Maker Pro
Maker Pro

Been doing it all wrong!!

Martaine2005

May 12, 2015
4,353
Joined
May 12, 2015
Messages
4,353
Hi guys, I am a newbie here but thought my formulas were correct....Danggg..
Ohms law. V / I * R or a variation of the same.
I just read the STICKY on L.E.D's and formulas to figure out Resistance and Current..
My brain tonight is like mushy peas.
I thought .. 12v supply.. LED Vf 2v @20mah
Was 12 v - 2v = 10v. So 10v * 20ma
= 200 Ohms
Am I being thick? Real thick?

The example really threw me..
12v supply, Vf 10.2v (3 * 3.4v leds) ..Difference is 1.8v. So 1.8v * 10.2 = 18.36?
Or 10.2 / 1.8 = 5.6666..?
Can somebody please put this into plain English..

I promise to have math lessons
I promise to have math lessons
I promise to have math lessons
etc etc etc
100 lines.:oops:
 

Gryd3

Jun 25, 2014
4,098
Joined
Jun 25, 2014
Messages
4,098
Hi guys, I am a newbie here but thought my formulas were correct....Danggg..
Ohms law. V / I * R or a variation of the same.
I just read the STICKY on L.E.D's and formulas to figure out Resistance and Current..
My brain tonight is like mushy peas.
I thought .. 12v supply.. LED Vf 2v @20mah
Was 12 v - 2v = 10v. So 10v * 20ma
= 200 Ohms
Am I being thick? Real thick?

The example really threw me..
12v supply, Vf 10.2v (3 * 3.4v leds) ..Difference is 1.8v. So 1.8v * 10.2 = 18.36?
Or 10.2 / 1.8 = 5.6666..?
Can somebody please put this into plain English..

I promise to have math lessons
I promise to have math lessons
I promise to have math lessons
etc etc etc
100 lines.:oops:
Voltage = Current * Resistance
Power = Current * Voltage

12V Supply - 2Vf = 0.020A * Resistance
(12V-2Vf) / 0.020A = Resistance

Additionally... 10V * 20mA = 200mW or 0.2Watts ... So you messed up on the equation, but also the math. There is no way 10 * 0.02 = 200 :s

So please read up : http://en.wikipedia.org/wiki/Metric_prefix
This is so you can understand the prefix for things... like mA compared to kA, or MW compared to mW

Then Ohm's Law :
Ohm's_Law_Pie_chart.svg

Voltage is usually represented by 'V' or 'E' (Units are the 'Volt')
Current is represented by 'I' (Units are 'Amps')
Resistance is represented by 'R' (Units is 'Ohms' or Ω)
Power is represented by 'P' (Units is 'Watts')
 

Fish4Fun

So long, and Thanks for all the Fish!
Aug 27, 2013
471
Joined
Aug 27, 2013
Messages
471
Martaine2005,

I think you and Gryd3 are on different pages...though he is obviously correct on the arithmetic side of your math problem(s). I think what you are asking is about subtracting the LED Vf from the supply Vf prior to figuring the resistance? ie:

Code:
Typical Specs for a 1W White LED:
Vf = 3.2V to 3.6V
Imax = 350mA

Supply = 12V +/- 1V

For a single LED:

13V - 3,2V = 9.8V  

9.8V = 0.300A * R ==> 9.8V/0.3 = 32.67Ohms 
P = I^2R --->  0.3^2 * 32.67 Ohms  = 2.94W

So, In this example the resistor would dissipate ~3W while the LED utilize 3.2V * 0.300A = 0.96W ..... an efficiency of ~25%...pretty horrible....

Now if we had three of the above LEDs in series then:

Code:
3 * 3.2V = 9.6V
13V - 9.6V = 3.4V

3.4V = 0.300A * R ==> 3.4V/0.3 = 11.33Ohms
P = I^2R --> 0.3^2 * 11.33 = 1.02W

So, in this example the resistor dissipates ~1W while the LEDs utilize 9.6V * 0.300 = 2.88W .....an efficiency approaching 75%.....still pretty bad, but better than 25%....The problem is that if we recalculate the results using the high-end of the LED Vf (3.6V each) and the low-end of the power supply voltage (11V) then we get the following results.....

Code:
3 * 3.6V = 10.8V
11V - 10.8V = 0.2V

0.2V = 11.33Ohms * I  ==> 0.0177A --> 17.7mA
P = I^2R --> 0.0177^2 * 11.33 =  0.0035W (3.5mW)

So, in this example the power supply drooped from 13V to 11V and the resistor only dissipates3.5mW while the LEDs utilize 10.8V * 0.0177A = 0.191W (191mW)....Our efficiency is a whopping 98%, but our light output will be negligible.....The point is that to maximize efficiency we need to minimize the power dissipated in the resistor....this is done by keeping the Supply Voltage as close to the Vf of the LEDs as possible; however, as the Vf of the LEDs approaches the Supply voltage and the current limiting resistor is minimized the circuit becomes increasingly sensitive to both changes in the LED Vf and power supply ripple.....As a general rule of thumb the Vf of the LEDs should be kept below 75% of the Supply voltage (for stable supplies) and below 50% for unstable supplies like automotive systems....

The real answer to driving LEDs efficiently is use of constant current power supplies (commonly called "LED Drivers"....In this case the voltage is left unregulated (within a defined range), but the current is kept constant....So, for an LED driver designed for operation between 3V and 37V with a constant current of 300mA, you could connect one of the above LEDs OR up to 10 LEDs in series and in each case the current would remain 300mA (+/- ripple). If you measured the Voltage across the LEDs it would read exactly what the Vf of the LED or Series of LEDs is @ 0.300mA.....

What makes driving LEDs difficult with resistors and "fixed voltage supplies" is that the Vf of the LEDs change with both current and temperature...... Hence the Vf "range" given in their specifications....Since neither a fixed voltage supply nor a fixed resistor can "change" based on the LED' s dynamic Vf, a great deal of energy is wasted in an attempt to protect the LED from over-current.....(which ends abruptly in catastrophic LED failure)....

So...that was prolly more than you were looking for....and as Gryd3 pointed out, you need to work on your math....hehe, and there is a lot more math involved in deigning efficient constant current drivers :) On the bright side, very good LED drivers are available for cheap (some as low as ~$1.50 on ebay....and NO math...hehehe) :)

Good Luck!

Fish
 

Kiwi

Jan 28, 2013
424
Joined
Jan 28, 2013
Messages
424
R = (Vs - Vf) / If

So your first example would be;
R = (12 - 2) / 0.02
= 10 / 0.02
= 500Ω

For multiple LED's in series, Vf is multiplied by the number of diodes.
Vf for your second example would be; 3 * 3.4 = 10.2
R = (12 - 10.2) / 0.02
= 1.8 / 0.02
= 90Ω
 

Martaine2005

May 12, 2015
4,353
Joined
May 12, 2015
Messages
4,353
WOW, Thank you all for the info. And FISH, you were right about being on different pages!!
Gryd3, Yes, I typed it wrong..Meant to type V over I*R like the ohms law triangle.
Fish, Yes Iwas refering to subtracting the Vf first.
12v.psu, diode Vf 3.2v, @350ma . So, 12 - 3.2 = 8.8.. So 8.8 / 3.2 = 2.75 ohms??
Blimey, hope I got that right. I need to learn the fundamentals first.
I think Kiwi is on my wave length hehe.
Thanks again for all your help. Martin.
 

Kiwi

Jan 28, 2013
424
Joined
Jan 28, 2013
Messages
424
Martin, Martin, Martin, I'm afraid you still haven't got it.

"12v.psu, diode Vf 3.2v, @350ma . So, 12 - 3.2 = 8.8.. So 8.8 / 3.2 = 2.75 ohms??"
Why did you divide by 3.2?
R = (Vs - Vf) / If , not (Vs - Vf) / Vf
 

Martaine2005

May 12, 2015
4,353
Joined
May 12, 2015
Messages
4,353
Crickey 'o' Reilley.. I really should read back what I type!!
That should have been /350ma
8.8v / .350ma = 25.14...?
I have printed all the replies here so I can chew on it until I understand.
Thanks Kiwi.
 
Top