What, only four decimal places of precision?! How crude!

I use no precision. "The power requirements of that particular part of the circuit is a metric buttload." It's sufficient for marketing meetings. So is, "You kidding me, the output of that motor will be puny; like an anemic little kitten." If they want more resolution, I might hold my hands up indicating a very tiny little kitten. That is, if the marketing folks are exceptionally smart. However, most marketing folks can't handle spoken word and gestures at the same time.... it confuses them.

 

I use no precision. "The power requirements of that particular part of the circuit is a metric buttload." It's sufficient for marketing meetings. So is, "You kidding me, the output of that motor will be puny; like an anemic little kitten." If they want more resolution, I might hold my hands up indicating a very tiny little kitten. That is, if the marketing folks are exceptionally smart. However, most marketing folks can't handle spoken word and gestures at the same time.... it confuses them.

Wow! Tell me what you really think.

That four decimal point spec has two things going for it. 1) It’s direct from the manufacturer and 2) Verified in an actual circuit which I built and measured.

You have to have specs for design. And you have to verify those specs by tests before and after your build. As President Ronald Reagan said, “Trust But verify!”.

 

Wow! Tell me what you really think.

That four decimal point spec has two things going for it. 1) It’s direct from the manufacturer and 2) Verified in an actual circuit which I built and measured.

You have to have specs for design. And you have to verify those specs by tests before and after your build. As President Ronald Reagan said, “Trust But verify!”.

My post was made with tongue firmly planted in cheek...

 

Just so you know, the circuit on the right has the potential for catastrophic failure. First, your voltages and resistances are wrong. I won't focus on how or why. However, the reason why the one on the right is in danger of complete failure is because one of the diodes will hog the current. Meaning if you pump too much current through the circuit you will blow that hogging diode. Then the next diode will then hog the available current and blow. Finally the last one will blow. And it will happen VERY fast. You would likely not have sufficient time to shut down the circuit.

Do you know how to calculate the correct resistor for the LED (just one of the three) on the left? An understanding of how they are calculated will reveal why I said the one on the left has the potential to work. At present, the way you designed it - it won't work. It won't fail but it will fail to work. What's the difference? Failure of the right side means the components are ruined. Whereas failure to work means it just won't do what you hope it will. The components themselves will be unharmed. Like trying to light a household lightbulb using a 9 volt battery.

Still, I have a doubt.
Assume all LED has the same specifications.
In the first circuit, the total current splits into 3 when it reaches the node and a current of 30mA flows through each LED, please don't consider the resistance 10K (what we need is 160Ohm resistor).
In the second case, the same happens a the junction where LED meets.
Then how the second will be a failure. I mean the same 30mA flows in both the cases and the voltage is brought down by sufficient resistance.

 

In the first case the 30 ma per LED is not correct. You did not subtract the LED voltage from the 5 volt supply.
Assuming a 2 volt drop across the LED then the current is 5v - 2v = 3v / 160 ohms = 18.75 ma which is OK.
In the second case using the same 160 resistor the current is the same but each LED (if perfectly matched) draws about 6.25 ma. If one LED opens then the other two now draw 9 ma each. If one LED shorts then they all go out.
SG