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.

I

_{tot}=I

_{1} + I

_{2} + I

_{3} + I

_{4} + I

_{5}I

_{tot}=21.33 + 21.33 + 21.33 + 21.33 + 21.33

I

_{tot}=106.67

Total circuit resistance R

_{eq} is 30 ohms.

1/R

_{eq}=1/R

_{1} + 1/R

_{2} + 1/R

_{3} + 1/R

_{4} + 1/R

_{5}1/R

_{eq}=1/150 + 1/150 + 1/150 +1/150 + 1/150

1/R

_{eq}= .00667 + .00667 + .00667 + .00667 + .00667

1/R

_{eq}= .03335

R

_{eq} = 29.985 rounded to 30 ohms.

Total circuit power is 341mW.

P=I

^{2}R

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/R

_{tot} = 1/R

_{1} + 1/R

_{2} + 1/R

_{3}1/R

_{tot} = 1/100 + 1/100 + 1/100

1/R

_{tot} = .03

R

_{tot} = 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=I

^{2}R

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