannot be demonstrated with a circuit that actually works.

author: hans
author_rank: Super Contributor
timestamp: October 11, 2021, 05:27:26 pm
content: My guess from an academic standpoint is that a solution should be able to provide the solution as derived by @Tomorokoshi. It is possible to calculate that with Ohm's law, KCL and the ideal diode equation.All the comments about the 2 parallel diodes should be matched or else they will burn out, is our practical thinking hat. There are always certain levels of theory vs practice you account for. If you're doing precision analog stuff, then you will also know that resistors have a certain tolerance, a temperature coefficient and even a voltage coefficient. Or if you learn about opamps, first you will learn how wonderfully excellent amplifiers they are, but then learn about offset voltage/current, a finite gain bandwidth, supply swing, and that the open-loop gain is finite, therefore even a unity voltage gain opamp stage is notexactly"unity"..

author: xrunner
author_rank: Super Contributor
timestamp: October 11, 2021, 06:22:52 pm
content: Quote from: lacek on October 11, 2021, 04:40:20 pmI would like to know how etnel came across this problem, if he was in a class where it was presented or just stumbled upon it somewhere. The forum indicates his account was active today, so I will PM him and ask him nicely to tell us.(of course nicely)

author: rstofer
author_rank: Super Contributor
timestamp: October 11, 2021, 06:47:16 pm
content: Quote from: lacek on October 11, 2021, 04:40:20 pmOr, this is a problem in early DC circuits (Kirchhoff's Laws) LONG before diodes are even introduced.I'm not convinced that applying the diode equation is appropriate at the level I assumed for the question.Yes, the circuit is an abomination but that's one of the reasons I chose the simple model (the LED Vf is always 2V regardless of current). Nobody would ever actually design something like this."All models are wrong but some are useful"https://stats.stackexchange.com/questions/57407/what-is-the-meaning-of-all-models-are-wrong-but-some-are-useful

author: lacek
author_rank: Regular Contributor
timestamp: October 11, 2021, 07:40:27 pm
content: Quote from: rstofer on October 11, 2021, 06:47:16 pmI am not so sure. For this problem diode equation is essential. Actually at my university there is an "electronics" course for physicists. There are literary 6 lectures, with crazy pace, and this problem would fit nicely to that course. The audience are people who know basic, ohmic circuits, at this course they are taught some practical aspects of pn junctions, bipolar transistors,mosfets, and some basic of signal processing. They are in their second year, so they have already learned the analysis, mathematical methods in physics etc.As for "Nobody would ever actually design something like this." this is never consideration for courses for non-engineering students. There are numerous problems on "infinite series of resistors" after all. For example Problem 2 from IPhO 1967.

author: lacek
author_rank: Regular Contributor
timestamp: October 11, 2021, 07:42:24 pm
content: The model "(the LED Vf is always 2V regardless of current)" is not bad as a "rough approximation". you get 10mA instead of 14mA.... which can be still usable for some power dissipation considerations. Just healthy safety margin should be used.

author: Logaritmo
author_rank: Newbie
timestamp: October 11, 2021, 08:13:26 pm
content: My practical theoretical guess is ..From the graph If x Vf mentioned by Dave in the video, a lookup table was produced, andusing successive approximation when the current in B must be the double of the current in C andthe voltage over the lower LED plus the voltage over the higher LEDs must be 4 v,the final result was:I B = 14.5301 mA ( V LED = 1.9168 v )I C = 29.0603 mA ( V LED = 2.0832 v )

author: SiliconWizard
author_rank: Super Contributor
timestamp: October 11, 2021, 08:26:53 pm
content: That's fun.Of course there are different levels of answers, depending on how realistic you expect them to be.For the first branch, iA = 20 mA. If we assume that Vf = 2 V@20 mA, then that's the equilibrium it'll reach. That'll be true neglecting other factors such as temperature and assuming Vf = exactly 2 V @exactly 20 mA, also assuming Vcc comes from an ideal voltage source.(Vcc - Vf)/R1= iAOf course the tricky part is with the right branch.Now, let's see for the case of ideal diodes first (so let's assume here Vf = 2 V for any current, and all LEDs having the exact same Vf). The right leg is going to be the equivalent of 2 diodes in series, with an equivalent Vf of 4 V, which is exactly Vcc. Now even in this ideal diode case, would the right leg actually draw any current? Or would it draw infinite current? It's already a trick question IMO. In any case, I do not quite agree with the 10 mA/20 mA answer, as that would assume the diodes are actually NOT ideal. If they have an ideal characteristic and there is no limiting resistance in series, then the current will be either zero, or infinite. Ideal diodes do not have current limits. At least, we should clearly agree on a definition of ideal diode here. Is that an I-V characteristic with an infinite slope at the "threshold" (which is what "ideal diode" means to me), or is that something approximating the Shockley equation, like a piecewise linear approximation?One of the issues with this question is that admittedly, the fact that we should consider "ideal diodes" is made unclear from the part which states "LED: Vf = 2v, 20mA", implying that Vf could depend on current, and thus the diode would not be ideal. Unless the question was explicitely asked to trigger an elaborate answer, this is indeed flawed.And now if we consider the non-ideal case, that still isn't likely to be iB=20 mA and iC=10 mA unless you got very lucky with the parts you have. We'll have to consider the real I-V characteristic - with Vf increasing with increasing current. Now we can say for sure that iC ~ iB/2 if the two LEDs in parallel are close enough in characteristics. As the latter will pass about half the current of the bottom right's LED, assuming all 3 are close enough in characteristics, that means the two LEDs in parallel will drop a lower voltage than the bottom right LED. For Vcc so close to twice the typical Vf, actual results will vary widely depending on the LEDs and temperature. One thing for sure is that as we increase Vcc, the bottom right LED will pass more current than the left LED. The opposite will happen as we decrease Vcc.

author: xrunner
author_rank: Super Contributor
timestamp: October 11, 2021, 08:43:14 pm
content: etnel has kindly replied to my PM and explained where he encountered the problem. He has replied to me in a PM so I can't reveal it myself. He has said he will reply tomorrow in the thread. I think you will find it interesting.

author: TrickyNekro
author_rank: Frequent Contributor
timestamp: October 11, 2021, 08:53:34 pm
content: Quote from: xrunner on October 11, 2021, 08:43:14 pmCrazy part of feedback loop or something? Cause something tells me it´s not just an exercise....

author: RDP
author_rank: Newbie
timestamp: October 12, 2021, 02:09:53 am
content: OK I just simulated this in an online simulator and this is what I gotHave I done something wrong?RDP

author: EEVblog
author_rank: Unknown
timestamp: October 12, 2021, 02:25:55 am
content: Quote from: mikeselectricstuff on October 11, 2021, 12:46:24 pmBut the question states 2V drop at 20mA, so given that's the only data you have, you have to assume 20mA is flowing.

author: EEVblog
author_rank: Unknown
timestamp: October 12, 2021, 02:27:18 am
content: Quote from: RDP on October 12, 2021, 02:09:53 amYes, you used a simulatorSorry, it was the obvious joke.Welcome to the forum.The simulator is going to go crazy with this depending on the diode model used.

author: RDP
author_rank: Newbie
timestamp: October 12, 2021, 02:36:35 am
content: Thanks Dave.... I suspected as much. I'm on holidays in Adelaide at the moment so don't have access to my bench equipment so I thought I'd just try out a simulator to see what it would produce. Obviously its way off even with the Standard default Diodes 2.2v that I used....

author: amyk
author_rank: Super Contributor
timestamp: October 12, 2021, 03:12:53 am
content: Quote from: lacek on October 11, 2021, 07:40:27 pmI agree this might be a theoretical question and not a practical one.... which then naturally leads to the question of what happens with an infinite series of diodes.

author: andy3055
author_rank: Super Contributor
timestamp: October 12, 2021, 04:04:52 am
content: I just watched the video on Youtube while having my lunch! When i first saw this a couple of days back, I thought it was a "trick question" like the Yanks say here.

author: DavidAlfa
author_rank: Super Contributor
timestamp: October 12, 2021, 04:17:52 am
content: I'll skip the Vf vs current curve and assume that it's always 2V drop.It states 20mA but not where.So the left led will draw (4-2)/100=20mA, and the right leds would draw nothing, as they see (4-4)=0V.They need anything over 4V to conduct.In real life they would draw some current, but that part seems to be ignored here.Is there any other tricking I might be missing?

author: thm_w
author_rank: Super Contributor
timestamp: October 12, 2021, 05:06:08 am
content: People going wild over connecting LEDs to a voltage source or paralleling them in the youtube comments. Try playing with some LEDs and a power supply, or low voltage COB LEDs (bigclive). Vf matched LEDs can be paralleled and share current. LEDs can be driven solely via a voltage source and not explode.Should it be done in production? Best practice is not to do it. Doesn't mean its not possible or there are no applications for it.Compared two red COB LED arrays and they were matched within 3mV. I don't feel like destroying them right now, but, at the individual die level I suspect they could be matched to <1mV.Also found someVf binnedSMD LEDs, but they offer nothing close to that level of matching.

author: lacek
author_rank: Regular Contributor
timestamp: October 12, 2021, 05:48:56 am
content: Quote from: EEVblog on October 12, 2021, 02:25:55 amNot at all. It nowhere says this problem is self-consistent in the sense of listing all formulas you are allowed to use. Student that got this problem, likely attended some lecture, which was full of other formulas unlisted here. In this case some "ideal diode" equation is useful, like that one: I=I_0 (exp(eU/2kT)-1) 20ma@2V allows to find i_0,assuming room temperature T.You also do not know if this is a closed form problem or an open one, which allows for presenting reasoning not simply the answer. To put bluntly what under that problem: a box where you type three numbers, or half of empty page where you show your solution?For super-realism: you can take any U(i) curve you like and solve the equation. You can also write that the current is I0, such that U(I0) + U(2*I0) = 4V and the exact details depend on U(I) curve. which can also factor in junction temperature (which also depends only on I0)

author: lacek
author_rank: Regular Contributor
timestamp: October 12, 2021, 06:07:43 am
content: Quote from: amyk on October 12, 2021, 03:12:53 amNice oneMy attempt to answer: voltage and current across the parallel diodes go to 0. The diode at the bottom gets voltage drop 4V and current from U(I) = 4V. In "practical terms" i will fry

author: etnel
author_rank: Contributor
timestamp: October 12, 2021, 07:11:59 am
content: Thank you, everyone. Especially Dave for even making a whole video regarding the problem. Very interesting to read your different approaches.So, the backstory of the problem. The circuit got handed to us in a high school entry course in electronics. The circuit was supposedly made by our teacher. Our teacher seems to have lacking knowledge regarding LEDs and current in a series circuit. By his means, the LED drops voltage AND current. I had to prove him wrong.We got to both calculate and elaborate. When elaborating, I got similar results as Dave in his video. But when I had to calculate the right branch, things got a little confusing, thus creating the thread. The circuit that stood out to me as my first thought was that insufficient information was given. But I made the assumption that Vf = 2.0 at all given currents(which would not be the case in reality). My conclusion was, half of the current goes through each LED which is in pairs. But the circuit seemed odd anyway.I see some of you made some good realistic calculations, but it is not near the level of this course. Ohms law and Kirchoff's are pretty much the only formulas used. If my teacher is watching this thread, maybe he could post the sole purpose of this circuit with calculations included. The circuit in my opinion is just a brain itch and would never be used in reality(hopefully). An LED circuit is never seen as a theoretical circuit for me.

author: not1xor1
author_rank: Frequent Contributor
timestamp: October 12, 2021, 07:39:47 am
content: Quote from: Capernicus on October 11, 2021, 01:24:59 pmIf the supply voltage were just 2V a few mA would go through the left LED while a few nA would go through the LEDs on the right one, more than that depending on temperature higher than 300K and environment light.

author: xrunner
author_rank: Super Contributor
timestamp: October 12, 2021, 11:10:32 am
content: Quote from: etnel on October 12, 2021, 07:11:59 amThat's what I was thinking all along. This wasn't some "physics introduction to electronics" advanced LED question and the student had been taught more theoretical equations than we had been told about in the initial post. Some teacher thought they were going to get cute and draw something apparently simple on the surface but as you understood (good for you!) it's not a simple sort of thing to understand and calculate properly without more knowledge. However, it is simple to actually breadboard and prove that more advance calculations are needed. LEDs are NOT like resistors and act very differently given different applied voltages.Again good for you to understand this.

author: StillTrying
author_rank: Super Contributor
timestamp: October 12, 2021, 11:27:30 am
content: It just goes to show that a little knowledge is dangerous, especially if you're a teacher.Of course we didn't doelectronicsat school, it was too long ago, but we did do science and physics.One day the part-time physics teacher drew that on the board to demonstrate how a simple diode works. Me and a mate just looked at each other, we just wanted to get home to continue working on our shed DIY X-ray machine.Everyoneelse wired the 4 D cells and the diode that way, and after 30 minutes the teacher still hadn't figured out why it didn't work, so it went down as a failed experiment, which was ideal as far as I was concerned because there was no write up to be done.

author: xrunner
author_rank: Super Contributor
timestamp: October 12, 2021, 11:33:47 am
content: Quote from: StillTrying on October 12, 2021, 11:27:30 amLOL.Yea ... but you know it actuallydidwork perfectly. It just didn't match the teacher's expectations.In a sense every circuit ever made works "perfectly", every single one performs exactly according to the laws of physics - but it may not be doing what the human being thought it would do.

author: Zucca
author_rank: Unknown
timestamp: October 12, 2021, 12:51:15 pm
content: Quote from: Tomorokoshi on October 11, 2021, 12:18:26 amThis is the correct math/physics answer.I would like to give a graphical view about this correct answer.The branch Ia with R1 is easy and just to determine if you should pass the exam or not, I will not dig into here.To have an idea on what happen to Ib and Ic it is better to split the second branch in two circuits:- V1, I1: 4VDC with a diode on the ground (VCC - Vd)- V2, I2: two diodes in parallel (basically one new diode with Vf=2V and If=40mA)Those are the resulting graphs:This branch is working at the point where the red line and the green line are crossing.Without knowing any details about the diodes, all we can say is:20mA < Ib < 40mAsince Ib = 2*Ic10mA < Ic < 20mA

author: Zucca
author_rank: Unknown
timestamp: October 12, 2021, 01:05:50 pm
content: Quote from: amyk on October 12, 2021, 03:12:53 ameasy if you think about V,I graph like I did above.

author: xrunner
author_rank: Super Contributor
timestamp: October 12, 2021, 01:11:51 pm
content: Quote from: Zucca on October 12, 2021, 01:05:50 pmYes and the bad thing is the "teacher" etnel had was clueless about it -Quote from: etnel on October 12, 2021, 07:11:59 am

author: Zucca
author_rank: Unknown
timestamp: October 12, 2021, 01:28:32 pm
content: Quote from: xrunner on October 12, 2021, 01:11:51 pmRegardless, it is a very good circuit to solve. In my electronic one exam at the university of Milan (third year of master in EE) they asked me to solve a circuit with a base floating BJT.I looked up the professor and he looked at me with a smile: "If you want all the points you need to answer this".Not practical circuit on paper are very good to test if you know electronics.

author: SiliconWizard
author_rank: Super Contributor
timestamp: October 12, 2021, 05:19:08 pm
content: Quote from: EEVblog on October 12, 2021, 02:25:55 amAs stated in my answer, I agree with mike here. Again, we need to agree on what an ideal diode is, and if the question actually assumes ideal diodes.So let's start assuming an ideal diode has a Vf that doesn't depend on current.(BTW, I do not agree with the definition that some have used in this thread, that an ideal diode would be one modeled by the Shockley equation. Certainly this equation is a *simplified* model of a real diode, but it is NOT an ideal diode by any means. But I've seen some confusing definitions all over the place, which is why I'm insisting on agreeing on a definition of ideal first.)In no case IMHO an 'ideal diode' could be assumed to have a fixed current passing through it. A diode is not a current source. This is what etnel eventually said: the teacher's assumption about this IS wrong.So as Mike and I stated, we have a limit problem here. Current should be either 0 or infinite in the 'ideal' case, depending on your exact definition of the diode threshold ('> Vf' or '>=Vf'), and assuming the ideal model with an infinite slope past Vf...And, in practice, the current in the right branch will be either lower, or higher than in the left branch. Slight changes in Vcc could lead to a very low current in the right branch, or a very high one, and anything in between.

author: Zucca
author_rank: Unknown
timestamp: October 12, 2021, 05:26:04 pm
content: Quote from: SiliconWizard on October 12, 2021, 05:19:08 pmIf we assume a ideal diode as you describe above the solution for Ib and Ic isIb = 2*Icfor any possible value of Ic >=0.I agree with Dave, 20mA@2V is given so it cannot be an ideal diode.

author: BeBuLamar
author_rank: Super Contributor
timestamp: October 12, 2021, 06:29:11 pm
content: Quote from: SiliconWizard on October 12, 2021, 05:19:08 pmThe diode doesn't behave like ideal as you describe due to its bulk resistance. So it's like an ideal diode with a small resistor in series. As I said in early post the bulk resistance is unknown so we can't calculate the current.

author: EEVblog
author_rank: Unknown
timestamp: October 13, 2021, 08:58:10 am
content: Quote from: SiliconWizard on October 12, 2021, 05:19:08 pmWell it comes down to the original intention fo the question. If it's an ideal diode trick question, and the 2V x 2 = 4V Vcc is deliberately designed as part of the trick, then you can chose your own adventure.But if it's seriously a (poor) legitimate question, then it becomes the 20mA/10mA answer. I suspect it's the later.

author: EEVblog
author_rank: Unknown
timestamp: October 13, 2021, 09:05:27 am
content: Quote from: etnel on October 12, 2021, 07:11:59 amThanks for clarifying.In that case (bold above), having a text box means that as I speculated in the video, the answer is designed to allow you to get multiple answers and explain why.Given that it's aiomed at high school students though, I still think the "correct" answer the teacher intended is the 10mA/20mA one. Anything else is way beyond a high school student electronics class.I also suspect the teacher may not fully grasp the consequences of the question.

author: Knith
author_rank: Newbie
timestamp: October 13, 2021, 12:20:49 pm
content: I think the key information from etmel is that the circuit was built by the instructor, which implies that it worked. I hope that he or she intended their students to learn the difference between a circuit diagram used to build a circuit and a circuit diagram used to analyze a circuit. They also seem to encourage students to make any reasonable assumptions, which is what I have done in my classes. Sometimes my colleagues and I do write nonsense problems, usually because of limited practical experience, and we can choose our words poorly so that the meaning of questions can be ambiguous. Often these broken questions lead to more learning than perfectly crafted academic questions.So, let us start by taking the question at face value and allowing for reasonable assumptions. The first thing I did was to simulate it using LTspice. Actually, I cheated and simulated an equivalent circuit using 1N914 diodes and a 1.4V power supply. The solution was very reasonable mainly because the diode model in LTspice includes resistance. Currents do go through the roof if the voltage is raised a bit. The 2V, 20mA assumptions for the series LEDs is quite good. The parallel diodes each carry about 10mA. An interesting variation from 'idealized' 2V 20mA conditions is that the voltage across the lower series diode is a little above 2V and the voltage across the parallel diodes is a little less. If students know anything about the nonlinear I-V curve for diodes, they will understand that this is because the operating point is at a lower current in the parallel diodes so the voltage is also lower. High school students generally learn to solve equations graphically, so the instructor could possibly be relying on that knowledge. I personally think that diode load lines make great problems for practicing graphical solutions because the answers can be checked experimentally.I also built the circuit using a 4V 'battery' that I configured using three somewhat depleted AA cells. This ends up being very much the same as driving a single LED with a 2032 coin battery. I have been having students build little LED circuits using these batteries for years. I mostly do outreach with kids as young as 6 and teach electronics to first year EE students and non-EE engineering students at the oldest continuously operating technological university in the English-speaking world. Actually, I retired 3 years ago, so I should really say taught. The circuit works just fine. Even with my old, under voltage batteries, it has been running for 24 hours and the voltage has dropped to 3.68V. I have used the coin battery experiment to get students to think about what is missing in the circuit diagram. Something is limiting the current. When they look up the properties of the battery online (everything requires online searches), they find that batteries have resistance. They also learn to look up component information and learn to ask simple, useful questions like 'do LEDs have resistance," from which the learn that they do, although not a simple linear resistance. I attached a photo of my setup. My office is also our guest room and often the only available horizontal surface for simple experiments is the bed.Bottom line, I think, as is shown in the many good responses to this posting, questions like this can lead students to explore the real world of circuitry, or any other concept, actually. They do not need a lot of math, but if they have good math skills, they can learn even more. Dave, you are absolutely right that this is a good question for job interviews. Such open-ended questions should help both the interviewer and job applicant to get to know one another.A recommendation for the great people who have written such interesting comments to this post ... find a good science or tech teacher and adopt them. Offer to help make their classes better by using your extensive experience to make student learning better. The best opportunities are based on mutual respect where both you and the teacher are good at what you do and are comfortable with your own limitations and talents. I am reasonably knowledgeable about a lot of practical EE stuff, mostly because I have done experimental plasma physics for nuclear fusion projects for more than 50 years. However, what has made me a much better teacher is that I have a sort of kitchen cabinet of practicing engineers who love having an impact on EE students so they share their thoughts and ideas with me on a regular basis. Our conversations are usually amazing. We definitely work in different worlds with different missions and schedules, but we have the same passions for understanding electronics and other related topics.

author: Vovk_Z
author_rank: Super Contributor
timestamp: October 13, 2021, 05:47:32 pm
content: I vote for 20/20/10 mA. But it depends, of cause. The task needs to ask more questions about some more assumptions from the examiner.

author: oolonthegreat
author_rank: Newbie
timestamp: October 13, 2021, 08:01:28 pm
content: LTSpice simulation with D(Vfwd=2) gives A=20mA (on the left branch) but no current at all on the right branch. I made sense of this in the following way: if voltage at node B is slightly higher than 2V, then the voltage across the parallel diodes is slightly less than 2V, meaning they are OFF, no current. Similarly if VB is slightly lower than 2V, than the diode at the B branch is OFF, thus no current too. So I guess the only solution is VB=2V exactly, and all 3 right diodes are OFF? Since for the diodes to actually switch ON, the voltage across must be actually slightly higher than 2V ?

author: xrunner
author_rank: Super Contributor
timestamp: October 13, 2021, 10:11:52 pm
content: Oh we're still going ...Once more with the actual components. To get a more reliable measurement I soldered up the circuit this time, because the breadboard connections were too unreliable. Still using red LEDs which all came to me in the same shipment. I didn't bother with the left side (A) because that's not an issue with anyone. So you can see the results of the actual components documented on the schematic.

author: rstofer
author_rank: Super Contributor
timestamp: October 14, 2021, 12:41:59 am
content: Nice lab setup!

author: gcewing
author_rank: Regular Contributor
timestamp: October 14, 2021, 01:10:51 am
content: Quote from: lacek on October 11, 2021, 07:40:27 pmAn infinite series of resistors is obviously an unrealistic situation, but to someone with no electronics experience this circuit is not obviously impractical. I don't think it's fair to mislead people with it even if they don't intend to take up electronics.

author: gcewing
author_rank: Regular Contributor
timestamp: October 14, 2021, 01:38:07 am
content: Quote from: SiliconWizard on October 12, 2021, 05:19:08 pmWhat constitutes an "ideal diode" depends on what you're using it for.If you're using it as a rectifier, an ideal diode would have zero Vf at any current, zero reverse leakage current and infinite reverse breakdown voltage.If it's part of a logarithmic amplifier or something like that, it would perfectly follow a Shockley-like curve with no variation with temperature.

author: gcewing
author_rank: Regular Contributor
timestamp: October 14, 2021, 01:48:08 am
content: Quote from: etnel on October 12, 2021, 07:11:59 amSo what did the teacher think the answer was? And were you able to educate him?

author: RBBlackstone
author_rank: Newbie
timestamp: October 14, 2021, 07:16:19 am
content: Since “A” in a circle usually denotes an ammeter,I would need the ammeter specs, too. Yikes!

author: rs20
author_rank: Super Contributor
timestamp: October 14, 2021, 09:21:22 am
content: Tomorokoshi's answer is perfect, but there's a much simpler argument that starts with just assuming that the V/I curve of the diode is exponential.If ID= 20mA --> VF= 2VAnd ID= 20mA / sqrt(2) --> VF= 2V - x (for some value x)Then ID= 20mA * sqrt(2) --> VF= 2V + xBecause that's how exponentials work. The sqrt(2) may seem to have randomly come from nowhere, but we know that we want the bottom diode to have twice the current as the top one, so we get that spread by multiplying by sqrt(2) one way and dividing the other.So we theorize that the top two diodes have 20mA / sqrt(2) each and drop 2V - xWe theorize that the bottom two diodes has 20mA * sqrt(2) and drops 2V + xKCL is satisfied because sqrt(2) * sqrt(2) = 22V + x + 2V - x = 4V, so 4V @ vcc is satisfiedSo we're done. Top diodes are 20mA / sqrt(2) = 14.14mA eachBottom diode is 20mA * sqrt(2) = 28.28mA.If found it rather odd, especially considering the emphasis on practicality and real-worldness, of Dave seeing a diode described as "2V @ 20mA" and just randomly declaring that wherever you see it in a diagram, it must be dropping 2V and passing 20mA.... Um, no!?!? It only drops 2V IF you force 20mA through it, and it only passes 20mA IF you force 2V across it. This is a common error that students make, rather than an accurate way of thinking about a device that you know A) is a diode with a vaguely exponential characteristic and B) we've been given the specs at one point on the V/I curve.

author: Logaritmo
author_rank: Newbie
timestamp: October 14, 2021, 10:26:20 am
content: Thats why I believe the V/I curve is the only possible usable source to find the correct answer.I extracted the V/I data from the TLUR6400 presented by Dave as a practical LED diode example.With that I calculated the currents and voltages that agree with the curve, using successive approximations.Of course extraction data from a image presented in a datasheet is not that precise, but the source of the curve could come from a better source.

author: xrunner
author_rank: Super Contributor
timestamp: October 14, 2021, 11:38:55 am
content: Quote from: Logaritmo on October 14, 2021, 10:26:20 amRight. Tomorokoshi's results do not match reality - they don't match the real components. He doesn't take into account an actual V/I curve, even if a generic one for a 20 mA class red led.Forget branch A that's not the issue.Quote from: Tomorokoshi on October 11, 2021, 12:18:26 amIf you look at the V/I curve for a real red LED (20 mA class wp1034hdt) below, you will see how it compares to my actual measurements. It matches them very well.Compare the measured results to a real red LED V/I plot -Branch C1 (left side) = 5.69 mA (1.97 V). Graph says ~ 5.9 mABranch C2 (right side) = 5.52 mA (1.97 V). Graph says ~ 5.9 mABranch B = 11.23 mA (2.1 V). Graph says ~ 12 mA

author: rs20
author_rank: Super Contributor
timestamp: October 14, 2021, 12:32:59 pm
content: xrunner, your LED has 2V @ 8mA, not 2V @ 20mA as shown in the question. OBVIOUSLY you're going to get much lower currents than Tomorokoshi's, because your LED is 2V @ 8mA, and Tomorokoshi's is working from the question's 2V @ 20mA. Just ridiculous to assert that the result "do not match reality" when your LEDs don't match the question. I mean, your bottom LED has MORE than 2V across it, but still has LESS than 20mA! Totally different universe to the OP's question.Reapplying the same technique to your LEDs would yield 8mA / sqrt(2) = 5.65mA for the top parallel LEDs and 8mA * sqrt(2) = 11.31 mA for the bottom one, which is marvellously close to what you measured. Actually confirming Tomorokoshi's result, not disproving it at all!

author: timenutgoblin
author_rank: Regular Contributor
timestamp: October 14, 2021, 02:20:00 pm
content: Thank you to EEVBLOG and Tomorokoshi.I don't think I have anything to contribute to the discussion other than thermal considerations relating to R1.The theoretical resistance of R1 can be calculated using R = R_0 ( 1 + a ( T - T_0 )).If the resistivity or temperature coefficient (tempco) of R1 is zero, then R = R_0 for all T including T = T_0.If the tempco is +ve, then R is directly proportional to T, ie. R >= R_0 when T >= T_0. Conversely, if the tempco is -ve, then R is inversely proportional to T, ie. R <= R_0 when T >= T_0.LED forward voltage (Vf) = VCC - VR1 where VR1 is the voltage across R1, V = I x R (Ohm's Law).The theoretical power dissipation in R1 can be calculated using P = I² x R (Ohm's Power Law). Internal power dissipation will cause R1 temperature to increase.If the tempco of R1 is +ve we can expect Vf to be <2V since R1 will increase in resistance due to P = I² x R.If the tempco of R1 is -ve we can expect Vf to be >2V since R1 will decrease in resistance due to P = I² x R.If R1 did have a +ve tempco the LED forward voltage and R1 voltage would reach an equilibrium and stabilize.If R1 did have a -ve tempco would thermal runaway ever occur due to LED forward voltage (Vf) increasing?Maybe? Probably not...I think the bonding wire in the LED would overheat, melt and fail open circuit first.

author: hans
author_rank: Super Contributor
timestamp: October 14, 2021, 09:52:28 pm
content: Quote from: lacek on October 12, 2021, 06:07:43 amI think I'd agree with that one. Back of the envelope reasoning:Let's define the current to the bottom as Ix = N * Id.Each top diode has to conduct Id = Ix / N current, where N is the number of diodes. Since we consider lim N->inf, we find that the current through each top diode will approach zero, and therewith it's voltage across will also approach zero.However since we an infinite amount of infinitesmall zero's (and both "infinities" are equally strong), all top diodes together will still be carrying the full Ix.Intuitively this also makes senses, because the more LEDs you put on top, the more the bottom LED will suffer.

author: SiliconWizard
author_rank: Super Contributor
timestamp: October 14, 2021, 11:35:22 pm
content: Quote from: EEVblog on October 13, 2021, 08:58:10 amYes, as we discussed.Quote from: EEVblog on October 13, 2021, 08:58:10 amWell, you're reading into a teacher's mind here.And yes, I'm pretty sure this was what they were expecting too.But I still don't agree with this answer, even if making dubious assumptions to fit the question's intent, and I'll try to explain why again.Assuming all diodes are identical, we can agree on the fact the two diodes in parallel will each pass half the current of the right bottom one.My problem here is that I just can't figure out what diode model it could be.* If the model is assumed to be: Id = 0 if Vd < Vf, and an infinite slope otherwise (to me that is the model of an "ideal diode" with a threshold), then the resulting current should be infinite. Now depending on whether the definition is a strict inequality or not, it could also be zero.* If the model is assumed to be: Id = 0 if Vd < 0 , and an infinite slope otherwise, then the resulting current should be infinite. (I discarded this ideal diode model as a Vf is provided.)* If we assume a model either based on the Shockley equation, or a piecewise linear approximation of it, then the issue is that the current in each of the diodes in parallel being half the current of the bottom right diode, then the latter will have a Vf that is greater than the Vf of the parallel diodes - whatever exact modeling you use. So we can expect a Vf in the bottom right diode greater than 2V, thus the current is not going to be 20 mA.I just have a hard time finding a model of the I-V characteristic that would have the point (2 V, 20 mA) and could still get you 20 mA in the right branch, because of the above. It doesn't seem to fit any reasonable model for a diode, even simplified. That's just my point.

author: Vincenzo
author_rank: Regular Contributor
timestamp: October 15, 2021, 12:33:19 pm
content: Assuming that the leds follow the model I=I0(eV/VT- 1) and VTand I0is the same for all, KCL and KVL can be used and lead to a quadtatic equation that ends up with the single diode voltage V=VTln((.25+2eVcc/VT).5-.5), then the parallel diodes voltage will be 4-(this value) and when all voltages are known, currents can be calculated using the model equation above. The problem is the assumptions that constants of the model equations are perfectly equal, but for a super theoretical question that is practically wrong, assumptions won't make the end of the world and at least we can assume that "their values are close enough to be considered equal" even though in a non-linear exponential-logarithmic curve "close enough" is much more complicated than in a linear curve (or a line)[ Specified attachment is not available ]

author: GrandTheftAuto4life
author_rank: Contributor
timestamp: October 16, 2021, 02:05:27 am
content: I have had a similar question back in college but we were calculating the reactance and something else I can't remember.This circuit had ideal capacitors all of the same value/variant and there was an ideal AC voltage source.In unison, we rejected the question in writing, stating multiple issues and the question was then marked as invalid.The teacher had just switched books and hadn't had a chance to get familiar with them just yet so this was one of a handful of issues that we later found.The circuit in a nutshell:AC source: 15VAC, 50Hz1uF capacitors, 300 ohm resistorsI can't remember the exact circuit but I'm attaching it drawn from memory (please excuse the mediocre quality, I've yet to find a good android app to draw circuits in)

author: rs20
author_rank: Super Contributor
timestamp: October 16, 2021, 07:55:20 am
content: @GrandTheftAuto4life, either your written rejection of the circuit was invalid, or that indeed isn't the circuit. Because if I'm interpreting the circuit correctly (attached), it's perfectly reasonable to simplify down using parallel and series formulae.

author: Vincenzo
author_rank: Regular Contributor
timestamp: October 16, 2021, 09:12:17 am
content: @GrandTheftAuto4life & @rs20even if the circuit is different, any series, parallel, delta-star, combinations of R's and C's is solvable even if the number of components is infinite (given a repetitive pattern). See the attachment to last post (mine) here:https://www.eevblog.com/forum/blog/eevblog-1399-electronics-fundamentals-voltage-dividers/msg3602488/#msg3602488In my college days, we had a teacher in the first semester of the first engineering year who gives midterm and final exam questions of only R's and dependent & independent DC sources connected in cool geometries that lead to answers that are either 0/0 or the square roots of negatives!!!My strategy for acing his class (and that was an absolute rarity) was to simply replace the numeric values given for the components with Latin and Greek letters and work out formulas/e that eventually show what the problem is, kinda like today's simulation software.

author: free_electron
author_rank: Super Contributor
timestamp: October 16, 2021, 05:09:17 pm
content: Since it is given that, for every led the current through is 20mA, then the only possible answer is that all leds carry 20mA. But only in a universe where the laws of physics allow for it of course.

author: EEVblog
author_rank: Unknown
timestamp: October 16, 2021, 11:07:26 pm
content: Quote from: SiliconWizard on October 14, 2021, 11:35:22 pmIt's easy, there is no diode model.You know this because of who the question is aimed at, high schoolers. This is why the 4V VCC matches 2x the given LED drop precisely. The question will not confuse someone with no in-depth knowledge of diodes. It's only us poor engineering schmuck's who get suckered into thinking it's more complicated

author: EEVblog
author_rank: Unknown
timestamp: October 16, 2021, 11:09:35 pm
content: Quote from: Logaritmo on October 14, 2021, 10:26:20 amYes, I think that it you want practical results to match up to theory, this is really the only way to do it.

author: GrandTheftAuto4life
author_rank: Contributor
timestamp: October 17, 2021, 01:21:29 am
content: Quote from: rs20 on October 16, 2021, 07:55:20 amThere is something missing but i can not remember it, sorry. (it's been almost 15 years)

author: timenutgoblin
author_rank: Regular Contributor
timestamp: October 17, 2021, 01:45:59 pm
content: The right hand side of this circuit has contradiction written all over it. It is also paradoxical, but there is a solution (*). The LEDs are stated to operate at 2V 20mA. The LEDs either operate under the stated conditions or they don't work at all.The top LEDs require 20mA each totalling 40mA in to the node. The bottom LED requires 20mA out of the node. This is a contradiction of Kirchhoff's Current Law.If the bottom LED is operating at 2V 20mA then the top LEDs will fail to operate at 10mA each being under powered. If the top LEDs are operating at 2V 20mA then the bottom LED will fail to operate at 40mA being over powered.The only way for the LEDs on the right hand side to operate properly is for the top half to operate and not the bottom half or the bottom half to operate and not the top half. This is a paradox. The right hand side of the circuit is invalid, contradictive and paradoxical.(*) The only solution to resolve the contradiction and paradox is to duplicate the right hand side of the circuit, flip it vertically about the node and superimpose it onto the original circuit. When the top LEDs are operating at 2v 20mA each, the bottom LED will also be operating at 2V 20mA. Also Kirchhoff's Current Law is obeyed since 3 x 20mA going into the node will be the same as 3 x 20mA coming out of the node. The extra 3 LEDs are redundant, but necessary to avoid contradiction and paradoxes.

author: Vincenzo
author_rank: Regular Contributor
timestamp: October 17, 2021, 02:24:57 pm
content: The question is in a book I once had (trying to remember the name, will post when I do) and the original question does not say that all LED's have 20mA and 2V at the same time because if that's true (presenting a contradiction 20mA+20mA=20mA which is impossible and I don't know why every commenter is going there!! probably original poster's fault) there is no question left and all voltages and currents are known. The 20mA/2V mentioned by original poster only meant that it is a point where LED's will be "lit" because part of the original question is which LED is lit!I gave an answer in page 4 of this thread.

author: lacek
author_rank: Regular Contributor
timestamp: October 17, 2021, 11:00:52 pm
content: Quote from: EEVblog on October 13, 2021, 09:05:27 amSo is this Ipho training material:https://www.ioc.ee/~kalda/ipho/e-circuits.pdfOn recent (69th) Polish National Physics olympiad (again, for high schoolers) there was problem centered around diode equation:https://kgof.edu.pl/archiwum/69/of69-1-2.pdfWhat the problem says (in Polish) is not crucial, but the diode equation is clearly visible.Not every high school problem is "take U/I = R and knowing two things calculate the third. There are Olympiad-oriented students. There are even high schools which have a special courses for people willing to participate in olympiad.The 4V = 2V *2 may be simply a "trap".Quote from: EEVblog on October 13, 2021, 09:05:27 amThere are different types of teachers: those that fail at doing even simplest non-standard problems, and those that excel at their job and their student win competitions.Both types are paid the same due to unions, "for fairness"There was even a study into that at my university at the education department and its results were very unpopular among teachers. The study clearly identified both groups...EDIT:well that is disappointing:Quote from: etnel on October 12, 2021, 07:11:59 amStill this is a problem a good teacher can work with. Too bad it was "an accident"

author: perieanuo
author_rank: Frequent Contributor
timestamp: October 18, 2021, 06:53:01 am
content: Quote from: etnel on October 10, 2021, 02:05:06 pmi'm surprised by the mass of those replies.this is a clear case of "school" type of question, there is no assumption the LED have a real-life characteristic, so the answer he wanted is imho 0.02/0.04/0.02, assuming the led doesn't change forward voltage when 40mAmps flow in.it was fabricated by a teacher he wants you to respond quick and make some further observations about B current, why etcaetera.the real result of course it's different, let's take the characteristic and put that curve equation to deduce what static DC point we get for B and C. i seriously doubt the teacher solve that equation in 10 minutes, put him to that challenge, but without spice simulator. so i added a little bit to this useless thing

author: lacek
author_rank: Regular Contributor
timestamp: October 18, 2021, 07:55:41 am
content: Quote from: perieanuo on October 18, 2021, 06:53:01 amQuote from: perieanuo on October 18, 2021, 06:53:01 amIn this case it seems that indeed this problem was interesting by mere accident, as OP claims his teacher has no clue about anything. But whatever the reason, it is a very good school problem for competetive students. If one considers schools where there teach about nonohmic characteristics U(I). Existence of such places very country-specific. AFAIK there are very different approaches across the world. I was told that in Italy "every students learns everything" meaning they have broad perspective on literature, biology even if they are interested in maths. I have no clue about France.In my country in high school you specialize - you pick "humanities/science/computer science/theater/....." class and this affects your curriculim. Also the exam after the high school comes in "basic" and "advanced" level - in practice. correlated with the class choice.In addition to existence of "extended science classes" (in almost every high school) , there exist public, state owned schools (one-two per large city) where the classes are even more advanced, and olympiad level problems are considered during normal classes and appear on tests, they also learn complex numbers, some differential calculus (at the price of literature, biology, geography, which will be severely downgraded in curriculum and eyes of every student in such a class) . Some even have classes (that result in grades in high school diploma) given by university employees. To be admitted to such a highschool you have to show interest in sciance and have top rate grade. Sometimes highresult in middle-school maths/phyisics/.... competition is a requirement.I think it was a valid assumption that this forum attracts people interested in science and I would not assume that they do just U=I*R in schools. I think there are equivalents here and there. For sure there is one in India and China.

author: Vtile
author_rank: Super Contributor
timestamp: October 18, 2021, 05:40:46 pm
content: This would be much more interesting example if the circuit would be powered from ideal current source and this circuit would act as parallel resistance/impedance of it.As of turning the world upside down from piecefull voltage controlled world as we / you usually approach every source and circuit in electrical and electronics networks.There a is reason why world is sourcing voltage, so boring isn't it.

author: free_electron
author_rank: Super Contributor
timestamp: October 19, 2021, 03:32:01 am
content: Has anyone considered that led's are light sensitive ? so , if they are arranged as in the diagram the center led gets more light than the outer ones. That has an impact on the led.calculators ready ? go !

author: dtmouton
author_rank: Contributor
timestamp: October 19, 2021, 10:20:34 am
content: A slightly expanded version of the solution posted above.

author: Vincenzo
author_rank: Regular Contributor
timestamp: October 19, 2021, 11:34:14 pm
content: Quote from: dtmouton on October 19, 2021, 10:20:34 amExactly,It was Sedra & SmithThat's where I saw the questions years ago, and exactly like my solution in page 4 of this thread

author: dtmouton
author_rank: Contributor
timestamp: October 20, 2021, 08:44:24 am
content: I fully agree with Dave that this circuit has major deficiencies when viewed from the practical perspective. For one, we saw how the current increased significantly as the junction temperature increased.The point is that this is a basic exercise in DC analysis that any third year electronic engineering student should be able to perform before moving on to analyzing more practical circuits. It's all about mastering basic concepts and techniques.Dave, your reply to what you think "the answers is that they wanted" is just fundamentally wrong. Please look at some of the solutions that were posted on this blog.

author: EEVblog
author_rank: Unknown
timestamp: October 20, 2021, 11:23:38 am
content: Quote from: dtmouton on October 20, 2021, 08:44:24 amNo it's not wrong, it's right.How do I know this? I suspected that the circuit was entry level stuff and required an entry level answer, and this was confirmed correct by the OP who said it was a question given to high school students. There is absolutely no way a question aimed at high schoolers requires any more complexity than my 20/10/10 answer. And especially when no characteristic curve is provided.

author: dtmouton
author_rank: Contributor
timestamp: October 20, 2021, 11:29:36 am
content: No its wrong. Where did you study engineering?

author: EEVblog
author_rank: Unknown
timestamp: October 20, 2021, 11:46:12 am
content: Quote from: dtmouton on October 20, 2021, 11:29:36 amIt's "right" for the intended audience. There is absolutely no way a simple high school question on LED's expected an answer based on the diode equation or an answer based on the (not supplied) characteristic curve. The "expected" answer is 20/10/10. It was presented as anintroductory electronicsquestion.Quote from: etnel on October 12, 2021, 07:11:59 am

author: dtmouton
author_rank: Contributor
timestamp: October 20, 2021, 11:57:24 am
content: I get the context.

author: dtmouton
author_rank: Contributor
timestamp: October 20, 2021, 12:11:31 pm
content: Maybe you can make another video about how to solve the problem correctly. I agree that its is far above the level of a high-school entrance exam. My apologies for the misinterpretation.

author: StillTrying
author_rank: Super Contributor
timestamp: October 20, 2021, 02:06:41 pm
content: How about a 3 X 6V bulb version.The second part of the question could be 'What could possibly go wrong?'

author: Ian.M
author_rank: Super Contributor
timestamp: October 20, 2021, 03:46:45 pm
content: Quote from: etnel on October 10, 2021, 02:05:06 pmIf the original question was "Solve .... for current in each branch, SHOWING YOUR WORKING." then its a nasty trick question designed to see if anyone has been reading ahead. The left (R + LED in series) branch checks the student's understanding of the course module to date, as it can be simply solved with the data given. The right branch(s) are unsolvable due to insufficient data, but its the working that will be interesting.If one assumes a sub-'spherical cow' grade LED model, namely that Vf remains constant for all useful values of If, (as commonly assumed for 'back of the envelope' LED resistor calculations), and that the impedance of the 4V supply is negligible, it is impossible to write a system of equations for the branch currents consistent with Kirchhoff's circuit laws - as its given that the LEDs have 2V across them when passing 20mA, either you throw out KCL or determine that the current is unconstrained.If you make the slightly more refined assumption that all the LEDs have non-zero internal resistance, you can determine that the upper pair will have less than 2V across them so will each pass less than 20mA, and that the lower one will have over 2V across it and will pass more than 20mA but less than 40mA, but will be unable to further refine your solution. I suspect that's the answer the teacher is looking for.If you attempt to apply the Shockley diode equation , you probably fail as its beyond the scope of the module, and you are forced to assume the ideality factor and saturation current, and attempting to fit a curve with two independent controlling constants to a single data point indicates a severe lack of scientific/engineering rigor, disqualifying you from any careers more advanced than burger-flipping except working as a sycophantistic yes-man with your nose so far up your boss's @rse, your socks are getting stained!

author: amyk
author_rank: Super Contributor
timestamp: October 20, 2021, 11:58:08 pm
content: It's been interesting reading through this thread and seeing the East European vs Western viewpoints on the appropriateness of the question and education level.Quote from: Ian.M on October 20, 2021, 03:46:45 pmIf you look at theoriginal solution using the diode equation, notice how it doesn't matter what the constants are, since they cancel out in the end. It could well be put in the form of a word problem: "Two diodes are connected in parallel, and they are connected in series to a third. Each diode has a forward voltage drop ofxvolts at a current ofyA. 2xvolts are applied across the whole assembly. What is the current that flows through each diode?"

author: Ian.M
author_rank: Super Contributor
timestamp: October 21, 2021, 03:34:23 am
content: Thank you for that. I failed to do the math and only skimmed the thread so must now eat crow!

author: dtmouton
author_rank: Contributor
timestamp: October 21, 2021, 05:58:57 pm
content: The one thing we don't know is the junction temperatures and this circuit would be sensitive to the junction temperatures. The bottom LED in the right-hand string will get hotter than the two above it.Once we know how to analyze the circuit we can do a sensitivity analysis and figure out ways to make the design less sensitive to variations in component parameters.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: October 22, 2021, 05:05:16 pm
content: In contrast with many on this thread, I think that this is an excellent question. It can be solved at many levels of approximation, each level indicating a greater understanding of both theoretical and real world answers.Those who actually breadboarded the problem and measured results got a rich opportunity to understand the results, which undoubtedly varied between the experimenters. Those variations are yet another opportunity for learning. The published results directly contradict the opinions expressed by some in this thread. Those folks also get an opportunity to understand where their analysis was wrong. Reality always trumps theory.I do have a comment about some of the opinions expressed here. There is a whole group saying that there is not enough information to solve the problem, identifying various things such as lack of diode curves, lack of temperature information and so on. I would ask this group if a question showing a single 1000 ohm resistor with 10 volts across it that asks for the current can be solved with the supplied information. Most would jump in with Ohm's Law and say sure. But what about the tempco of the resistor and the thermal impedance to ambient and the ambient temperature? Or any of several other things that might matter. Almost all problems lack sufficient information if enough thought is put into the issue. A good student will show how the problem was addressed, and a good teacher will grade the results as they compare to the desired performance in the course. Radically different in high school and grad school.

author: AG6QR
author_rank: Frequent Contributor
timestamp: October 24, 2021, 11:49:21 pm
content: Quote from: CatalinaWOW on October 22, 2021, 05:05:16 pmBut the difference is that, with a resistor, the secondary effects such as tempo, ambient temperature, or amount of ambient light shining on an LED typically won't affect the first or second significant digits. With the LED problem as stated, you can't even determine the first digit of the currents in the right branch without knowing those types of details.Part of good design practice is to avoid designing circuits where the secondary effects affect the desired behavior so much.

author: szir
author_rank: Newbie
timestamp: October 26, 2021, 08:53:27 pm
content: If I would give my job interview answer then it would look like something like Tomorokoshi’s after pointing out that this should not be built like this in practice....except (1.3 should be) I=Is*(exp(V/VT/n)-1)Since the current at zero voltage should be zero (otherwise I could just get a bunch of diodes and generate current… though I would need a lot since Is is really small)Since we assume ideal and identical diodes we can use n=1 and calculate Is from the given data point Vf=2V, If=20mA and rearranging (1.3)(1.5) Is = I/(exp(V/VT)-1)Where(1.6) VT = k*T/qk=1.380649e-23; // J/K = C*V/Kq=1.602176634e-19; // CT=300; //KNote: Because we don’t know anything about the temperature or cooling it’s easy to assume constant identical temperatures (T=300), but it will not be the case in real world.From 1.3 we can calculate V by rearranging the terms and taking the natural logarithm of both sides(1.7) I/Is+1=exp(V/VT)(1.8 ) V=ln(I/Is + 1)*VTThere are multiple ways solving this.The quickest to do without a computer is to take the voltage difference between diode B and C(3.2) dV = V2-V1 = VT*(ln(I2/Is+1) - ln(I1/Is+1)) = VT*ln((I2/Is+1)/(I1/Is+1))Because we know that I1>>Is (I1 is much greater than Is) and I2>>Is, therefore I1/Is>>1 so we can simplify with little error(I2/Is + 1)/(I1/Is + 1) ~~ (I2/Is)/(I1/Is)=(I2/I1)and we get(3.1) dV = VT*ln(I2/I1)Because I2=2*I1,(3.3) dV=VT*ln(2) = 17.919241 mVWe know that V1+V2 = Vcc and V2=V1+dV (from 3.2), so V1+V1+dV=Vcc and therefore(3.4) V1=(Vcc-Vd)/2 = 1.9910404 V(3.5) V2=(Vcc+Vd)/2=Vcc-V1= 2.0089596 VSubstituting the results to (1.3) we get the currents:IC=I1=Is*(exp(V1/VT)-1)=14.142136 mAIB=I2=Is*(exp(V2/VT)-1)=28.284271 mASince we calculated the currents from voltages it’s not surprising that V1+V2=Vcc with zero error (V1+V2-Vcc=0). The only error of our calculation we can notice is in the currents. While I2/I1=2.0000000 is true, I2-2*I1 should also be 0, but it is not. It’s -3.504e-16, but it is close enough to 0, it’s unmeasurebly small (-0.0003504 pA). The error is likely comes from floating point precision rather than the simplification we made at (3.1). I1/Is = 2.805e+33 so adding 1 does not make a difference, cannot be represented in float or double precision. (I1/Is+1)-(I1/Is) does give 0 as result (instead of 1).I used SciLab to solve the problem in other ways.We can get an idea by graphing either the voltages over currents or the currents over voltages.Or we can solve it numerically by either way: reducing the voltage error or the current error.First define the diode functions:Code:function I=diodeI(V)I=Is*(exp(V/VT)-1);endfunctionfunction V=diodeV(I)V=ln(I/Is + 1)*VT;endfunctionThen we can graph the Voltages over an interval (10mA..20mA) of currents of C. The solution is when VB = Vcc-VC (while IB=2*IC)Code:I=linspace(10,20,4000)/1000; // 10mA to 20mAfigure;plot(I*1000, diodeV(I*2) ); // VB =plot(I*1000,Vcc-diodeV(I) ,'r'); // Vcc-VCOr we can graph the currents (IB and 2*IC) over a range of voltages.Code:V=linspace(1.9,2.1,4000);figure;plot(V,diodeI(V)); // IC =plot(V,diodeI(Vcc-V)*2,'r'); // 2*ICTo solve it numerically we can also use these two approaches:We minimize the voltage error as a function of current:Code:function Verr=diodeVErr(I)Verr = diodeV(I)+diodeV(I/2) - Vcc;endfunctionIsol = fsolve(20e-3, diodeVErr); /// start from 20 mAIsol =0.0282843Or we minimize the current difference as a function of voltage:Code:function Ierr=diodeIErr(VB)Ierr = diodeI(VB) - 2*diodeI(Vcc-VB);endfunctionVsol = fsolve(Vcc/2, diodeIErr); /// start from 2VVsol = 2.0089596Thankfully the results match the one we got earlier. (Which does not mean it’s correct, but at least consistent ; )It’s also nice that it matches the simulation result of hadibaria and Wytnucls.I was curious how would temperature change effect the result. It’s likely that higher current will result higher temperature (unless there is significant different in cooling or environment between B and C). Obviously we don’t have any data on this, so it’s complete guesswork…A diode at 2V and 20mA consumes 40mW, some of it will become heat (60%..95% of it). Checking the datasheet referenced in the video has 60mW power dissipation (I guess assuming non typical forward voltage and calculating it with the max 3V) but more importantly 500K/W thermal resistance. That would mean 20K increase @ 40mW heat generation. Obviously we don’t know how well the diode is insulated from the ambient environment so lets just assume total 500K/W resistance and that all the dissipated power becomes heat eventually. So around 2V operation voltage dT=Rth*P=Rth*U*I=500*2*I, or dT/dI=1 K/mA. Which means there should be around 14K difference between the parallel and series diodes on the right side.We can still solve it with ease (to not have to recalculate Is at a higher temperature to make it easier to compare with previous results, we will still assume 2V, 20mA @ 300K operation, which would mean a bit cold 280K ambient temperature)Code:function V=diodeVT(I)T=300+(I-20/1000)*1000; //1 K/mAVT = k*T/q;V=ln(I/Is + 1)*VT;endfunctionfunction Verr=diodeVErrT(I)Verr = diodeVT(I)+diodeVT(I/2) - Vcc;endfunctionIsolT = fsolve(20e-3, diodeVErrT); /// start from 20 mANow we get26.888 mAinstead of 28.284 mA around 1.4 mA difference.Note that this is still using an ideal diode model. And it’s clear from the pdf that real LEDs are not ideal because I-V log-lin graph is not linear. They have at least serial resistance.We could use a model that includes serial resistance like this:Code:Rs=10;function V=diodeVR(I)V=ln(I/IsR + 1)*VT+Rs*I;endfunctionAnd solve it like this:function Verr=diodeVErrR(I)Verr = diodeVR(I)+diodeVR(I/2) - Vcc;endfunctionIsolR = fsolve(20e-3, diodeVErrR);But here we have to recalculate the Is for Vf=2 @ 20mA to remain true.Code:function Is=diodeIsR(V, I)Vd=V-Rs*I;Is = I/(exp(Vd/VT)-1);endfunctionIsR = diodeIsR(Vf,If);With this we get a similar curve as in a real LED in the pdf.Code:I=linspace(0.1,100,4000)/1000;plot("ln",I*1000, diodeVR(I));The solution gives26.846 mAcurrent.Using both temperature and resistance could give us more realistic result (26.769 mA), but it’s still not gonna be perfect. Even if we use real word measurement data of I-V curve of a LED it's not going to solve everything. Because of manufacturing variances the 2 diode in parallel will not be perfectly matched, so the currents will not be evenly distributed. We haven’t mentioned resistor tolerances and their temperature dependency…...Or the voltage supply tolerance. The difference between currents we got was small at 4.00V Vcc (within 2 mA), but they get much larger as Vcc changes.At Vcc=4.2 (+5%) we get1353 mA for the ideal diode44 mA with T39 mA with Rs34 mA with T and Rs.It is interesting how complicated calculations can get with a seemingly simple circuit like this.

author: dtmouton
author_rank: Contributor
timestamp: October 30, 2021, 09:00:04 am
content: I watched the video again and would definitely not have hired Dave if that was a job interview. The "context" is not explained in the video. Based on the video, my conclusion is that Dave does not know how to do a standard DC analysis of a circuit that contains diodes or transistors.This is highly concerning since Dave pretends to be a "Professional Engineer". Yet, there is no indication of an accredited Engineering degree on his LinkedIn page. Being a senior member of the IEEE is not a qualification in itself.Dave, I think you have to spend a bit of time thinking about the legal implications of what you are doing with your YouTube channel. It looks like you are providing professional advice without any credentials. What degrees do you have and are you registered as a chartered engineer in Australia?

author: Nusa
author_rank: Super Contributor
timestamp: October 30, 2021, 09:26:17 am
content: Quote from: dtmouton on October 30, 2021, 09:00:04 amPerhaps you should educate yourself on Australian law before choosing to climb on a high horse and suggesting legal consequences.Dave has mentioned in videos that "engineer" is not a protected title in Australia.

author: dtmouton
author_rank: Contributor
timestamp: October 30, 2021, 09:46:19 am
content: I am fully aware of the Australian system. However, the channel is watched all over the world. In South Africa, the term "engineer" is not protected, but "professional engineer" is.Let's hear from Dave what his qualifications are. Then the viewers can decide if is fair for Dave to refer to himself as a "professional engineer".I am sitting on a Zebra and not a horse.

author: dtmouton
author_rank: Contributor
timestamp: October 30, 2021, 09:54:06 am
content: According to:https://www.engineersaustralia.org.au/Membership/Occupational-Categories#:~:text=Engineering%20Associate-,Professional%20Engineer,Apply%20leadership%20%26%20management%20skillsProfessional Engineers hold an Engineers Australia accredited or recognised four-year professional engineering degree.

author: dtmouton
author_rank: Contributor
timestamp: October 30, 2021, 10:19:39 am
content: Furthermore, according to:https://www.engineersaustralia.org.au/News/registration-engineers"The title “engineer” is not protected in Australia, therefore anyone can claim to be an engineer and practice without the necessary competencies, understanding of standards, or in compliance with a code of ethics. This exposes our profession to poor quality engineering, a lack of public trust, and increased general risk – be it financial, safety or environmental.Most people are surprised to learn that the engineering profession does not have this technical assurance in place, compared with other professions such as medicine, law, accounting, teaching, architecture etc. which all require registration, either on a state or national level.It has been strongly advocated by Engineers Australia for many years that as part of maintaining professional standards we need to ensure that our engineers are properly qualified, assessed, responsible for their own ongoing continuous professional development (CPD), and are accountable to a code of ethics. For EA this is known as the Chartered Engineer process.Furthermore, Australia should have a national register of engineers whose competency and integrity have been assessed against a national benchmark standard of professionalism.As a result, the National Engineering Register (NER) was launched by EA in 2016. There are currently over 16,000 people registered, and it is publicly available and searchable online. The EA chartered engineer route is one way (and by far the most common) onto the register, and the other is by individual assessment."I could not find Dave on this register.

author: Jwillis
author_rank: Super Contributor
timestamp: October 30, 2021, 12:09:43 pm
content: I had my fill of so called "registered " "engineers" in the field of civil works I'm in . I won't have a lot of them design a cardboard box let alone any structure or highway . To tell the truth , I'll take one Dave over any "registered engineer" that boasts about the fancy ring on their finger .

author: dtmouton
author_rank: Contributor
timestamp: October 30, 2021, 12:42:48 pm
content: I like watching Dave's channel. But when it comes to the more subtle aspects of electronic engineering he often gets it wrong or just fudges it. This particular video infuriated me since he criticizes lecturers who give this kind of problem to their students without knowing how to solve it himself.Unfortunately I can no longer refer my students to his channel. In myopinionhe has remarkable knowledge with respect to technical aspects of engineering. However. it is clear that he doesn't have the rigorous formal training to be classified as an engineer in many countries of the world.As I previously pointed out: You need to have a four year accredited engineering degree to call yourself a "professional engineer" in Australia.

author: dtmouton
author_rank: Contributor
timestamp: October 30, 2021, 01:09:59 pm
content: As an Electronic Engineer, I don't care about the "chartered" part at all. My employer forced me to register. However, I think the degree is very important. I had an electronic design company before I studied engineering and already knew a lot about design. The formal training really took me to a much higher level. It was a true eye opener.

author: Jwillis
author_rank: Super Contributor
timestamp: October 30, 2021, 01:41:54 pm
content: I agree 100% that an education is important . But you already had many years of practical experience before a formal education. Many engineers coming out of school have no practical experience at all . They are taught how things should work from a book but are never taught what could go wrong. In my own experience many of these engineers are so arrogant that there might be one in five that isn't likely to get someone killed. I've personally had to fire some of these so called professional engineers simply because they would put people lives in danger because they have no practical experience . Book smart but totally useless in the field.Being forced to get a degree from your employer sounds more like protection from possible litigation towards the company because now if something goes wrong the responsibility falls squarely on you and not the company. But that's totally different discussion on it's own .I would far sooner trust someone that makes a mistake when teaching to show what could go wrong or a possible work around than someone that thinks they know it all .

author: dtmouton
author_rank: Contributor
timestamp: October 30, 2021, 02:03:59 pm
content: Just to clarify one point: I decided to study engineering myself. My employer forced me to register as a chartered engineer.In South Africa, you need to have an accredited degree and a few year of work experience before you can register as a chartered engineer. You have to give a detailed account of projects you worked on. Nonetheless, I don't care for it and signing the Washington accord was the worst decision our Engineering Council ever made. We had to decrease the content of our program to meet the Washington accord.It sounds like your students don't gain enough practical experience during their studies. Maybe you should lean on your universities? I did a postdoc at Laval University in 1992. Their electronics laboratories looked quite advanced.

author: dtmouton
author_rank: Contributor
timestamp: October 30, 2021, 02:30:29 pm
content: I forgot to mention that in South Africa, you have to be a charted engineer and have a "Government Ticket" if you do work that is potentially life threatening. The "Government Ticket" also requires a few years of part-time study and is mainly about safety and industrial law.

author: jesuscf
author_rank: Frequent Contributor
timestamp: November 08, 2021, 09:06:20 am
content: With two test points the problem is easy to solve, because it allows us to use the diode equation. In my opinion two test points should have been provided from the beginning. Attached image shows how I solved the problem assuming the four diodes are identical, and two test points are available. I used XCAS to solve the non-linear system of equations.

author: jesuscf
author_rank: Frequent Contributor
timestamp: November 08, 2021, 03:19:11 pm
content: Quote from: dtmouton on October 19, 2021, 10:20:34 amFor LEDs n is often larger than 2.