Hi. It's infuriating. Exam question time. This one.

It comes from the Eev blog forum. I'll link it in down below if you want to join the discussion and you should be on the Eev blog forum. Um, this question Uh, comes courtesy of Etmel, Uh, from Sweden? Hi to all my Swedish viewers and I presume Etn is doing some sort of course on electronics and um, this looks like an exam. Some sort of exam question? I don't know.

I don't think they've put more details in here. Anyway, seems pretty simple. It's got four leads in here. Le or Leds.

For those who don't like me calling it leads lead, it's easier. Hello, How would I calculate the current at the different points in this circuit? Now, the only information we're given is that Vcc equals four volts. So we've got a four volt rail up here. We've got uh, the forward voltage of the lead is two volts at 20 milliamps.

so they only give us the forward voltage at the 20 milliamp current which is very typical uh for a red lead and R1 is 100 ohms. So we have to calculate the current through the Led resistor with the resistor dropper here and then the current in one branch of this circuit here, and the uh of one of these leads in two leads in parallel and then then the other lead down here. So we've got, um, a series parallel combination string here. Now, the first time you see this, you just go.

why on earth are they like? This is a completely flawed question and it's infuriating. And if I was, I've had similar actual questions like this in exams and I don't care if I get that Well, I would try and get the right answer that they're expecting. But then I would also put a huge diatribe of how this is like complete and utter full, impractical. Um, so it's one of these differences between uh, theorem, you know, theory and uh, practice of course.

But um, yeah, it doesn't even make sense sort of in theory. Really, it's well, it does. but it's It's dumb. But anyway, actually, I really think this could be a potentially good question for a job interview candidate.

If you're looking for an employee or something like this, throw something incredibly simple question like this and see what the see the response. It's for me, A question like this I would like. rather than just have a multiple choice answer or just you know, have like, just tell me the current I would want like a big description of what's going on here. Sort of like you know, explain the issues with this circuit or something like that.

So I think this could be a really interesting job interview question just to see how people respond to it. Do they just go in and blindly try and calculate stuff? Or do they put their thinking cap on and go, uh, this doesn't really make sense because Xxx x But I'll still try and answer your question. but you know, like if I got this question in a job interview, I'd immediately want to go to the whiteboard. This usually, uh, often you haven't have your job interviews in a, like, a boardroom, uh, kind of thing.

At a company, there's usually a whiteboard in there. Um, yeah, I'd go to the whiteboard and explain. Um, it's a classic little, uh, job interview. Tip there.

If you can, go to the whiteboard, get up, be animated, and start doing stuff. So impressed, they'll highly on the spot anyway. So let's look at the issues here. And there are quite a few uh, responses down here by the ways, but I haven't actually read them all yet.

But anyway, let's think about this. Okay, the first thing is is that you never, ever put a voltage source, presumably a low impedance voltage source. You have to assume that because it's driving 20 milliamps, it's at least a reasonably low impedance, right? Um, that you know you don't put a voltage source directly across a non-linear element like an Led. Here, It's just.

it's just ridiculous. So yeah, you wouldn't do that. Okay, but we are told that the forward voltage is two volts here. Okay, so there's two volts drop across this one and there's two volts drop across these two in parallel because they're in parallel.

so they'll have the same voltage drop. Okay, so this mid point here, um, is going to be like two volts. It's half rail, so it all kind of makes sense. So you know you can sort of brush that the practical aspect of that away and go.

Okay, there's four volts Vcc and we're going to have 20 milliamps. That means the answer. Um, for B here, it must be 20 milliamps. Okay, there must be 20 milliamps flowing through B here.

But then you engage your brain for two seconds and you suddenly realize, well, there's 20 milliamps through here. Kirchhoff's current law says, um, that you know the current entering the nodes must equal the current leaving the nodes. I've done a video on that. I'll link it in, up there, and down below.

If you haven't seen that, Kirchhoff's current law and you'd explain this. If you're in a job interview and this was an exam question, you go. Oh, and Kirchhoff's current law, right? The current. If you've got 20 milliamps like coming out of this node here through this lead, which we know right, is it the voltage is going to be split across these, then you must have 10 milliamps each coming through these leads assuming that they're matched.

of course. And and but here's the problem. Like they only tell us 2 volts at 20 milliamps. They only give us the one data point of course.

But of course a Led is a non. Led is a non-linear element. So let's just go to a data sheet here. This is literally a universal Led, right? A five millimeter universal Love it.

Um, anyway, this so happens to be, you know, typical voltage. two volts there. We go forward voltage at 20 milliamps. So this is bang on.

it. Literally is a universal Led. Okay, but of course you go down here. And here's the characteristic curve.

Of course it is. It's non-linear forward voltage. Sure enough, forward voltage. 2 volts at 20 milliamps there.

But if you've got 10 milliamps, it's instantly. which we know is in both of those legs. right? Must be due to assuming they're matched due to Kirchhoff's current laws. Then you look at it like 1.8 volts.

Maybe 1.85 or something like that. Like it. It's just. it's completely screwed up.

anyway. So the answer for a Um C here is of course 10 milliamps. And of course then the answer for a down here. Then you have to do your lead dropper thing.

So let's go through that calculation. we know there's two volts. Uh, drop across the Led here. Okay, because that's the only information we're given.

so we have to run with that. So that so that means 4 volt rail minus 2 volts means that must be 2 volts across the 100 ohm resistor. And of course 2 volts across 100 Ohms is 20 milliamps A is 20 milliamps B is 20 milliamps C must be 10 milliamps. I mean that's the answers that they're obviously after.

But yeah, I would just write a whole like I'd really go to town. I wouldn't even care if I got this answer wrong. Damn it, I'd I would want to give him hell in the uh. comments.

You know I. So Anyway, let's see what some other people here say. It's just a dumb question. It doesn't take a practicality into account at all.

Um, because you know Leds are non-linear devices. Um, that's just like we. We know this. That's actually information that we're giving where we're actually.

even though we're giving given these figures up here, we are given the schematic symbol for an Led and we know that every Led is going to have a, you know, a non-linear response like this. So it just it. It just makes no sense. It's dumb so you could build that up and I think somebody uh did down below and you could see the results.

but the results are going to be on the type of Um Led that you have. Okay, uh, Copernicus says, uh, you don't have to think the exact numbers when you do this. I think that people do too much math and not enough real thinking when they do electronics. This circuit is so basic, why bother calculating any number at all? But Copernicus is saying so.

The right is definitely a lot brighter than the left because it has a resistor in series. But no, there's actually 20 milliamps flowing through. um, this one as well. And of course, the brightness is not a function of the voltage as a function of the current.

There diodes are essentially including Leds light emitting diode it's in the name is essentially a Um a current driven device. Uh, Basically it just the voltage drop happens to be a result of Uh driving it at a particular current. But they're not voltage driven devices. That's not their intention.

which is why it's stupid to not have a current limiting resistor or some sort of constant current circuit driving an Led. Don't do it. But then there's people who say, oh, those little keychain flasher things. They put the Led directly across the battery and well, batteries have internal resistance, so you do effectively have a serious resistance in there.

So here comes the tricky bit right. which are not one X or one uh picks up. Uh, So you would get Uh 20 milliamps through the left branch with with the resistor of course, but he's saying you would get something more than 20 ambulance through the right one. Seen as that there are two leads in parallel.

That means less current, less foliage drop with real components, etc you'd get on the left side and larger variations of burnt leads on the right. Well see. here's the thing right. If you rigidly stick to that Vf 20 milliamps, you'd have 20 milliamps through here.

20 milliamps through here. and then you'd have 40 milliamps. Kirchhoff's climate Current Law must hold. you'd have 40 milliamps flowing through this one down here.

But then, and like in the non-linear nature up here, 40 milliamps? You're up to, you know, 10, 20, 30, 40, right? You're up to like 2.2 volts per lead. What's that? just? no. It's dumb and marush here. if I'm pronouncing that correctly.

Um, so if we assume an ideal circuit, the leads are fully open at 2 volts and the wire has no resistance, then those Leds would simply blow up because there's nothing to limit current going through them. Well, you know that's another sort of like theoretical question. It's like, oh god, I hate this question. This is dumb.

But here's the Uh point. As Jay Melson uh, points out, the current in the right branch is uncontrolled. Very slight variations in Vcc are Led forward voltage records. Huge changes in lead current.

That's why all practical lead circuits use a series resistor or a current source. Yes, the current. But as a homework exercise, the current in the right branch cannot be calculated unless you assume the leads draw exactly 20 milliamps at 2 volts. It is a pathological circuit and should not even be given as a homework question.

It's pathological. Love it. Even with that assumption, it doesn't work unless they assume Vs is absolutely constant from 10 to 20 milliamps, which is not and it's just Ag6qr says, uh, the two leads in parallel on top half the right side must each be getting a portion of the current that goes into the bottom lane on that side, so they can't possibly all three be conducting 20 milliamps at 2 volts. That's the thing.

But because this is like a theoretical question, you have to just make the dumbass impractical assumption that, well, Kirchhoff. Well, I would I would say above everything else, you have to assume that Kirchhoff's current law holds. If you're going to learn basic electronics like this, you've probably already. Oh, would you have been taught? You'd probably have been taught Kirchoff's current law at this point.

And well, essentially, based on my Um. Dc circuit, fundamentals are series. Yeah, this would be. you know.

It's one of the first things that you actually, uh, learn. So you know it. it must hold. And you've been stated that the forward voltage is 2 volts at 20 milliamps.

So you have to assume that You know. Look, because these are in parallel. So you have to assume it's just one right? in terms of like a voltage divider thing. So you have to assume that it's two volts.

Two volts. You'll have to assume two volts at that midpoint junction. There, It's just. it's just something that you, even though it's infuriating, you have to assume that.

And once you assume that, and you assume that Kirchhoff's current laws hold, you've been told 2 volts at 20 milliamps. Therefore, the current must evenly split. And because you're making these ideal assumptions, Yeah, C up here must be 10 milliamps. But it's infuriating So I would never write down just the numbers for this.

um, answer here. I don't know. No. Rebel against the system you've got.

You've got to write down an explanation of why you made various assumptions and things like that. and I would never say somebody is wrong or one's and if they explain why they came up their reason, I would never say that they're actually wrong unless they were unless the answer from their assumptions was actually, um, wrong. But yeah, that's this is why I think this makes a really good interview question because it it provokes, you know, heated discussion, right? And it shows what the person knows and they're You know, it shows that they know Kirchhoff's current law. It shows that they're willing to put their thinking cap on.

and no, no, this, this can't be right, You know, and and how to analyze the situation. Um, it it. It might be deliberately designed like this. I don't know if the question actually has like a a text box.

you know, explain your answer. Um, as a lot of questions are and geez, I'd need a whole page to. you know, I'd really let them have it anyway. X Run has got it all.

Bread boarded up. Yep. the right side is not acting predictable as suspected because the two leads are not identical. exactly.

They're not matched. However, the left side pretty much, uh, matches calculations. Yes, it will because you've got a low impedance current source. Your power supply is a very low impedance output.

It's like millions, right? It's it's really small, so it's going to force unless you've got the current limiter. Uh, you know. set. Um, then it's going to it.

It's going to force 4 volts to this point. And these diodes on the right hand side. They've just got to sort themselves out. The poor bastards, but this one over here, nicely designed with this current limiting resistor.

Then yes, it's going to get the proper 20 milliamp treatment there, assuming you've got the exact lead that you know gives you that exact value. But even then you're going to have temperature and process, uh, variations, manufacturing variations over here. This is why over here in the data sheet. look forward.

Voltage: 20 milliamps, right? It's right. And can you see that? Yeah, there you go. It's typical of: sorry, um, yeah. no maximum of two.

Unit Three: What's wrong with it? Why is that Unit Three? What does that mean? Beulah Mueller, Mueller? Um, I don't get that. So here we go. X Run has gone to town here. Hats off.

Uh. set the input voltage to four volts at the breadboard measure. Resistance of R1 was 98.7 good enough for Australia Path A measure. Current was 70 milliamps of Vf or 2.14 and you could like you could like go in there and like trim things and stuff.

But yeah, the uh, left hand side branch doesn't uh matter because it's it's gonna do its own thing. It's the right hand, uh path that you're gonna, um, worry about. So here we go. Path B current was 10 milliamps.

There you go. So it did actually split and Vf of uh Lead B was 2.1 volts. So it did actually share the current because Kirchhoff's current laws got A. You know, though, those Leds will sort themselves out.

Now here's where it differs in practice: Path B current was 10 milliamps and Vf across. Lead B was 2.1 volts. and if we go up to the top, this is actually 10 milliamps through here at 2.1 volts. So if we've got 2.1 volts across here, we're going to have like 1.9 volts left across here like this.

So you know these things have sorted themselves out and they've determined they've talked among themselves and going right. We're going to have 10 milliamps through here. So the current assuming these leads was like from the same batch and recently split. So I uh, expect C to be roughly half of that or 5 ish milliamps.

And sure enough, Path Uh C was 1.93 volts. The current through each of the Leds in parallel was very touchy. I got from two to four milliamps depending on the stability of the connection I made with the probes or tapping on the desk. In fact, any slight push and evil lead result in both Led variations of light output, yet due to slight resistance changes in the breadboard connection.

Um, so this is all sorts of stuff that you could talk about in your job interview. Uh, you know, if you got given this question or something like that, you can really go to town on the whiteboard like you know talking about this sort of stuff In in other words, it's very unstable as as was, uh, suspected. But that's only because of the physical breadboard. and uh, stuff like that.

Um, wait, if you actually soldered them, Yeah, you'd So yeah, I think you'd actually have to have multiple meters in there. You would have to have like three different meters in there, all measuring that. But suffice to say, and like if if we had this over here was branch D, if we had a meter there, measuring branch D, current C plus D would always equal B even if they're you know, changing over time due to the connections or thermal and the leads are just going like this. Um, yeah, it's at any instant in time the current C plus the current D will equal the current B.

Kirchhoff's current law must hold. and then Tomarowski here has gone to town. Given the pathological nature, I love that term nature of the question, along with the limited amount of data, we can approach it with a certain ideal conditions. All diodes are exactly the same temperature.

All diodes are identical. The ideal diode equation applies going into the diode equation. Now, so we're getting into the physics side of things. Uh, look up the various terms, don't.

Yeah, I'm Look, this video is not going to go into details and I I haven't gone through his calculations down here, so I got you know I did. Look, I I don't know. But yeah, you can treat it sort of more mathematical like this. And um, it hasn't given a summary like that.

So anyway, I'll leave it up there. There we go. There's a, Um homework for those. uh, playing along at home to, uh, give a reply.

Go go to the Forum and give a reply to uh tomorrow? Yeah, um. great work. I think everyone is trying to add reality to a homework problem. No issues with the well, it's not.

You've got to add reality because the question itself is completely dubious. Um, because it's telling you it's a physical lead, right? It's an early use as the symbol and that just tells you right away. This is a practical circuit problem and it's just yeah. It's just infuriating.

So yeah, our staffer basically comes to the same conclusion: I did that: Current A is 20 milliamps current B is 20 milliamps current C is it must be half 10 milliamps due to Kirchhoff's current law. No, if no other information was presented, I think the ideal component is all you can use how you justify the 10 milliamps across the parallel Leds. And yes, the 10 milliamps is correct. Whether that would actually produce 2 volts isn't given in the problem statement.

So yes, So yeah, I put my response down here. and then I just went, ah, no bugger and I'm shooting this video. So there you go, one of those stupid theoretical questions, and if you ever see something like this in the exam, let them have it. Damn it.

Do not stand for these impractical problems in these exams. But of course it's obvious what answer they're after. But yeah, I really hope a question like this would have a please explain your answer uh box I would never like. Seriously, that's a serious fail of a lecturer.

If they produced this as an exam question and did not give you a please explain your answer box then that is a failure on their part and you should tell them that, um, seriously might get thrown out. but you know that's worth it. But I'll say if they did get give that, please explain your answer box then you know hat tip because this is. this is an interesting one.

And as I said, great job Interview question because it just promotes discussion and and you can really extract a lot out of an interviewee if you, um, you know, actually provoke them with a question like this. So anyway, I hope you found that interesting If you did leave your thoughts and comments down below and I'll link to the Euv blog forum. If you're not on the Eev blog forum, join in the discussion down below. That's where everyone hangs out and uh, we can see who's there at the moment.

There you go. X Runner Amy K Me: um Rfx asked offer Enters: Seven guests are viewing this topic probably a lot more after this video, but there you go. I I thought I'd share. thank you very much Etna.

So Edna has not, uh, actually replied to this uh yet? but ten posts so you know not. Uh, not a complete newbie. so I assume Edna will come back and join back. Join the discussion.

But yeah, please join the discussion in this one, let us know how you would solve this. and if you think this is an insulting question without having to please explain box because I think it is, I think this question. yep, it's just anyway. Thoughts and comments down below catch you next time.

Oh, all right, I know people aren't going to be happy unless they actually build this thing up. So I've actually managed to find after some searching um, four red Leds that are pretty damn bang on to 2 volts at 20 milliamps and trust me, it wasn't easy to find, but I found them anyway. Um, so I'm using my Um external Keithley current source here, so I've set that to um, you know, pretty close to 20 milliamps. It's only three digit adjustment unfortunately, but you know that's good enough for Australia, right? 20 milliamps? Let's uh, measure each lead.

uh, the voltage drop across each Led at 20 milliamps to make sure they're reasonably matched. And wouldn't you know it, I've actually blown one. Um, I must hooked it up backwards. and then my compliant.

I thought my compliance voltage was as low as possible was down at 10 volts. but still. Dawn? what? Anyway, Lead Number One: 1.98 Not too shabby. There you go.

It's dropping. She's actually, uh, warming up a bit. Lead Number Two: 1.94 volts. Not much, but you know, good enough for Australia.

Number Three: 1.99 volts. There you go up. No. 1.98 right? So anyway, I'm going to use those three matched and this one, um, is the dyed.

I don't know if there's a physical fire or whatnot, but um, look, I'll go find another red lead and I'll use that in the uh. left-hand circuit because it's it's not going to matter. I mean, the whole point is what happens on that right-hand current circuit with the three Leds in series parallel? Look at that. I was able to find another one.

2.02 That's not too shabby. Five millimeter jobby. Okay, so I've just got the single resistor on the Uh left hand side of the circuit. so that's our four volts input.

There you go and across the resistor. This is bang on 100 Ohms. so 1.96 so I know 19.6 milliamps is flowing through there. Ohm's law.

I don't even need to get the current meter out for that. So here you go. I'm now going to move it over to put the other Leds in series. Let's see what happens.

I'm going to apply Four volts. Tada. Yep, they all come on. Okay, so I've actually set 50 milliamp current limit on my power supply and they've all come on.

I'll put it right there. Yeah, we still got four. There we go. Four volts.

now. let's see what we get across. the center tap of this string here. Will it be two volts? I don't know.

There we go. that's not too. Wow, that's that's pretty close to bang on. Isn't it interesting? There you go.

So let's measure the top one. of course we must have 1.98 across there. Well, 1.9 Well, yeah, you can see yeah, like it's It's varying a little bit and that's going to have to do with the dickey connections on the breadboard so it's going to be jumping, jumping a bit all over the shop there. But as you can see, uh, it? yeah.

we've got two volts across there. How much current's actually flowing through down there? Well, we'll have to measure that now of course. Uh, burden voltage of your multimeter could get out the microcurrent stuff. couldn't be bothered.

So what I'll do is I'll just whack it on amps range and we'll put it on amps jack here and that will give us, um, a 0.1 milliamps resolution. But really like quite low burden voltage. Okay, yeah, you can see those Leds flicker when I actually move that resistor in there. I don't Well, I, I think you can.

Yeah, that's just like maybe dirty, crappy contacts on the on the resistor, Led and or breadboard. But anyway, what I'm gonna do is put in there ready. I'll turn it on and let's see what we get tonight. 24 milliamps at 25 milliamps 28.

30 more. There you go. So we are actually getting in this particular case. we are getting uh, higher current.

It's backwards. all the electrons are falling out the wrong way. But there we go. All right now.

now it's down to 25.. there you go, 26. But it is certainly higher than the other branch. So there you go.

That's interesting. Just give that a few wiggles in there just to clean those contacts. This is an old resistor from my junk bin, so you know, bear with me. Okay, so what I'm going to do now is this resistor.

I don't know if it's left or right. Doesn't matter. Take your pick and there you go. It is half.

It's roughly half. We've got our 10 milliamps flowing through the lid. There you go. Winner winner chicken dinner.

So because these three leads are all reasonably matched, then they're going to roughly share the current. But of course, that our total current through there you saw that was higher could be up to 30 milliamps something like that. But definitely when you put two in parallel like this. Um, and they're reasonably matched, then you're going to get a reasonable uh, 50 50 split in the current between there.

So, and of course, the remainder of the current must be in the other Led, so I don't need to. I could put multiple, like meters and stuff. but then you have to like, get like, you know, like solder them into proper banana jacks so that you've got, you know, proper cabling and everything set up. And it's all like, I couldn't be bothered.

That just shows that the current does split between those two, uh, upper ones. and, uh, the combined current. Kirchhoff's current law. That one, even though I can't really see any difference in the brightness between these two and this, it's not just the camera, it's really hard to, uh, actually discern that difference.

And this is a thing. Even though this has more current or twice the current as these two, here, they appear similar apparent brightness. Uh, to the eye. That's just the trick on the eye, so you know there's no need to run Um leads.

At like 20 milliamps, you can easily half your current there. And if anything, those two actually look a bit brighter than that one. I don't know, it's just because I wasn't looking straight on. But yeah.

anyway, yeah, that's just a uh function of our eyes. really. If you actually got proper, uh, measurement instruments that were measuring the, uh, milly candelas or the lumens out of this thing, then you'll find that, uh, yeah, the brightness would be proportional to the current because that's pretty linear. Having said that, diodes and leads are non-linear devices.

They are in terms of voltage drop. But in terms of output, Uh, lumens output per current per milliamp through them? Uh, then that's a pretty linear relationship for the most part. So there you go. I hope you enjoyed that bit of pointless fun.

If you found it interesting though, give it a thumbs up. And as always, discuss down below, catch you next time you.