Calculate Current

[Logaritmo]

   My practical theoretical guess is ..


      From the graph If x Vf mentioned by Dave in the video, a lookup table was produced, and
      using successive approximation when the current in B must be the double of the current in C and
      the voltage over the lower LED plus the voltage over the higher LEDs must be 4 v,
      the final result was:

                   
                   I B = 14.5301 mA ( V LED = 1.9168 v )
                   I C = 29.0603 mA ( V LED = 2.0832 v )

      

[SiliconWizard]

That's fun.
Of course there are different levels of answers, depending on how realistic you expect them to be.

For the first branch, iA = 20 mA. If we assume that Vf = 2 V@20 mA, then that's the equilibrium it'll reach. That'll be true neglecting other factors such as temperature and assuming Vf = exactly 2 V @exactly 20 mA, also assuming Vcc comes from an ideal voltage source.

(Vcc - Vf)/R1= iA

Of course the tricky part is with the right branch.

Now, let's see for the case of ideal diodes first (so let's assume here Vf = 2 V for any current, and all LEDs having the exact same Vf). The right leg is going to be the equivalent of 2 diodes in series, with an equivalent Vf of 4 V, which is exactly Vcc. Now even in this ideal diode case, would the right leg actually draw any current? Or would it draw infinite current? It's already a trick question IMO. In any case, I do not quite agree with the 10 mA/20 mA answer, as that would assume the diodes are actually NOT ideal. If they have an ideal characteristic and there is no limiting resistance in series, then the current will be either zero, or infinite. Ideal diodes do not have current limits. At least, we should clearly agree on a definition of ideal diode here. Is that an I-V characteristic with an infinite slope at the "threshold" (which is what "ideal diode" means to me), or is that something approximating the Shockley equation, like a piecewise linear approximation?

One of the issues with this question is that admittedly, the fact that we should consider "ideal diodes" is made unclear from the part which states "LED: Vf = 2v, 20mA", implying that Vf could depend on current, and thus the diode would not be ideal. Unless the question was explicitely asked to trigger an elaborate answer, this is indeed flawed.

And now if we consider the non-ideal case, that still isn't likely to be iB=20 mA and iC=10 mA unless you got very lucky with the parts you have. We'll have to consider the real I-V characteristic - with Vf increasing with increasing current. Now we can say for sure that iC ~ iB/2 if the two LEDs in parallel are close enough in characteristics. As the latter will pass about half the current of the bottom right's LED, assuming all 3 are close enough in characteristics, that means the two LEDs in parallel will drop a lower voltage than the bottom right LED. For Vcc so close to twice the typical Vf, actual results will vary widely depending on the LEDs and temperature. One thing for sure is that as we increase Vcc, the bottom right LED will pass more current than the left LED. The opposite will happen as we decrease Vcc.

[xrunner]

etnel has kindly replied to my PM and explained where he encountered the problem. He has replied to me in a PM so I can't reveal it myself. He has said he will reply tomorrow in the thread. I think you will find it interesting.

[TrickyNekro]

etnel has kindly replied to my PM and explained where he encountered the problem. He has replied to me in a PM so I can't reveal it myself. He has said he will reply tomorrow in the thread. I think you will find it interesting.

Crazy part of feedback loop or something? Cause something tells me it´s not just an exercise....

[RDP]

OK I just simulated this in an online simulator and this is what I got   

Have I done something wrong?

RDP

[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.

[EEVblog]

OK I just simulated this in an online simulator and this is what I got   

Have I done something wrong?

Yes, you used a simulator 
Sorry, it was the obvious joke.
Welcome to the forum.
The simulator is going to go crazy with this depending on the diode model used.

[RDP]

Thanks Dave.... I suspected as much. I'm on holidays in Adelaide at the moment so don't have access to my bench equipment so I thought I'd just try out a simulator to see what it would produce. Obviously its way off even with the Standard default Diodes 2.2v that I used....

[amyk]

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.

[andy3055]

I just watched the video on Youtube while having my lunch! When i first saw this a couple of days back, I thought it was a "trick question" like the Yanks say here.

[DavidAlfa]

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

[thm_w]

People going wild over connecting LEDs to a voltage source or paralleling them in the youtube comments. Try playing with some LEDs and a power supply, or low voltage COB LEDs (bigclive). Vf matched LEDs can be paralleled and share current. LEDs can be driven solely via a voltage source and not explode.

Should it be done in production? Best practice is not to do it. Doesn't mean its not possible or there are no applications for it.

Compared two red COB LED arrays and they were matched within 3mV. I don't feel like destroying them right now, but, at the individual die level I suspect they could be matched to <1mV.
Also found some Vf binned SMD LEDs, but they offer nothing close to that level of matching.

[lacek]

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

Not at all. It nowhere says this problem is self-consistent in the sense of listing all formulas you are allowed to use. Student that got this problem, likely attended some lecture, which was full of other formulas unlisted here. In this case some "ideal diode" equation is useful, like that one:  I=I_0 (exp(eU/2kT)-1)   20ma@2V allows to find i_0, assuming room temperature T.

You also do not know if this is a closed form problem or an open one, which allows for presenting reasoning not simply the answer. To put bluntly what under that problem:  a box where you type three numbers, or half of empty page where you show your solution?

For super-realism: you can take any U(i) curve you like and solve the equation. You can also write that the current is I0,  such that U(I0) + U(2*I0)  = 4V   and the exact details depend on U(I) curve. which can also factor in junction temperature (which also depends only on I0)

[lacek]

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

[etnel]

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.

[not1xor1]

I know with 2 leds in series, you need 4 volts to power it.   I swear the initial post was 2 volts in,  and he changed it!!!

If the supply voltage were just 2V a few mA would go through the left LED while a few nA would go through the LEDs on the right one, more than that depending on temperature higher than 300K and environment light.

[xrunner]

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.

That's what I was thinking all along. This wasn't some "physics introduction to electronics" advanced LED question and the student had been taught more theoretical equations than we had been told about in the initial post. Some teacher thought they were going to get cute and draw something apparently simple on the surface but as you understood (good for you!) it's not a simple sort of thing to understand and calculate properly without more knowledge. However, it is simple to actually breadboard and prove that more advance calculations are needed. LEDs are NOT like resistors and act very differently given different applied voltages.

Again good for you to understand this. 

[StillTrying]

It just goes to show that a little knowledge is dangerous, especially if you're a teacher.

Of course we didn't do electronics at school, it was too long ago, but we did do science and physics.

One day the part-time physics teacher drew that on the board to demonstrate how a simple diode works. Me and a mate just looked at each other, we just wanted to get home to continue working on our shed DIY X-ray machine.

Everyone else wired the 4 D cells and the diode that way , and after 30 minutes the teacher still hadn't figured out why it didn't work, so it went down as a failed experiment, which was ideal as far as I was concerned because there was no write up to be done.

[xrunner]

Everyone else wired the 4 D cells and the diode that way , and after 30 minutes the teacher still hadn't figured out why it didn't work, so it went down as a failed experiment, which was ideal as far as I was concerned because there was no write up to be done.

LOL. 

Yea ... but you know it actually did work perfectly. It just didn't match the teacher's expectations.  In a sense every circuit ever made works "perfectly", every single one performs exactly according to the laws of physics - but it may not be doing what the human being thought it would do. 

[Zucca]

1. Given the "pathological" nature of the question, along with the limited amount of data provided, we can approach it with certain ideal conditions:
1.1. All diodes are exactly the same temperature.
1.2. All diodes are identical.
1.3. The ideal diode equation applies: i = I_Se^(v/nV_T)
1.4. I leave it to the student to look up the various terms, derivation, etc.
....

This is the correct math/physics answer.

I would like to give a graphical view about this correct answer.
The branch Ia with R1 is easy and just to determine if you should pass the exam or not, I will not dig into here.

To have an idea on what happen to Ib and Ic it is better to split the second branch in two circuits:
- V1, I1: 4VDC with a diode on the ground (VCC - Vd)
- V2, I2: two diodes in parallel (basically one new diode with Vf=2V and If=40mA)



Those are the resulting graphs:



This branch is working at the point where the red line and the green line are crossing.

Without knowing any details about the diodes, all we can say is:

20mA < Ib < 40mA

since Ib = 2*Ic

10mA < Ic < 20mA

[Zucca]

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.

easy if you think about V,I graph like I did above.

[xrunner]

easy if you think about V,I graph like I did above.

Yes and the bad thing is the "teacher" etnel had was clueless about it -

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.

[Zucca]

Yes and the bad thing is the "teacher" etnel had was clueless about it -

Regardless, it is a very good circuit to solve. In my electronic one exam at the university of Milan (third year of master in EE) they asked me to solve a circuit with a base floating BJT.
I looked up the professor and he looked at me with a smile: "If you want all the points you need to answer this".

Not practical circuit on paper are very good to test if you know electronics.

[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.

[Zucca]

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.

....

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,

....

ideal model with an infinite slope past Vf...

If we assume a ideal diode as you describe above the solution for Ib and Ic is
Ib = 2*Ic
for any possible value of Ic >=0.

I agree with Dave, 20mA@2V is given so it cannot be an ideal diode.