Obviously you didn't have the resistor in series with the LED.

If that were the case there would have been no voltage accross the resistor. See #4

 

Voltage source: 9 Volts DC (battery)
Forward Voltage Drop: "Unspecified"
Designed Current: 50 mA

Solution:
9V - unspecified = Unresolved voltage
Unresolved voltage ÷ 50 mA = Unresolved resistance

Without knowing the forward voltage it's not possible to determine the proper resistance.

ASSUME:
9V DC
3.1 Vf
50 mA

Solution:
9V - 3.1Vf = 5.9V working voltage
5.9V (wv) - 50 mA (0.05A) = 118Ω

OBSERVATION:
50 mA is too much for conventional LED's. By conventional I mean those that typically have two leads and are soldered into a board (or SMD type). Typical MAX current for those is recommended at 30 mA. 20 mA is a far better current, and the LED's don't burn as bright and hot - but still VERY bright.

Suggested solution:
9 -3.1 = 5.9
5.9 ÷ 20 mA (0.02A) = 295Ω

Suggest trying a 300 Ω resistor.

Thank you for that. I will go through it and see what pops out the other end.

 

What do you mean "Not at DC"? Do you mean that capacitors don't work in DC circuits?

For DC, the capacitor charges so quickly that any effect won't be noticeable. No more than the parasitic capacitance across a resistor.

 

Would anyone care to comment on the original question? The calculation of R in the circuit. The required resistance to produce a 5.6V voltage drop in that circuit. Some advice perhaps along the lines of

You select an LED current and get the approximate forward voltage from the datasheet. Then you calculate the resistor value using KVL and Ohm's Law.

You can refine the resistor value after you have a better idea of the diode voltage at the desired current. In most cases, using typical forward voltage is sufficient because the human eye response to light is logarithmic, so small changes in current won't give a noticeable difference in brightness.

For better brightness control, you use a current source.

 

Voltage source: 9 Volts DC (battery)
Forward Voltage Drop: "Unspecified".
Designed Current: 50 mA
(all LED's have a specific forward voltage - red is typical but not always 2 volts, green and blue are around 3 volts. Others can be higher (or lower).)

Solution:
9V - unspecified = Unresolved voltage
Unresolved voltage ÷ 50 mA = Unresolved resistance

Without knowing the forward voltage it's not possible to determine the proper resistance.

ASSUME:
9V DC
3.1 Vf
50 mA

Solution:
9V - 3.1Vf = 5.9V working voltage
5.9V (wv) ÷ 50 mA (0.05A) = 118Ω

OBSERVATION:
50 mA is too much for conventional LED's. By conventional I mean those that typically have two leads and are soldered into a board (or SMD type). Typical MAX current for those is recommended at 30 mA. 20 mA is a far better current, and the LED's don't burn as bright and hot - but still VERY bright.

Suggested solution:
9 -3.1 = 5.9
5.9 ÷ 20 mA (0.02A) = 295Ω

Suggest trying a 300 Ω resistor.

Thanks Tony. That worked fine.

Vs : 8.96V
Vr : 5.989V
Vled : 2.97V
Current draw : 18 mA

It seems what you have done there is set a target current of 20 mA instead of what I did which was to use the current reading from the power source when the LED was the only item in the circuit. Is that a general rule? Pick a current that everything in te circuit is going to be happy with? Sounds like a method problem on my part. Could you possibly pin point where I went wrong?

 

For DC, the capacitor charges so quickly that any effect won't be noticeable. No more than the parasitic capacitance across a resistor.

Ok. Gotcha.

 

Is that a general rule?

For powering LED's it's necessary to know their forward voltage and their max current capabilities. Then the user decides what amperage they want to push and from what voltage source. There are different methods for ascertaining the proper resistance, the one I posted is the one I typically use. It's how I've done it all these years. Take the supply voltage and subtract the forward voltage. Now you know what kind of voltage drop you're going to need and can calculate the proper resistance to achieve that voltage drop.

The current through the entire circuit is the same since your circuit consists of a series battery, resistor and LED. YOU decide what amperage you want or need depending on the brightness you desire. And as so many have said, the difference in what your eye can perceive for brightness is highly relative and you're not likely to see small differences.

 

Ok. Gotcha.

Don't over complicate things. First order approximations are often sufficient.

If you try to consider all parameters, you'll go crazy. Junction capacitance, junction resistance, bond wire resistance, bond wire inductance, lead resistance, lead inductance, capacitance between leads, etc.

Most aren't even specified in the datasheet; because they almost never matter.

 

One caveat that should be mentioned is the LED's reverse voltage breakdown. IF you power an LED from an AC source, the voltage must be lower than the reverse breakdown voltage. Exceed that and the LED will pop. To use a higher voltage AC source you'll need a series diode that can handle the voltage and will protect the LED from the reverse voltage.

 

For powering LED's it's necessary to know their forward voltage and their max current capabilities. Then the user decides what amperage they want to push and from what voltage source. There are different methods for ascertaining the proper resistance, the one I posted is the one I typically use. It's how I've done it all these years. Take the supply voltage and subtract the forward voltage. Now you know what kind of voltage drop you're going to need and can calculate the proper resistance to achieve that voltage drop.

The current through the entire circuit is the same since your circuit consists of a series battery, resistor and LED. YOU decide what amperage you want or need depending on the brightness you desire. And as so many have said, the difference in what your eye can perceive for brightness is highly relative and you're not likely to see small differences.

This is what I see when I just have a blue LED accross the source. 61 mA. I used a white one earlier. They are different as you say. Anyway thanks for your help with this. I'm going to cut and paste this and have a play around. Do some calcs.

 

Thanks for the input folks. It's been very helpful.

 

Good luck with your project. Still concerned with 61 mA. But if it's working and not blowing anything out - - - .

 

Good luck with your project. Still concerned with 61 mA. But if it's working and not blowing anything out - - - .

Cheers. I think I got it now. Economise on the current. That's the message I take away from this.

 

Economise on the current.

You shouldn't use more current than you need.

This curve for L934HD shows the relative brightness versus current; normalized to 10mA:


The LED has an absolute maximum continuous current of 30mA. Operating this LED at 30mA gives you a 50% increase over the brightness at 10mA; a difference a person can't discern.

If used as some sort of an indicator, a millamp or two might be sufficient.

 

You shouldn't use more current than you need.

This curve for L934HD shows the relative brightness versus current; normalized to 10mA:
View attachment 199175
View attachment 199174
The LED has an absolute maximum continuous current of 30mA. Operating this LED at 30mA gives you a 50% increase over the brightness at 10mA; a difference a person can't discern.

If used as some sort of an indicator, a millamp or two might be sufficient.

I have given it a lot of thought now. It makes sense because it's the current that does the work and the current that does the damage from a power dissipation viewpoint. When I used the 51 mA from the reading on my power source in the calculation I determined a value of 100 ohms. At 3.4V / 51 mA the power dissipation in the LED is 170 mW. At 9V it is 459 mW. It blew at about 5V ie: 255 mW. Reducing the current to 20 mA the 9V power dissipation is 180 mW. Which is why I was able to crank the circuit up to 9V without blowing the LED. I am still a little bit puzzled though because according to an LED datasheet I have read the power dissipation is 100 mW. See link. I don't know what the tolerance is. I have no idea what make my LEDs are so they are not the type in the data sheet. So I think the way forward is to look for the smallest functioning current and work on that basis whatever you are building.

http://www1.futureelectronics.com/doc/EVERLIGHT /334-15__T1C1-4WYA.pdf

 

At 3.4V / 51 mA the power dissipation in the LED is 170 mW

You're abusing the LED. Maximum power dissipation is 100mW. That's an absolute value and devices aren't guaranteed to survive anything above (or at) absolute maximum ratings.

Absolute max continuous forward current is 30mA, yet you chose to operate at 51mA.

Absolute maximum power dissipation is 100mW, yet you chose to operate at 170mW and higher.

I don't know what the tolerance is.

There is no tolerance for above absolute maximum ratings; even briefly.

I have no idea what make my LEDs are so they are not the type in the data sheet.

Then why bother giving us an irrelevant datasheet? If you don't know what they are, you should assume that maximum current is 10-20ma. If you want to push the specs, you need to know what they are.

 

Put a 3.4V supply across the circuit. LED shines. Supply unit tells me the current through the circuit is 51 mA.

So, I deduce that 51 mA makes my LED shine brightly.

Yes and it may shine brightly for a few min or till death. The LED should have a Vf (Forward Voltage) and an If (Forward Current). If my LED has a Vf of 2.0 volts and a If of 20 mA the formula is Vsupply - Vf LED / If. The LED is a current and not a voltage device.

This is what happens when you buy 100 LEDs for $1.00 which have no data sheet. Place a high value pot like 10K in series with the LED and start at max resistance slowly reducing the resistance until the LED glows what looks to be average and measure the current and voltage drop across the LED. Even that only comes with a "maybe" you will get close to what you actually want.

Ron

 

Cheap no-name-brand LEDs look bright because the case focusses the light into a very narrow beam.
The datasheet posted shows a narrow beam of only 15 degrees total.

 

Could you possibly pin point where I went wrong?

I see two fundamental misunderstandings in your comments so far.

The first is that you expect a predictable resistance, or ESR, from the LED. That's not gonna happen. Forget thinking about the LED in terms of equivalent resistance. Whatever resistance you calculate in any given driving condition will change as soon as you change the applied voltage, current, etc.

I'm on my way out the door, but tonight I'll upload a few sim images to illustrate my point... better yet, do your own experiments to learn yourself. Run the LED at a bunch of different current values from 1mA or less up to 20mA or more, measuring current and Vf along the way, then calculate the apparent resistance for each scenario. Then you'll understand why we tend to describe diodes in terms of forward voltage instead of resistance. The forward voltage is *relatively* stable, although it also varies as a function of current (and temperature, etc...)

The second is wattage/dissipation calcs. You don't multiply current times supply voltage, but rather against voltage across the LED. So, you never had a 9V, 20mA dissipation of 180mW as you suggested earlier. Your Vf would've been in the 2-4V range somewhere, with other voltage dropping across the resistor, so 20mA might be 40-80mW, but not 180.

 

So I think the way forward is to look for the smallest functioning current and work on that basis whatever you are building.

As @dl324 says, don't over complicate things. Make some simple assumptions.
Assume the maximum current is 20mA.
Assume an operating current of 1-10mA for starters.
Assume forward voltage is 2V for red LED and 3V for blue LED.

Thus for 9V supply, required series resistor for red LED @ 1mA is
R = (9V - 2V) / 1mA = 7kΩ

Required series resistor for red LED @ 10mA is
R = (9V - 2V) / 10mA = 700Ω

Hence start with R = 7kΩ and gradually decrease the value until you reach a suitable brightness. Do not reduce R below 700Ω.

See my blog.