LEDs and Resistors

[Jdog]

If I have many LEDs in parallel in a circuit, why do I need a different resistor, rather than 1 before the LEDs? For example isn't
B+--r----------------------
                |          |
                0          0
                |          |
B-  -----------------------
the same as                             (r= resistor 0= LED)
B+------------------------
             r            r
             0           0
            |            |
B-  ----------------------

[Soeren]

Hi,

In your first example, the resistor (R1) has to carry the current for 2 LED's (eg. 2 x 20mA = 40mA).
In your second example, the resistors (R2 and R3) each has to carry only the current for one LED (eg. 20mA).
In both examples the voltage drop over the resistors (U_drop) are the same (given same supply voltage and same LEDs).

So
R1 = U_drop/40mA
and U2 = U3 = U_drop/20mA


Try looking at it through Kirschoffs glasses   

[Jdog]

So are you saying that as long as the resistor in the first example can take the extra current load, it would have the same results?

[Soeren]

Hi,


Nope, I'm saying that Ohms law is really hard to break and even trying to will result in immediate punishment 

If the two LEDs are the same type, preferably from the same production batch, they might have a close enough voltage drop over the range of currents a sensible person will subject them to and in that case, you might get away with parallel connection, at least for a part of their life expectancy.
The separate resistor for each led approach is the safest way to go however.

Say you have two blue LEDs and wanna run them in parallel at 12V:
U_drop_LED = ~3.6V
U_drop_Resistor = (12-3.6) = 8.4V
If I_LED_total=40mA
R = 8.4/0.04 = 210 => use 220 Ohm
Actual current will thus be: 8.4/220 = 38.2mA (i.e. 19mA for each)
The power rating of the resistor needs to be: P=U*I [Watts Law] = 8.4*0.04 = 321mW use 1/3W or higher rating.

In the above theoretic case, it would be better to connect both LEDs and the resistor in series, making U_drop_Resistor = (12-2*3.6) = 4.8V
Resistor will then have to be: 4.8/0.02 = 240 Ohm. It will be more likely for you to find a 220 Ohm resistor (from the E12-Row) so making the current 21.8mA (going through both LEDs).
Power dissipated in the resistor is then: 4.8x0.22 = 106mW, so no problem there, not even with 1/8W resistors.
It saves power, it saves LEDs and it saves worries, but you shouldn't use a to low V_drop_Resistor without calculating the consequence if used eg. in a car, where the voltage varies between 12.6V and up to 14.8V under normal operating conditions with a sound battery.
In the latter example, it would mean that R_drop_resistor varies between 5.4V and 7.6V and the current will vary with it, if you use a simple resistor to control the current.

[Jdog]

Okie Dokey, thank you.