Calculate Current

[etnel]

Hello, how would I calculate the current at the different points in this circuit?

Vcc = 4v
LED: Vf = 2v, 20mA
R1 = 100 ohms

[Capernicus]

You dont have to think in exact numbers when you do this... I think that people do too much maths and not enough real thinking when they do electronics.

This circuit is so basic, why bother calculating any number at all.

So you can see we have 1 led on the left with a resistor,  and 3 leds on the right, 2 in parallel 2 in series, with no resistance,   so the right is definitely alot brighter than the left, and if you used a small enough battery then you wouldnt be able to even see anything on the led with a resistor, regardless of the resistance.

Thats worth more any amount of pages of confusing mathematics, and gets it over and done quicker, and you dont have to worry about pesky details that dont matter.

There is more to conclude here and know than just that,  like the led voltage drop, you need to be able to handle that as well...  but that probably can be "read" as some approximate laws and the exact numbers of things isnt so important,  the general concepts are way more important.

[not1xor1]

Hello, how would I calculate the current at the different points in this circuit?

Vcc = 4v
LED: Vf = 2v, 20mA
R1 = 100 ohms

Supposing ideal (identical) LEDs you wold get 2V/100Ω = 20mA through the left branch and something more than 20mA through the right one (since there are 2 LEDs in parallel and that means less current and less voltage drop).
With real components you might get 15-25mA on the left side and larger variations or even burnt LEDs on the right side 
LEDs should always be connected in series to a constant current source/sink or resistor.

[bdunham7]

You dont have to think in exact numbers when you do this... I think that people too much maths and not enough real thinking when they do electronics.

Actually, in physics and EE the rule is that if you can't do the math, then you don't really understand it.  But...

This circuit is so basic, why bother calculating any number at all.

Probably because it is a homework problem or something like that.  But it is a very bad example of a problem because although perhaps you can 'calculate' currents, your calculations won't have any meaning in the real world the circuit design fails to accommodate the non-ideality of the components.  And they give V_f of the LEDS only for 20mA, so it can be argued that there is not enough information to solve the problem.  So,  you're right in this instance--the circuit is so flawed that there's no point in calculating anything. 

[mariush]

You can split that circuit in the middle.

On the left side you have the led and you have the resistor.

Voltage = current x resistance  according to ohm's law

So input voltage - (number of leds x forward voltage)  = current x resistance

4v (vcc) - (1 x 2v ) = current x 100 = > current = (4-2)/100 = 2/100 = 0.02A

on the right branch, you have two leds in parallel, then one in series with the two in parallel.
So your current would be input voltage - (1 group of leds in parallel x forward voltage)  - (1 led in series x forward voltage) = current x resistance
If you had a resistor to limit current on this branch, the section of circuit with 2 leds in parallel will have same current as the single led experiences, so the two leds in parallel will have half the current.

As it is ... you have equivalent of 2 leds in series, so around 4v voltage drop and you don't have any resistance, unless you start to account for the wire resistance as current limiter.
So if we assume ideal circuit where the leds are fully open at 2v and the wire has no resistance, then those leds would simply blow up because there's nothing to limit current going through them.

[Terry Bites]

If you're not making circuits before you know the maths you're going to end up a tedious accademic. True.

[Capernicus]

If you know whats going to happen,  the exact mathematical intricacies didnt stop you from guessing it.  But on the other hand its easy to get it wrong so I do still believe in alot of experimentation.

[xrunner]

Someone should breadboard it and check what really happens ...

Me right? 

[jmelson]

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

[Capernicus]

anything above 2 will light up the leds,   below that they cut out because of the voltage drop.  its 4 volts for 2 in a row.  Xrunner - test it out if we are right!!!!!

If its true, all you need is 120 leds and you can have a full 240v connection and the leds are fine with a 100 ohm resistor.

[mariush]

Someone should breadboard it and check what really happens ...

Me right? 

Sadly they will probably use a couple CR2032 batteries or a 9v battery (and a linear regulator) to power the circuit and you wouldn't get the expected results... because they won't consider the internal resistance of the battery as a current limiter.

So, it may not be quite the desired lesson.

[xrunner]

Sadly they will probably use a couple CR2032 batteries or a 9v battery (and a linear regulator) to power the circuit and you wouldn't get the expected results... because they won't consider the internal resistance of the battery as a current limiter.

So, it may not be quite the desired lesson.

I'll do it later I think I have time. The LED Vf won't be 2.0 V I'll have to use red, more like 2.1 or 2.2. 

That's a textbook problem someone made up as we all realize. I'll be back later ...

[bdunham7]

cannot be calculated, unless you assume the LEDs draw exactly 20 mA at 2.000 Volts.  This is a pathological circuit, and should not even be given as a homework problem.

Even with that assumption it doesn't work unless they assume V_f is absolutely constant from 10mA to 20mA.

[AG6QR]

The current in the right branch is uncontrolled.  VERY SLIGHT variations in Vcc or LED forward drop will cause HUGE changes in LED current.  That's why all practical LED circuits use a series resistor or a current source.  Anyway, as a homework exercise, the current in the right branch cannot be calculated, unless you assume the LEDs draw exactly 20 mA at 2.000 Volts.  This is a pathological circuit, and should not even be given as a homework problem.

Agreed, except that the right branch is ill-defined even if you DO assume the LEDs draw exactly 20mA at 2.000 Volts.

The two LEDs in parallel in the top half of the right side each must each be getting a portion of the current that goes into the bottom LED on that side.  So they can't possibly all three be conducting 20mA at 2.000V!

In a simplified classroom model of an LED, you might be able to assume that vF is 2.0000V, completely independent of current, but that's going to be a false assumption if you ever try to test it in the real world.  If you make that false assumption anyway, you'll find that the bottom LED conducts 20mA at 2.000V, so each of the top LEDs would conduct 10mA, if you assume they're completely identical.  The assumption that LEDs in parallel split the current symmetrically is another one that won't hold up to real-world experimentation, but might be an assumption that's used in a theoretical classroom exercise.

So you get an answer where the voltage across each of the top two LEDs on the right side is equal to the voltage across the bottom LED, despite the fact that the bottom LED is carrying twice the current.  That's preposterous, but it follows from the preposterous assumptions made.

The circuit won't work in practice.

[xrunner]

Agreed, except that the right branch is ill-defined even if you DO assume the LEDs draw exactly 20mA at 2.000 Volts.

The two LEDs in parallel in the top half of the right side each must each be getting a portion of the current that goes into the bottom LED on that side.  So they can't possibly all three be conducting 20mA at 2.000V!

In a simplified classroom model of an LED, you might be able to assume that vF is 2.0000V, completely independent of current, but that's going to be a false assumption if you ever try to test it in the real world.  If you make that false assumption anyway, you'll find that the bottom LED conducts 20mA at 2.000V, so each of the top LEDs would conduct 10mA, if you assume they're completely identical.  The assumption that LEDs in parallel split the current symmetrically is another one that won't hold up to real-world experimentation, but might be an assumption that's used in a theoretical classroom exercise.

So you get an answer where the voltage across each of the top two LEDs on the right side is equal to the voltage across the bottom LED, despite the fact that the bottom LED is carrying twice the current.  That's preposterous, but it follows from the preposterous assumptions made.

The circuit won't work in practice.

I got it all breadboarded up. Yep, the right side is not acting predictable as suspected because the two LEDs are not identical. However the left side pretty much matches calculations. Will be back with more results. It's a textbook problem that can't really be solved without more information.

[CaptDon]

Xrunner, that theoretical problem goes against every practical design
of a proper LED circuit and sets such a poor example of a real world
practical circuit. If your teacher laid that out as a learning example
he/she is an idiot and I would hate to even think that circuit was published
in a training textbook. If that was in a published book no wonder all
of my recent engineering applicants leave me wondering what they
actually taught in these engineering classes. The answers I got to
basic questions and examples on our entrance exam tests make me
only want to hire engineers over the age of 60. If this was a textbook
example circuit it would even be hard to use the old cliche "Book smart
but real world stupid". Anyway, best luck in your studies and hope you
have a rewarding career path! I dearly miss my old mentors many of
which learned electronics during WWII and then worked at places like
Raytheon, Westinghouse, GE, Martin Marietta and others. I was very
lucky to have been mentored by them and did very well in my career
path from knowledge gained from the old school guys!!!
 

[xrunner]

Xrunner, that theoretical problem goes against every practical design
of a proper LED circuit and sets such a poor example of a real world
practical circuit. If your teacher laid that out as a learning example
he/she is an idiot and I would hate to even think that circuit was published
in a training textbook.  ... Anyway, best luck in your studies and hope you
have a rewarding career path!

I've already had a career. 

It's not my teacher or problem it's the OP's (etnel) 

[xrunner]

Refer to the diagram.

I set the input voltage to be 4.00 V at the breadboard. The measured resistance of R1 was 98.7 ohms.

Path A measured current was 17.1 mA and Vf of the LED was 2.14 V. Calculated current would be 18.5 mA. Reason for a little lower measured current is probably resistance in the breadboard clip connections.

Path B current was 10.5 mA and Vf of LED B was 2.1 V

Path C Vf (across both LEDs) was 1.93 V. The current measured through each of the LEDs in parallel was very touchy. I got from 2 to 4 mA depending on the stability of the connection I made with the probes or tapping on the desk. In fact, any slight push on either LED resulted in both LEDs variation of light output (due to slight resistance changes in the breadboard connection). In other words it's a very unstable situation as was suspected. Best way would have been to solder in all the components and leads for that kind of measurement. So no, you cannot calculate the solution for current through path C from the information given.


Equipment used

hp 34401A
hp E3611A
Rigol DM3058E

[BeBuLamar]

What you don't know is the bulk resistance of the diode. This is the determining factor of how much current will go thru the right branch.

[mariush]

Also, measure the resistance of each wire from the power supply to the prototyping board, and the resistances of the thin wires connecting the proto board "bus bars" to the leds.

Your circuit is basically

[ + psu ] ---[ R - red wire from psu ] --- [ R - thin wire ] --- [ led 1 and led 2 in parallel] - [ led 3 ] -  [ R - thin wire ] --- [ R - black wire to psu ] - [ - psu ]

You will have around 0.1-0.2 ohm of resistance in the wires.

[xrunner]

Yep guys all true. Where's etnel? Seems like he'd want to say something about all this by now. 

[Tomorokoshi]

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.

2. Calculations of the "A" circuit
2.1. We are given that the forward drop is 2.0V at 20mA.
2.2. The resistor is 100 ohms and the power supply is 4V.
2.3. The current is conveniently (4V - 2V) / 100 ohms = 20mA, which matches the diode.
2.4. The current through point A is 20mA.

3. Calculations of the "B" and "C" circuits
3.1. With some math we can derive that for the same diode we have this relation: (v₁ - v₂) = nV_T*ln(i₁ / i₂)

4. Relating the "B" circuit to the "C" circuit
4.1. i_B = 2i_C
4.2. v_C = 4 - v_B
4.3. v_B - v_C = nV_T*ln(i_B / i_C), or:
4.4. v_B - (4 - v_B) = nV_T*ln(2)
4.5. 2*v_B - 4 = nV_T*ln(2)

5. Relating the "B" circuit to the "A" circuit
5.1. The voltage on diode B is greater than the voltage on diode A.
5.2. v_B - v_A = nV_T*ln(i_B / i_A), or:
5.3. v_B - 2 = nV_T*ln(i_B / 0.02)

6. Combining the results
6.1. Multiply equation 5.3 by 2:
6.2. 2*v_B - 4 = nV_T*2*ln(i_B / 0.02)
6.3. Note that equations 4.5 and 6.2 have the same left side. Make an equality from the right sides:
6.4. nV_T*ln(2) = nV_T*2*ln(i_B / 0.02)
6.5. ln(2) = 2*ln(i_B / 0.02)
6.6. ln(2) / 2 = ln(i_B / 0.02)
6.7. e^(ln(2) / 2) = i_B / 0.02
6.8. i_B = e^(ln(2) / 2) * 0.02, or:
6.9. i_B = 28.28 mA
6.10. i_C = 14.14 mA.
6.11. i_A = 20 mA.

[rstofer]

I think everybody is trying to add reality to a homework problem.  No issues with the left LED, it gets 20 mA, no doubt.
The two LEDs on the right feed into a series LED that will drop 2 volts at 20 mA and each of the parallel LEDs drop 2V at 10 mA and everything balances.  Just as long as you don't think about real components.

Current A = 20 mA
Current B = 20 mA
Current C = 10 mA

Given that no more information was presented, I think the 'ideal' component is all you can use.  How you justify the 10 mA through the parallel LEDs, I don't know.  The 10 mA is correct, whether that would actually produce 2V drop isn't given in the problem statement.

Not a particularly good example.

[EEVblog]

Hello, how would I calculate the current at the different points in this circuit?

Vcc = 4v
LED: Vf = 2v, 20mA
R1 = 100 ohms

It's one of those questions that actually makes no sense in practice. You wouldn't put a supply (presumably low impedance) of 4V across non-linear LED's in series like that.
But even theoretically the question is still stupid.
If vf is 2V @20mA, that makes 2V/20mA throguh the lower LED on the right, but then makes no sense for the upper two LED's which must be 10mA each (assuming even distribution), so the Vf will no longer be 2V.

[EEVblog]

Could help myself, video coming shortly!