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### Topic: Trying to understand current dispersal control across a parallel circuit  (Read 3222 times)previous topic - next topic

#### Frostline ##### Jan 28, 2016, 05:36 am
First let me start off by saying I am just trying to understand a concept.
This is not for a specific project or problem.  There is no xy-problem involved.  The schematic below is just for illustration of what I am trying to ask.

Second I apologize if it takes me too long to explain the questions I am having.
I am going to try to be as through as possible so hopefully I will not be accused of wasting people's time for omitting important information or being unclear as to how I arrive at certain values.

Formulas used in my example can be found here.

Let us assume for the example that the power supply is a 5v 1A switch-mode power supply.
The LEDs are 3mm red LEDs with a voltage drop of 1.8 volts.
The resistors are all 1/4 watt 5% metal film resistors.

There are 3 circuits in the following schematic.  From top to bottom they are labeled in red 1,2 and 3. Now circuit 1 is how I normally would approach constructing a circuit of this type.

Each branch of the circuit, due to the individual LED current limiting resistors is using 21.33mA.
V/R=I
( 5-1.8 )/150=I
.02133 = I

Total circuit current is the sum of the branches of the circuit or 106.67mA.
Itot=I1 + I2 + I3 + I4 + I5
Itot=21.33 + 21.33 + 21.33 + 21.33 + 21.33
Itot=106.67

Total circuit resistance Req is 30 ohms.
1/Req=1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5
1/Req=1/150 + 1/150 + 1/150 +1/150 + 1/150
1/Req= .00667 + .00667 + .00667 + .00667 + .00667
1/Req= .03335
Req = 29.985 rounded to 30 ohms.

Total circuit power is 341mW.
P=I2R
P=( .10667)2*30
P=.34135W

Question 1a: Do I calculate the total circuit and branch values correctly for circuit 1 or if not where am I making errors?

Now circuit 2 is where I attempt to create a circuit with similar total circuit values to circuit 1.
This is based on the idea that in circuit 1 with the resistance of each branch being equal the current through each branch was also equal.
The current through each branch was 1/5th of the total current.

So total circuit resistance is 33 ohms.
1/Rtot = 1/R1 + 1/R2 + 1/R3
1/Rtot = 1/100 + 1/100 + 1/100
1/Rtot = .03
Rtot = 33.333

Total circuit current is 96.97mA.
I=V/R
I= ( 5 - 1.8 )/33
I= .09697

Power of the circuit is 310.3mW.
P=I2R
P=( .09697)2*33
P= .3103

Given the near 1/3rd watt of power is why I used 3 resistors to dissipate the current  instead of a single 30 ohm resistor.  Otherwise I felt I might be accused of overloading my imaginary resistor.
I have seen it suggested several times both here on this forum and other sources on the internet that this is an acceptable practice.
But this concept of the power dissipation being spread out equally over multiple paths is why I wondered if current might follow a similar logic.

Question 2: Would the current through each branch in circuit 2 be equal to 1/5 of the total current or 19.39mA?

Now on to the final circuit.
As can be seen the circuit is incomplete with an unknown resistor ready to be placed into the circuit to complete the circuit.
So say it was desired to lower the current through each of the LEDs to 15mA and just simply lowering the supply voltage was not an option.  Also replacing each of the 150 ohm resistors with 220 ohm resistors was not an attractive option either.
If we pretend that we cut the circuit so that everything to the right of LED 12 and R11 is gone we are left with a simple series circuit.
A 68 ohm resistor could be put in place of the unknown resistor giving a total resistance of 218 ohms and 14.8mA of current for the LED.
But this is a parallel circuit with 5 LEDs and it does not seem to work out even though each LED branch looked at individually is a series circuit.
For total circuit resistance I would need to take the equivalent resistance of the 5 resistors already in the circuit plus the new resistor since it is in series.
So that would be 98 ohms.
Total circuit current based on the new total resistance is 32.65mA.
I=V/R
I=3.2/98
I= .03265
Which is a far cry from the desired 74mA total circuit current.  So clearly this seems not the way to figure it.
So then I thought of idea two which was simply calculating a needed total circuit resistance to get the desired total circuit current.
I=V/R
.074=3.2/R
R=3.2/.074
R = 43.24

And since 30 ohms is already in the total circuit an additional 13 ohm resistance would be sufficient to replace the unknown resistor and complete the circuit.  Thus total circuit current is lowered so available current for each branch is lowered.

Question 3: Am I even close with the second idea and if not how would the unknown resistor be calculated correctly(or is that not possible in this type of circuit)?

Assuming the above calculations were correct in circuit 1, total circuit voltage for circuit 1 would be 3.2
V=IR
V=.10667*30
V=3.2
which is the same as the source voltage minus the voltage drop of a single LED.
So although the voltage is dropped 5 times in the circuit by the 5 LEDs the total voltage effect is equal to a single LED.
Almost as if it is the average voltage drop per branch.
This is why I used 3.2V in my calculation for my second idea before question 3 as well as the calculations for question 2.

Let us then pretend LED5 is replaced with a blue LED that has a 3V drop and R5 is now a 94 ohm (just to keep branch current about the same).
Now the average voltage drop per branch is 2.04V

Question 4: What effect if any would differing branch voltage drops have on total circuit voltage compared to a circuit where the branch voltage drops were all the same? Or asked another way, for the calculation using Ohm's Law for question 3 when trying to calculate R what value would be used for V? 3.2 or 2.96 or something else if LED5 and R5 were replaced as noted above?

Thanks

#### larryd #1
##### Jan 28, 2016, 06:09 am
Circuit 1
5V / .1066A= 46.7 ohms

5V * .106.66 = .533W
No technical PMs.
If you need clarification, ask for help.

#### larryd #2
##### Jan 28, 2016, 06:18 am
Question 2
Total Current is 96ma
In practice you will only know what the current in each leg is by measuring it.
No two LEDs have the same forward voltage drop.
No technical PMs.
If you need clarification, ask for help.

#### larryd #3
##### Jan 28, 2016, 06:26 amLast Edit: Jan 28, 2016, 06:42 am by LarryD
Question 3
If total resistance is 46.7 use ohms law to calculate the needed resistance.

My eyes are tired, reminds me of school days.
No technical PMs.
If you need clarification, ask for help.

#### Frostline #4
##### Jan 28, 2016, 06:45 am
Circuit 1
5V / .1066A= 46.7 ohms

5V * .106.66 = .533W
So Ohm's Law and not equivalent resistance formula?

#### larryd #5
##### Jan 28, 2016, 06:47 amLast Edit: Jan 28, 2016, 06:51 am by LarryD
Total voltage / total current gives resistance Assuming the battery resistance is 0 ohms.
No technical PMs.
If you need clarification, ask for help.

#### Frostline #6
##### Jan 28, 2016, 06:53 am
Question 2
Total Current is 96ma
In practice you will only know what the current in each leg is by measuring it.
No two LEDs have the same forward voltage drop.
But to be clear the current will be balanced between each branch(LED) based on voltage drop?
That is LED 6 won't get 90mA due to lacking an individual current controlling resistor while the other 4 split the rest of the current if the forward voltage drops are near the same?

#### larryd #7
##### Jan 28, 2016, 07:02 am
You never put LEDs in parallel.
No technical PMs.
If you need clarification, ask for help.

#### dwightthinker #8
##### Jan 28, 2016, 07:16 am
1a You forgot to include the power lost in the LEDs.

2. This circuit would not work well.
LEDs voltage and efficiency drops with temperature, like any diode.
As they begin to heat up, one will begin to draw away more and
more of the current, most likely burning up.
If the LEDs shared the current evenly, you would be right but they wouldn't.

3.
drawing 3
The total voltage around any loop is 0. 1.8V for LED. 15ma through a 150 resistor
if 2.25V.
2.25 + 1.8 = 4.05V.
5V - 4.05 = .95V to be drops on the unknown resistor.
The total current is 5 * .015 = .075ma
.95/.075 = 12.6... ohms.
4.

I don't quite follow your question. The LED is a relatively constant voltage device.
It would stay around 1.8V but its current would change if you changed the individual
resistor. If using circuit 3, changing any one resistor would have some effect on
all the others.
Still one can figure things out based on the sum of the voltage around any loop
is 0 and the sum of the current going in and out of any point is also 0.
That with ohms law will give you the results.
If you are worried about the fine details of the slight change in voltage of the
LED, the problem could be more difficult to solve for a circuit of type 3 as
it would need to be solved for several simultaneous equations with the non-linear
LEDs. Such equations are best solved with successive approximations.
Of course the fine result would only be good for the particular LEDs you had.

When looking at total power, remember, anything with both voltage and current
will use power.
Dwight

#### Frostline #9
##### Jan 28, 2016, 09:15 am
1a You forgot to include the power lost in the LEDs.
How do I include that?

2. This circuit would not work well.
I agree and your explanation makes  sense to me.

The total voltage around any loop is 0. 1.8V for LED. 15ma through a 150 resistor
if 2.25V.
2.25 + 1.8 = 4.05V.
5V - 4.05 = .95V to be drops on the unknown resistor.
The total current is 5 * .015 = .075ma
.95/.075 = 12.6... ohms.
It took me awhile, but I finally got what you are saying here.
And while your method is different than how I was trying to do it, had I used 75mA for total current instead of 74mA I would have arrived at the same answer.  Even though my method was more luck based I think now.

Agree it is not very clear, more so since it was based on my incorrect assumption on how I first answered question 3.
In my example schematic all of the branches were the same, so my question was basically what if one of the branches was different(using a different LED with different drop voltage and replacing the resistor to keep that particular loop current near the same as the other 4 so total circuit current would remain the same).  Could a single resistor be added in series with the 5 parallel branches to lower the current of each branch to 15mA if one branch had a different voltage drop?
So four of the loops would have been just like you explained for question 3.
1.8V for the LED
15mA across a 150 ohm resistor for 2.25V
For a branch(loop) total of 4.05V
So 5V - 4.05V = .95V to drop across the unknown resistor
R=.95/.075
R=12.66
And that would work for all 4 red loops.
But with loop 5 being blue the value would be different
3.0V for the LED
15mA across a 94 ohm resistor for 1.41V
Makes the loop total 4.41V
So 5V - 4.41V = .59V to drop across the unknown resistor
R= .59/.075
R=7.87 ohm

So one resistor will not equally solve all 5 loops unless all 5 loops are the same.
Which answers question 4(I think anyway).

#### aarg #10
##### Jan 28, 2016, 10:07 am
You never put LEDs in parallel.
Sorry, that's not completely true. LEDs from the same fabrication process and batch can be, and are, safely paralleled in some LED modules. But unless you have matched LEDs, it's not a good idea.
... with a transistor and a large sum of money to spend ...
Please don't PM me with technical questions. Post them in the forum.

#### Paul__B #11
##### Jan 28, 2016, 12:20 pm
Sorry, that's not completely true. LEDs from the same fabrication process and batch can be, and are, safely paralleled in some LED modules.
Very commonly.  Universal in those arrays rated at 10 W and above.  Thermal runaway is limited by mounting on a common heatsink.

And in all the multi-LED torches you see.  Because the LED also has an internal resistance.

Mind you, you notice the differences as the batteries go flat.

#### MarkT #12
##### Jan 28, 2016, 01:05 pm
For all of these you actually have a non-linear set of equations to solve as the LED V/I curve is logarithmic.
However if you know the LED current is going to be close to a particular value (20mA) then you can approximate
by saying the voltage across the LED is fixed, and the rest is Ohm's and Kirchoff's laws (from which the
series and parallel resistance formulae are derived).

In reality Q2 will not work well as explained above, and noticably unequal current sharing is likely.

The constant voltage approximation for the LEDs is good enough - and at the end of the day a 10%
difference in current from what is intended cannot be detected by the human eye, and is less than
the variability in optical performance of LEDs anyway I suspect.  You usually should try various
resistor values by hand to get the brightness level you actually want with the particular LEDs you
are using.
[ I will NOT respond to personal messages, I WILL delete them, use the forum please ]

#### 68tjs #13
##### Jan 28, 2016, 02:01 pm
Quote
Sorry, that's not completely true. LEDs from the same fabrication process and batch can be, and are, safely paralleled in some LED modules. But unless you have matched LEDs, it's not a good idea.
Sorry, that's not completely true.
This is true if Leds are close on the same Wafer.
There are differences between distant areas on the same wafer.

Only one rule = never put LEDs in parallel.

#### dwightthinker #14
##### Jan 29, 2016, 10:34 pmLast Edit: Jan 29, 2016, 10:35 pm by dwightthinker
How do I include that?

Power on each LED is simple. If the current through it is 20ma
and the voltage is 1.8V:

1.8 * .02 = .036 watts.
if you have five bulbs:
5 * .036 = .18 watts
Just add this to the watts in the resistors.
Power is power. It doesn't make any difference if it is a resistor
or LED diode.
Dwight

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