EEVblog 1427 - An INFURIATING Electronics Exam Question! 

There is a reply from the OP: "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 8). 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."

The only correct answer is A = 20mA, B=C=0mA because that LED no longer has any magic smoke in it.

I’ll never forget in the first assignment of my electronics degree I was asked to find a suitable value power resistor to limit the current in an LED in a big, convoluted yet massively oversimplified circuit of a solar panel directly charging a battery (zero protection). Of course the LED was of the magical 3.65 V ideal type so, instead of doing a big massive current analysis for all of the insignificant wire resistances and unrelated circuit branches they gave I assumed ideal wires and just used the voltage of the power source (an ideal constant voltage battery) and a single ohms law calculation to get the value of 12 Ω, a nice E12 value. But of course I was marked wrong for not solving it ‘correctly’ with three pages of working to get exactly 11.908 Ω. From there I realised that the electronics class I was doing was actually just an electronics themed math class detached from any concept of practicality. There were plenty more massively over simplified circuits in that class. I really do agree that those questions are “pathological” and should be avoided as I have had a number of people who should know better ask how I’m running a 3.3 V white/blue diode off a 3.3 or 3.0 V supply.

I got this question in an interview and ended up getting arrested for assault, under the instructions of Dave I let them have it. I didn't get the job.

Dave muttering to himself: "Right, that's question 1 done. Question 2..." Examiner: "Okay, pens down please."

Entry-level hobbiest here and having you analyze questions like this is very satisfying. I get to think my thoughts and then learn the truth! Thanks!!!!

Answering test questions like this is an entirely different skill than real-world applications, mostly about understanding the answer they're looking for and assumptions they're making.

A is easy. B and C you would need a characteristic curve to solve.

I love this problem! This is a classic Analog IC design question. The intuitive understanding of semiconductor junctions give s a quick answer to this problem. Because the nature of the junction is logarithmic, the two parallel diodes (LEDs) will have a decreased drive exactly cancelling the increased drive to the bottom single diode (LED), a ~30mA current will result in the second path due to the 1.5X effective junction area spread over the 3 diodes in the second path. The kT/q ln I1/I2 is typically18mV for a diode at room temperature, so for a small signal analysis for the LEDs, the same relationship should hold true. I am a former Analog IC designer and Analog guys LOVE the nature of logging characteristic of diodes, using them in multiplying, RMS, logging and other current mirror applications in designing ICs.

Kirchoff's current law is litterally the first law you learn after Ohm's law....

I found I learned more from the 'bad circuit' ideas in the early versions of Horowitz & Hill than i did from many other texts

The question "How would I calculate ..." appears to provide a philosophical opportunity! First, I'd power up this circuit to let the magic smoke out of that right side of the circuit. Once the smoke had cleared, the answer on the left branch of the circuit would be 20mA and the answers on the left would be a peaceful ZERO. I would choose to ignore the temporary and catastrophic vaporization of material on the right side and only consider this calculation after this circuit was at steady state. It's akin to ignoring what occurred in our Universe during its first few moments before baryonic matter condensed and only pondering life after photons were able to be emitted from these Universal LEDs.

Good point about the whiteboard. Before Covid I asked all software job candidates to draw some architecture on the whiteboard. That quickly sorts out the also-rans or the bluffers.

I went through the Tomorokoshi's calculations and everything I can say about that is that they are perfectly correct and IMO it is the only correct answer to that question. It's a little bit "speedy" to me, there is no explained how the (3.1) equation was obtained from (1.3), but I didn't have to take a paper and pencil to prove it is correct, so on the other hand, it is not so hard. If you look at the (1.3) equation, there are a lot of variables, but according to (1.1) and (1.2), the η and Vt are constants, because that diodes are the same, they are only at different points of its V-A characteristics, so only the V and I varies. At the next step (1.3) was splitted to two equations with Vb Ib and Vc Ic, the rest variables was untouched because they are constants and that equations was substracted one by another which resulted in (3.1) equation. The rest is simple and easy to understand. In my opinion, this is much beter question than something like "write the Ideal Diode Equation", because this proves that student is capable of using that equation. In fact, I have no experiences in teaching, but it looks like a good idea to me. Maybe someone experienced in teaching (like @TheSignalPath :3 ) could tell us whether exam questions like this are good or bad idea.

My biggest take-home from this was "always bring whiteboard pens to any technical job interview from now on". I feel like a knobber for never having considered that before.

At a job interview once, I was given a multiple choice test they had ‘borrowed’ from Microsoft where one of the questions famously had the wrong answer... ...what did I do?... ...I was congratulated for being their first candidate to ever get 100% 🤣 (I’ve felt vaguely unclean ever since).

I don't understand why people seem to think the problem is not suitable for hand calculation or not solvable without knowing the LED specs, because a hand calculation works fine for a decent estimate (assuming matched LEDs), and isn't even difficult. Assuming that the LEDs follow the familiar Shockley diode equation, the I-V relation is I = I0*exp((V - V0)/(n*VT)), where I0 = 20 mA and V0 = 2 V. If the voltage of the center node is 2V + Delta, KCL applied at that node gives 2*I0*exp(-Delta/(n*VT)) = I0*exp(Delta/(n*VT)), or exp(Delta/(n*VT)) = sqrt(2). The current through the bottom LED is then sqrt(2)*I0 = 28.3 mA, and the current through the upper two LEDs is half that: 14.1 mA.

At a job interview for a proof-reader/editor position at a CD-ROM publishing company, I was given a written exam with nonsense questions, so I proof-read and corrected their exam paper... ...I didn’t get the job! Some people have no sense of humour ☹️

Once passed by the EE lab and saw a friend in there crying. Her lab was to build a DAC with resistors, but it wasn't giving her the right result. Lab design showed the drive circuits providing 0 and 5V for the logic levels, but the lab provided TTL chips. No 5V out on those, so no accurate to predicted DAC output. I often wondered if any of these professors ever designed a real circuit.

I had an old school EE prof for my first circuits class. Had him in just about the last term before he went emeritus. He was very practical. He would say that if you couldn't understand whether the answer your calculator had given you was reasonable or not, it wasn't worth the electrons used to calculate it. I'm pretty sure this question would send him into an apoplectic fit.

The only answer that can be given from the stated data is: A: 20mA B: >20mA and equal to 2xC C:

Diodes obey Ebers-Moll: I(V)~I0 exp(V/V_T) and V=V_T ln(I/I0) where V_T=kT/q. The left side does balance; since (100)(0.02)=2V, point A is at 2V at 20mA so the LED voltage is 2V at 20mA... this is the given point. But the right side does not balance. Since I_B=2I_C, we can find the difference between the two LED voltages: V_B-V_C=V_T [ln(I_B/I0) - ln(I_C/I0)] =V_T ln(I_B/I_C) = V_T ln(2) Kirchoff's loop also gives V_B+V_C=4V Adding these we see 2V_B=4V+V_T ln(2) and V_B=2V + V_T ln(2)/2 The node voltage is elevated... Now, use V_B-2V=V_T [ln(I_B/I0) - ln(20 mA/I0) = V_T ln(I_B/20 mA) Combine these V_T ln(I_B/20 mA) = V_B-2V = [2V + V_T ln(2)/2] - 2V = V_T ln(2)/2 Cancel V_T to see I_B = 20 mA * sqrt(2) = 28.3 mA Identical diodes will divide this current equally so that I_C=14.1 mA. Without Ebers-Moll there would be no current mirrors, no log amps, no band gap references, etc. So it is NOT a silly question although small fluctuations will indeed be amplified in the real world. I0 and V_T are TINY, so any ammeter voltage will reduce the current significantly. This, in fact, is what causes the fluctuations to be amplified.

Back in the old day's LEDs would actually explode when overcurrent was applied. So those that branch with no resistors would be all kinds of fun, when the supply voltage drifted over 4V. A software engineer wrote that question.

That table at 15:01 is seriously broken. The headings don't match the collumn data and there are TWO collumns called MIN. I suspect the headings should be Symbol, Minimum, Typical, Maximum, and Unit.

Had what I think was a question like this in my physics final. The problem with letting them have it in the answer is that it's usually some postgrad working from an answer sheet that's marking it and they won't care or be able to do anything about it. Spend the time improving your answers on other quesitons.

The guy with the long mathematic proof actually had an amazing answer that wasn't super complex

This is a Voight-Kampff test for determining whether or not an EE is a replicant.

I'd say that we can only introduce reality when it does not conflict with the given information. The total voltage is exactly 4v and each LED across 2V passes exactly 20mA. It's _impossible_ for current B to be 20mA because that would mean the LEDs up top see exactly 2.0V and thus would, by the nature of the given information, each draw 40mA. That obviously isn't the case because of Kirchoff's Current Law. We know in reality that there is a positive slope relating forward voltage and current and this does not conflict with the given information. The answer ends up being somewhere between the two conflicting "ideals", specifically that B is 20-40 and C is 10-20. This is the answer lacek gave on the forum. Tomorokoshi's solution assumes the diodes follow the ideal curve but this may not be the case, we only know the parameters at one specific point. We don't even know if the LEDs are matched if operating at something other than 2V. Tomorokoshi's solution _may_ be valid, but possibly not because it assumes information not given. Naturally it _does_ does fall into my given range of possibility.

Led characteristics also change depending on the amount of light *hitting* the led itself. Infact, you can use a lit led as a light sensor with sensitive current sensing. I imagine the changes from touching the led from the breadboarded circuit included some of this effect.

The idea of analyzing a simple circuit like this is very helpful and makes me wonder whether you could do a series on well known electronic circuits we all should know. You could go through a circuit per video explaining exactly what is happening. For example, so many examples of "good" circuits in the "Art of Electronics". You could pull circuits from there that you know well and that you think we should all know and understand. I would watch them for sure!

When I was an instructor, I could always jig the curve by making about 65% of the questions easy, about 30% challenging, and 5% ridiculously hard.

Actually Tomorokoshi did it right and his calculations would probably match the experiment. But anyway you can actually easily calculate the approximate values, A brunch current is 20 mA because 2v(LED Vf @20 mA) + 100Ohms + 0.02 = 4V. And using simple logic you may estimate the approximate values for the current in branch B like bottom LED voltage drop should be higher than upper parallel LEDs, because their individual currents are less than the bottom LED current (KCL). And if we assume the top led voltage is about 1.8 volt and the bottom is 2.2 volts we would probably be close to the right answer, next we just estimate the current through the bottom led knowing it's exponential nature at 2.2 v it is higher than 20 and is probably around 28. Then the current through the one of the top LEDs is approximately 14 mA. Even with the slight parameters variation in practice due to temperature and manufacturing process the circuit will still be close to these values, given that the LEDs are of the same type from the same manufacturer. I think an experienced engineer who designed LED drivers a lot and holds the picture of the exponential curve for a "universal LED" in mind could give even more precise estimate for the branch B.

Question on exam: Draw up a circuit to light 2 LEDs at full brightness and 2 LEDs at half brightness using only one 100 ohm resistor.

I would use a substitute circuit for each of the LEDs consisting a voltage source and a small resistor. A linear current / voltage relationship can be assumed within a limited range. The voltage sources can be ignored, so the circuit to solve becomes two resistors in parallel with one in series. This gives current B is 26.6667 mA and current C is 13.3333 mA.

I'm glad that I took a glance at the thumbnail of the video before listening to you, Dave. I'm especially glad that I was able to roughly estimate the values, but for me it was even more important that I could realize the stupidity and uselessness of such a circuit on an exam. If only it wasn't an intentional trap, of course. And thank you so much for another great video!

The resistance of any meter connected at points B and C would greatly effect the current.

the correct answer is : all leds carry 20 mA of current and have 2v volt across them , but only in a universe where the laws of physics of that universe allow it.

This gives me flashbacks to high-school. We had to solve theoretical questions like these for two years, without once holding a real component. When we were allowed to build our first breadboard setup after those two years, the question included a similar illogical theoretical schematic... Not that it mattered anyways... Besides being able to name the symbols on an electrical schematic, most people in our classroom didn't know how the real component looked like anyways... Then our government and industry wonders why there is a lack of technical skilled people...

Totally love your videos! Thanks for this video! Cheers!

I had exactly one college course in basic electronics. Kirchoff’s current law was definitely in the course. As soon as I looked at that right hand branch I looked for any current limitation at the input, and seeing none, realized that this circuit would be, at best, highly dependent on small variations and, more realistically, likely to blow out the bottom right LED.

My hardest professors I had for my electronic courses were either from India or Sweden. As soon as I was done with those classes I just mind dumped everything, I dont know why they pushed so hard.

My answer would have been A = 20mA, B = C = 0mA. Without even thinking.After a second of thinking, right side blown.

Finally something I can understand, love this video.

Theoretical electronics is all about the theory. At the uni we were forced to do calculation on absurdly complex schematics using variety of methods. Basically it took us 30-40 minutes to calculate something a computer does in less than a millisecond.

One thing I'd point out is that some circuit that seem stupid if you imagine them on a PCB can make perfect sense if you think about them as multiple devices on the same wafer in a chip ... very different conditions and the devices / diode / transistor end up very closely matched (because they're right next to each other on the same wafer).

If they give you the forward voltage at a specific current it means that the LED has to be modeled as a real component with a curve that has a unambiguos voltage/current characteristic. The circuit would work. The current will be equally shared between the two parallel leds. Thus it is impossible that all leds work at 20 mA, than the full curve is needed to determine the actual work point.

Thanks Dave for motivating me into getting my neurons off their duff and working again. Your videos are the best! I thought to play a little and wired up the circuit. My instruments aren't top quality but what I've measured was current thru ammeter A was approximately 19mA and for ammeter B, 6mA. Keep up the good you do for all!

Agree it would be a fantastic employment question . Designed to see if the one person questioned know the difference from real world and text book . Enjoyed the video.

Thought: '... ah, you're leading me down the garden path... or are you?'. One of those questions where I'd end up possibly using it as leverage to suss out how competent the interviewers are (if I knew their claimed backgrounds were relevant, not HR or they were hiring because they had no idea about the field). Depending on the circumstances, it might be the case that you find out the real reason why they have a position open and maybe that position isn't for you even if you are offered it.

It's notable that these comments include a wide spread of confident answers ranging from total ignorance about diodes to neat solutions using Shockley's diode equation. But its rather sad to see how many of the former there are on the EEV channel where I would have expected a better educated audience especially considering Dave's previous videos on fundamental principles. I don't expect the average electronics enthusiast to know the diode equation but they should at least understand the general shape of a diode V-I curve and be able to see that the current C will be less that 20mA and current B will be greater than 20ma.

I think this schematic was drawn from a previous test where the LED's were originally incandescent bulbs. I think some guy decided making it with LED's would make it more "modern". Edit: I think the original schematic also assumes the bulbs are a constant resistance.

I love these type of videos Dave. Please do more....

Would it be fair to say the result is 3 out of four burned out LEDs?

If you have V I graphs in hand you can, assuming all LEDs are identical, multiply by two the current on the graph (for the parallel branch), sum the voltages (for the series branch) and then intersect that with 4V. That will give you the current for that branch, divide by two to get the current for the parallel branch.

I was just practicing my nodal analysis yesterday. It’s refreshing to see that even the electronic veterans use KCL.

"Good enough for Australia" That time they tried to legislate Pi to 3 digits.

Yep we learned Kirchhoffs law in the first few months, that and ohms law is the first thing that is learned

Worse, I've recently seen a design which drives a couple of LEDs in series with an npn to ground(open collector, driven by a pwm signal). No series resistance whatsoever. Guess what, the LED Type changed due to availability and now the current draw as well as the light output dropped by a factor of ~11.

My first thought was: I assume that the power supply doesn’t have any particular output protection and delivers an over voltage spike that kills the right branch. From there, either the left branch is fine or the right branch fails closed and possibly pulls down the Vcc or causes a cascading failure.

This is actually an amazing question and one I find very typical. As an instructor, I may actually use this question in my Analog Electronics course where we cover semiconductor based components. To "properly" analyze this, we would have to do iterative analysis using ideal assumptions to get a ballpark answer, similar to as Tomo* in the video did. Assume 2V @ 20mA for the single LED. KCL says 10mA through the parallel LEDs. Then using the 10mA, calculate the voltage drop across an ideal diode. Then use the voltage drop to calculate a new voltage drop for the single LED. Then calculate the current through that LED with that voltage drop. Then divide that current and calculate the voltage drop across the parallel LEDs. Repeat this process a couple more time to allow the variations to stabilize.

"The poor bastards on the right will have to sort them selves out..." Haha, loved that one.

OMG! I remember a very similar question from my students' days. (I am from Sweden)

enjoyed this videos thanks for posting.

Reminds me of the questions in the GAT my son did recently. Ridiculously convoluted maths questions that, once you sorted through all the BS, boiled down to the most basic multiplication of three or four numbers.

The parallel LEDs must be half the bottom. Hence the bottom will have more than 20mA (2.1V?). The top two will be less (1.9V?). This would give currents either side of the datasheet figures, giving a very unpredictable current down the right leg.

Assuming ideal diodes RHS will burn out (either the leds or wires) and become an open circuit leaving you with just the LHS with the led and current limiting resisitor. easy q 🙃

I just discover your channel and I feel immediately identified…. I have lost soo many friendships for this kind of discussions… but I consider my own sanity above any temporal assumption

My answer would be something like "Because LEDs are not constant voltage devices, the two LEDs in parallel would have something less than 2V across them; therefore, the fourth LED would have something greater than 2V across it. B would be >20ma and C must be B/2."

Although a diode is an active device, operated on DC it can be treated as a resistor in idealized case. 2v/0.02a = 100 ohms. Just replace the LED's with 100 ohm resistors and give the idealized answer.

Fun fact: semiconductors are sensitive to all kinds of energy, not just heat. Theoretically, the LED is reacting to incoming light too. Obviously, the effect is negligible, probably unmeasurable.

Long ago I calculated the effective series resistance of a common red, orange or yellow LED to be 100𝛀 at 20mA. It's been a useful bit of knowledge whenever I've needed to quickly approximate component values for a circuit.

Hello back to you Dave from another Swedish fan. Great channel and content.

I don't know what it is with the "educator" class and coming up with really goofy contrived scenarios (that are often "trick questions" to some extent) when there are plenty of actually useful real life design patterns to analyze and you'll regularly encounter in the future. Goes for really any type of educator.

I need to spend more time in the forum....

The theoretical questions in school often use ideal components, eliminating confusion about the various components in the real world.

I suspect xrunner also had some measurement error in the parallel LED pair as a result of the shunt resistor of the multimeter.

Reminds me of a question in electronic music class in high-school: Considering perfect human hearing can hear from 20hz to 20khz, is it worth the extra money to get an ideal amplifier (no sloped roll-off) with a frequency response 20hz to 30khz compared to one 20hz to 20khz? Expected answer was yes as you'd want to pass harmonics. But I always argued you can't hear any harmonics past 20khz. I couldn't find a way to articulate that the ear itself would be a bandpass filter just like the amps.

This is a theoretical question. Schema parameters are chosen so that the result does not depend on the nonlinearity of the diodes. Ia = 20mA, Ib = 40mA, Ic = 20mA

A homework question for future designers of Harbor freight flashlights.

demosntration of RS20 in the forum is awsome, using sqrt(2) to find 28mA... really great response.

Step 1:Build the circuit Step 2: Probe the multimeter wherever you want

They shouldn't use this kind of circuit in an exam, because that would teach the student to expect something that doesn't hold in real life, and a circuit that is a very bad idea to design like that. At least do not simply ask for the current to be calculated, which would be an impossible task with real-world components. Instead ask if the current can reliably be calculated on each of the test points, and justify the answer. Just by looking at the datasheet on the video (which the exam doesn't even provide), it says that a the LED has a typical Vf of 2v, and a max Vf 3v (if i understood the mess they made to the columns right). And what's the point of using ideal (or better call them magical) components? Just use resistors instead of leds on the right branch, and nobody's hurt. On the right branch, assuming identical leds (and good luck finding that!! :P), then the current thru the upper leds has to be half of that on the lower led, which means that the Vf on the upper leds is going to be less than 2v, and on the lower led it's going to be higher than 2v, and the current thru the lower led is going to be higher than 20 mA, but exactly by how much? damn if i know! By the way, spanish speakers always call them led instead of L.E.D., as the pronunciation of any letter combination is obvious in spanish, and a lot shorter to speak "led" than "L.E.D." In fact, we always use letter sounds in acronyms, instead of letter names, probably because letter names are never shorter than letter sounds, usually around 2 or 3 times the lenght, and most letter names include E, so the result of using all that letter names put togheter would sound funny. For example: "ele.e.de.", instead of "led".

You can use the lineal approximation of the diode, Id=(20ma/(2-vo))×(vd-vo) where vd is diode voltage, and v0 is the voltage where Id=0. This v0 is unknown and we can give some values a little less than 2v. If we call V1,I1 the voltage drop and current in the upper diodes and V2,I2 in the bottom diode, we have: V1+V2=4 and I2=2×I1. Because we have 3 equations and 4 unknown values, if we give v0=1.8v, we can find the voltage in the upper diodes is 1.93 and in the bottom 2.07, the currents are 13.3 and 26.6ma.

Using Ohm's Law, each LED has an apparent resistance of 100 Ohm. Note that the left leg of the circuit and the right leg of the circuit could be separated into two separate circuits and the results would be the same for meters A, B, C. Solving for the series and parallel resistance circuit values yields: A = 20.0 mA, B = 26.6 mA, C = 13.3 mA

It's not about theory vs practice... Those who insist that the two are in opposition have not much idea about either. It's flawed theory vs good theory

At best; mildly infuriating. EEV: "These diodes on the right side they just gonna hafta sort 'em selves out... the poor ba**ards" Douglas Adams: Marvin trudged on down the corridor, still moaning. "I've got this terrible pain in all the diodes down my left hand side..." Me: *hard chuckle* while eating, checks myself... ok, ok... *swallows again* ...I'm good =o)

What constitutes an correct answer to this question depends on the context in which it's being asked. In many situations, Dave's take on it, that it's nonsense, is reasonable. However, at a certain level, it is certainly a reasonable question, calling on the examinee to identify what, if any, additional information would be necessary to solve, and what assumptions can be made to guess at it. Simply saying that the Diode is nonlinear and giving up on going further is literally saying to an interviewer that you know enough to make a certain amount of money and don't care about being worth more than that. If I were using this question in an interview, it would be because I would be looking for whether the candidate understands that the nonlinearity of a diode's IV curve is exponential, and how one works with exponential functions. Tomorokoshi gives the best reasonable final result, but still doesn't quite get full marks, relying too much on equations rather than a good intuition to get there. A solid A, but not quite A+. The answer I would be looking for is: 1) Assume the diodes are all matched, and there is no thermal runaway between the two parallel diodes. 2) In the absence of more detailed information of the LED's IV characteristic, assume it's in the exponential region of it's IV curve (e.g. series resistance isn't yet significant) From 1), we have right away that the top diodes each pass an identical amount of current, and the bottom diode, passes the sum of that, or twice the individual top current: Ib = 2*Ic If we hold Vcc at 4V and the node where all three diodes are connected together at exactly 2V, then from the diode spec given, all three diodes would be passing 20mA. There's an imbalance of 20mA that needs to be provided to make this happen. If you don't provide (sink) that 20mA, the voltage of that node will be pulled to a higher voltage by the two diodes being stronger at pulling the voltage up than the bottom diode trying to hold the voltage down. As we let the intermediate voltage deviate from 2V, the currents respond exponentially. The top diodes will see their current reduce by a factor of k=e^((Vmid-2V)/nVt) where Vt is the thermal voltage (~26mV), but n isn't given. Conversely, the bottom diode will see its current increase by the same factor, assuming it's an identical LED (same colour) to the others. We're not given the wavelength, hence n, so can't figure what Vmid ends up being, but do have that the two top diodes pass 2*Ic=2*20mA/k, while the bottom diode passes 20mA*k, and that these are the same. Taking out the common 20mA factor, we have 2/k=k, or k=sqrt(2). From this, Ic = 20mA/sqrt(2) ~= 14.1mA and Ib ~= 20mA*sqrt(2) * 28.2mA

The column headers on that datasheet are pretty much all screwed up. The first "MIN" is the shorthand name of the parameter. The next "TYP" should probably be "MIN". Then TYP, then MAX, and finally UNIT.

I like you calling it a led (rhymes with dead) Dave! Growing my dad would always say it like that, and I cringe every time I hear L-E-D... It's a led!

You know what's worse? Being penalized for questioning such a stupid question in the margin. Years later I am still being thrown these types of idiotic questions even at work where all answers are wrong and you have no recourse to point this out, however, you have to complete the multiple choice "training" in something not related to my actual job because compliance says so. It is so infuriating.

In school I had a physics teacher that was a physics doctor (don't know if that's the right term) and he would have loved that problem. He wasn't a very good teacher (didn't know how to tell the stuff to the students) but he was really good at math and physics. And he was also interested in questions that were a bit off-topic or only topic-related. So if you didn't understand something or had a question to another topic, he took the time in his break or even after school to calculate the answer or to explain it with really good drawings on the board. If he didn't know something, he told us he didn't know instead of trying to find an easy answer to not "look bad" in front of the class. But he then took the time to look the question up and explain it in detail in the next lesson.

Great video Dave

Oh no, parallel diodes and no current limiting resistor on one side! Thermal runaway, here we come! How can anyone calculate what's going to happen here, or is it that you calculate "ideal" values?

oh dave. You silly man, thinking university teaches anything practical. Its pretty unbelievable I have senior EE undergrads coming to me for an internship and they have never even shopped digikey let alone looked at a datasheet.

I was a professor of electrical / electronic technology for thirteen years while still running my engineering firm. Based on my years in public education, questions like this hardly surprise me and are more a commentary on the competency of the instructor than the student's lack of understanding. It's unfortunate that students are subjected to these kinds of confusion but sadly, it is not that uncommon to find instructors over their heads teaching courses that they have no expertise in, largely because administrators simply need a body at the head of the class to keep the financial wheels turning, irrespective of whether or not the chosen instructor is the best person suited to teach that class. At a time when we need STEM fields to be rigorously taught more than ever, here in North America (Canada and US), we're seeing a rapid decline in the quality of public education at all levels, including colleges and universities. This is attributable in large part to cultural shifts and political forces that no longer value true knowledge or skills and their importance to society. The physics of the universe remains nevertheless immutably complex and we either rise to its level or seek employment elsewhere.

Greetings Dave, Well, this is an interesting problem because practically it doesn't make sense if a constant LED voltage drop model of 2V is used. However, if I were interviewing for an analog INTEGRATED circuit designer position at a semiconductor company, this is a BEAUTIFUL problem. That's because in the integrated world, we quite often put PN junctions in series or parallel. For example, if I were to make one BJT twice as WIDE as another, then it would have TWICE the current handling capability for the SAME VBE voltage drop (does that make sense?). If I asked this question during an interview, I would expect the potential hire to say, "Well obviously, we can't use a simple constant voltage drop model here - we need to lean on William Shockley's Diode Equation and calculate V's and I's more precisely." ME: Ahh...Music to my ears!! Keep going... OK, now, for the circuit on the left... If we sketch the exponential vi characteristic with the given point (2V, 20mA) indicated, it turns out that a LOAD LINE drawn for the linear circuit (4V source and 100 ohm resistor) PASSES EXACTLY through the same point. Therefore, the voltage across this LED is EXACTLY 2V and the current through it is EXACTLY 20mA. Done with that! ME: Great! You've already done better than 90% of the people I've interviewed and I can guarantee you a LEVEL 1 designer's position. Keep going... Shockley's Diode Equation must now be applied to the LEDs on the right to determine unique voltages and currents, but first we must determine three unique constants - the Saturation Current (Is), the process parameter (n), and the Thermal Voltage (Vt). I will assume that n = 5 (typical for many LEDs) and Vt = 25mV at 20 deg C (room temp). Also, given our data point (2V, 20mA) we can determine (Is). I calculate (Is = 2.25nA approx.) ME: Where did you go to school? Do you have friends I can interview? It's worth $5k per hire for you. You're now at LEVEL2 (+$10k)... Keep going... OK, so now we can write two Shockley equations - one for the parallel LEDs (which must have the same voltage across them and current through them) and one equation for the bottom LED that must have twice the current of one of the LEDs above. Taking the ratio of these two currents yields that the difference between the voltage across the parallel LEDs and the bottom LED must be 86.64mV, and solving for the currents yields: 14.14mA through each parallel LED and 28.28mA through the bottom LED. In summary: LED on left: 2V across it , 20mA through it Parallel LEDs: 1.96V across, 14.14mA through each Last LED: 2.04V across, 28.28mA through it ME: BRAVO!! Level3 Designer (+20k per year and a hiring bonus!!)! When can you start ??? Let's go to lunch! Hey Bill, I'd like to introduce you to our new Level3 integrated circuit designer. Would you like to join us?

I built a circuit like that recently with some solid LED's and some flashing LED's. When the flashing ones went on and off, they changed the current and got an interesting flickering type effect on the solid color ones.

GREAT STUFF DAVE! 🤔💡💻 EEVblog 😁

You summed it up in the first two minutes.

ASSUMING LEDs exactly on spec, dropping 2V at 20mA, the current through B is trivially between 20mA and 40mA. Exact value depending on the characteristic curve. Assuming resistors in place of the LEDs for the same voltage/current operating point given, ~26.7mA, giving an upper bound estimate for current on assumptions given. Real world components will be less predictable.