# Thread: Estimate LED limiting resistor - Ohm's law at work

1. ## Estimate LED limiting resistor - Ohm's law at work

Given
power supply specification
Nominal ( per label ) spec 9V /200mA Wallwart transformer / DC supply

LED "spec"
forward voltage 3 V
max current 30 mA

Currently used "limiting resistor " of 500 Ohms

3 LED's connected in series and 10 of these serial connections in parallel

Measured
Transformer Idle output approximately 16V
Transformer load output @ array of LED's approximately 8 V
LED array draws / is limited to approximate load current of 10mA
Voltage drop across limiting resistor - approximately 0.5V

Estimate NEW limiting current resistor value to control the intensity of the
LED array. ( I do not have suitable variable resistor to test with )

The load current cannot exceed 30mA and assuming the wallwart will actually supply
full voltage of 9V @ load of 30 mA.  Reply With Quote

2. ## are we allowed to write a program to work this out?  Reply With Quote

3. ## Since the voltage drop remains the same across an LED bulb as the current through it changes, you subtract the LED voltage drop from the supply voltage to obtain the voltage on the resistor. (and thus the current via ohms law). Three 3v LED's in series would be a total of 9v across the string. At 30mA per series LED string, you pull 300mA having 10 of those in parallel. In other words, you have a total of 9v drop and 300mA draw through the entire network of lights. That is too much for that power supply. Given that the addition of a limiting resistor to further lower the brightness will increase the loss, its likely that trying to limit the current further with a resistor will simply shut the lights off. Step one, get a more powerful supply.

For simplicity sake, lets do the math on just one 3v 30mA LED connected to a 9v supply. 9v - 3v = 6v across the load resistor. if we want 30mA through a resistor with a 6v drop, 6v / .030A = 200ohm. And the power rating of that resistor is 6v x .030A = 0.180W. This means it needs to be at least a quarter watt resistor as an eighth watt (.125w) would overheat.

Expanding that math to your current LED array of 3 (series) x 10 (parallel of the series) LED's creates a 9v drop at 300mA. If you had a more powerful 9v supply, say 500mA rating, you would not need a limiting resistor at all as the LEDs will self-limit due to the absence of excess voltage (in terms of burning out). Any attempt to limit this current with a resistor would create additional voltage drop and getting a dimmer effect would be extremely difficult as you would be basing the resistance needed on the turn-on voltage curve and not current alone. You need more voltage.

Now lets assume you happen to find an old 12v 500mA wall wort... For a 12v supply, you will have a 3v drop across the limiting resistor and have plenty of room to play around without losing the forward bias on the LED's. Lets say you wanted to give the array half the rated power (150mA) at the same forward voltage drop (9v). Now we need to find a resistor that will have 150mA through it at 3v across it. 3v / 0.150A = 20 ohm resistor (with a rating of at least 3v x 0.15A = 0.45W ~ a half watt 20 ohm resistor).

In the end, the better way of controlling brightness (simply) is with a current mirror circuit using matched transistors, or (a bit more complicated) a PWM circuit to turn them on and off at full power faster than the eye can see. The latter method is far superior and much more efficient in the end.  Reply With Quote

4. ##  Originally Posted by brandon lind Since the voltage drop remains the same across an LED bulb as the current through it changes, you subtract the LED voltage drop from the supply voltage to obtain the voltage on the resistor. (and thus the current via ohms law). Three 3v LED's in series would be a total of 9v across the string. At 30mA per series LED string, you pull 300mA having 10 of those in parallel. In other words, you have a total of 9v drop and 300mA draw through the entire network of lights. That is too much for that power supply. Given that the addition of a limiting resistor to further lower the brightness will increase the loss, its likely that trying to limit the current further with a resistor will simply shut the lights off. Step one, get a more powerful supply.

For simplicity sake, lets do the math on just one 3v 30mA LED connected to a 9v supply. 9v - 3v = 6v across the load resistor. if we want 30mA through a resistor with a 6v drop, 6v / .030A = 200ohm. And the power rating of that resistor is 6v x .030A = 0.180W. This means it needs to be at least a quarter watt resistor as an eighth watt (.125w) would overheat.

Expanding that math to your current LED array of 3 (series) x 10 (parallel of the series) LED's creates a 9v drop at 300mA. If you had a more powerful 9v supply, say 500mA rating, you would not need a limiting resistor at all as the LEDs will self-limit due to the absence of excess voltage (in terms of burning out). Any attempt to limit this current with a resistor would create additional voltage drop and getting a dimmer effect would be extremely difficult as you would be basing the resistance needed on the turn-on voltage curve and not current alone. You need more voltage.

Now lets assume you happen to find an old 12v 500mA wall wort... For a 12v supply, you will have a 3v drop across the limiting resistor and have plenty of room to play around without losing the forward bias on the LED's. Lets say you wanted to give the array half the rated power (150mA) at the same forward voltage drop (9v). Now we need to find a resistor that will have 150mA through it at 3v across it. 3v / 0.150A = 20 ohm resistor (with a rating of at least 3v x 0.15A = 0.45W ~ a half watt 20 ohm resistor).

In the end, the better way of controlling brightness (simply) is with a current mirror circuit using matched transistors, or (a bit more complicated) a PWM circuit to turn them on and off at full power faster than the eye can see. The latter method is far superior and much more efficient in the end.
Sorry , your very informative analysis and actual measurements do not agree.
But I do appreciate you spending your time on it, for real.

I think the real problem is figuring out how "hard" these DC supplies are, AKA how does the output hold under load.
Also - LED blows very fast when the current is not limited IMMEDIATELY !

I have replaced limiting resistors - two 1k resistors in parallel with four resistors in parallel.
Now I am reading almost 9V across the array and almost 20mA of total current.

I can accept the 9V , but 20mA is way off the scale !
I am not sure if that is normal behaviour of LED's connected in parallel, after all it is a diode.

It should be at least 200 mA !

Keep n in mind that I am using cheap VOM and most of mine measurements are not very accurate.

I do have other "surplus" DC power supplies with more current output, but I have burned several series connected LED's before I dusted of my VOM and started measuring.

At this point I do not want to risk burning out the array , it took some soldering to get it mounted...

I will go back and wire 3 LED's in series and find some variable resistor to do some real testing,
however this discrepancy between the current draw in theory and practice got me puzzled and needs at least
theoretical explanation.

Cheers  Reply With Quote

5. ## I realize I am getting ahead and answering / posting before my posts are approved.
I run another test to get better feel for the DC supply.

My ultimate goal is to pass air over the array.
I am currently using 12V DC fan and the DC supply output drops to steady 13V @ 55mA
I think I will size the limiting resistor to pass 150 mA to the array.  Reply With Quote

6. ## Admittedly, i made two critical errors in my first response. The first error was suggesting a single resistor for the entire array. Diodes in parallel to other diodes have the strange ability to unevenly pass current. This means every parallel path of LED lights will need an individual limiting resistor. The other error i made was making the assumption your supply was regulated. Being it is typical for wall worts to be rectified sine waves with no regulation, the unloaded peak voltage and the loaded RMS voltage prior to rectification will be different ~ hence the 16v you see with nothing connected

This needs to be taken into consideration when designing the system.  Reply With Quote

7. ##  Originally Posted by brandon lind Admittedly, i made two critical errors in my first response. The first error was suggesting a single resistor for the entire array. Diodes in parallel to other diodes have the strange ability to unevenly pass current. This means every parallel path of LED lights will need an individual limiting resistor. The other error i made was making the assumption your supply was regulated. Being it is typical for wall worts to be rectified sine waves with no regulation, the unloaded peak voltage and the loaded RMS voltage prior to rectification will be different ~ hence the 16v you see with nothing connected

This needs to be taken into consideration when designing the system.
I woudl not say you have made an error / wrong assumption when it comes in connecting serial / parallel LED array.
You are just working what is known / given , same as I.

This whole exercise did not started as theoretical application of Ohms law, but as an experiment with real limitations, including wallwart DC supply.

I need to point out that the specification for the LED is not written in stone, just my best guess.
I really do not have real data what is maximum current before LED became "fast blow fuse".

Here is the latest.
With the fan running @13V , drawing 55mA I added 100 Ohms limiting resistor connected to 3 LED's in series.
The LED's draw 22mA, hence 10 of these woudl draw approximately 200mA .

The combination of LED array and fan does exceeds the load spec ( 200mA) of the DC supply , and I could use other, heftier supply,

My "scientific guess" is to include two 22 Ohms current limiting resistors to feed the entire array.

Three "reasons"
22 Ohms 1/8 w are in my junk box
paralleling 1/8 w resistors will help to keep them cool ( will recalculate if my guess is OK later, before I do whole system "stress / smoke test " )
having 11 Ohms instead of 10 gives some "protection".

The LED array "intensity" is important, but again, I have no real "measuring stick" to evaluate it , just my eyeballs. .  Reply With Quote

8. ## With 11 Ohms limiting resistor the array drawa 90 mA and there is 1 V across the resistor.
Currently running "smoke test".  Reply With Quote

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