Calculate Current

[BeBuLamar]

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.

But the question states 2V drop at 20mA, so given that's the only data you have, you have to assume 20mA is flowing.

As 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.

The 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.

[EEVblog]

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.

But the question states 2V drop at 20mA, so given that's the only data you have, you have to assume 20mA is flowing.

As 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.

Well 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.

[EEVblog]

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.

Thanks 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.

[Knith]

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.

[Vovk_Z]

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.

[oolonthegreat]

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 ?

[xrunner]

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.

[rstofer]

Nice lab setup!

[gcewing]

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.

An 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.

[gcewing]

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.

What 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.

[gcewing]

Our teacher seems to have lacking knowledge regarding LEDs and current in a series circuit. By his means, the LED drops voltage AND current.

So what did the teacher think the answer was? And were you able to educate him?

[RBBlackstone]

Since “A” in a circle usually denotes an ammeter,
 I would need the ammeter specs, too.  Yikes!

[rs20]

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 I_D = 20mA  -->  V_F = 2V
And I_D = 20mA / sqrt(2)  -->  V_F = 2V - x    (for some value x)
Then I_D = 20mA * sqrt(2) --> V_F = 2V + x

Because 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 - x
We theorize that the bottom two diodes has 20mA * sqrt(2) and drops 2V + x

KCL is satisfied because sqrt(2) * sqrt(2) = 2
2V + x + 2V - x = 4V, so 4V @ vcc is satisfied

So we're done. Top diodes are 20mA / sqrt(2) = 14.14mA  each
Bottom 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.

[Logaritmo]

   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.

[xrunner]

Thats why I believe the V/I curve is the only possible usable source to find the correct answer.

Right. 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.

6.9. i_B = 28.28 mA
6.10. i_C = 14.14 mA.

If 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 mA
Branch C2 (right side) = 5.52 mA (1.97 V). Graph says ~ 5.9 mA
Branch B = 11.23 mA (2.1 V). Graph says ~  12 mA

[rs20]

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!

[timenutgoblin]

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.

[hans]

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.

I 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.

Nice one My 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

I 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.

[SiliconWizard]

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.

But the question states 2V drop at 20mA, so given that's the only data you have, you have to assume 20mA is flowing.

As 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.

Well 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.

Yes, as we discussed.

But if it's seriously a (poor) legitimate question, then it becomes the 20mA/10mA answer. I suspect it's the later.

Well, 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.

[Vincenzo]

Assuming that the leds follow the model I=I₀(e^V/V_T - 1) and V_T and I₀ is 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=V_Tln((.25+2e^V_cc/V_T)^.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 ]

[GrandTheftAuto4life]

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, 50Hz
1uF capacitors, 300 ohm resistors

I 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)

[rs20]

@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.

[Vincenzo]

@GrandTheftAuto4life & @rs20
even 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/#msg3602488

In 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.

[free_electron]

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.

[EEVblog]

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.

It'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