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Maybe a simple question about calculating resistors for LEDs?

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atmdc

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I am trying to add some extra smd LEDs to a ps3 controller. Only issue I am having is that I dont know what supply voltage to calculate the resistors with because it is a wireless controller so the voltage will change as it is used right?

the battery is rated at 3.7v with a maximum charge rating of 4.2v. I measured the fully charged batter with a dmm and it was 5.2v

Which of the 3 voltage numbers should I use to calculate resistors for my LEDs for them to be as close to maximum lumination without over powering them?

if it makes a difference the smd LEDs are
3-3.2v
20-30mA

Thanks alot, I have been trying the entire day to figure this one out
 

Use some 51 ohms resistors/0.25W (if not available try the 47 ohms value).
 

Looks like you over-charged your battery too much. !!

Calculate LED resistance as (Vmax-Vled) / I led.. so 4.2-3.2 = 1V and choose 10mA i with 100Ω instead of 50Ω for 20mA
Add a red LED (1.6V )in series with White or blue LED and use (5.2v - (3.2+1.6 ) / 0.02mA = 50Ω for red LED.

Now the White LED will be 10mA at 4.2V and (5.2-3.2V) / 100Ω = 20mA when over charged and Red LED turns on at 3.2+1.6 V= 4.8V ( actually slightly 200 mV or so less V with dim light) and is full brightness when over charged at 5.2V on both LEDs using 20mA through separate resistors. i^2*R power values appear safe. for SMD < 40mW.

The ESR of the 20mA LEDs will be in the 10~20Ω range.
 

Looks like you over-charged your battery too much. !!

Calculate LED resistance as (Vmax-Vled) / I led.. so 4.2-3.2 = 1V and choose 10mA i with 100Ω instead of 50Ω for 20mA
Add a red LED (1.6V )in series with White or blue LED and use (5.2v - (3.2+1.6 ) / 0.02mA = 50Ω for red LED.

Now the White LED will be 10mA at 4.2V and (5.2-3.2V) / 100Ω = 20mA when over charged and Red LED turns on at 3.2+1.6 V= 4.8V ( actually slightly 200 mV or so less V with dim light) and is full brightness when over charged at 5.2V on both LEDs using 20mA through separate resistors. i^2*R power values appear safe. for SMD < 40mW.

The ESR of the 20mA LEDs will be in the 10~20Ω range.
Combining additional red LEDs? Why, where the red LEDs?

And '(5.2v - (3.2+1.6 ) / 0.02mA = 50Ω' ? 5.2-4.8= 0.4 and 0.4/0.02= 20 :-D

There’s a Li-Ion battery, as full-charge per-cell is usually around 4.2-4.3v. If he need maximum illumination choose at least 20mA and not 10mA.
To much theory for a simple series resistor… :roll:
 

There’s a Li-Ion battery, as full-charge per-cell is usually around 4.2-4.3v.
This is a concern, though:
I measured the fully charged batter with a dmm and it was 5.2v
If the battery voltage can be as high as 5.2V, a 100 Ohm resistor would be safer. That will give about:
20mA with battery = 5.2V
10mA with battery = 4.2V
5mA with battery = 3.7V

If he doesn't want the current to change with battery voltage, a low drop-out constant current source will be needed.
 

C'mon, have you ever found any Li-Ion battery charged for more than 4.5V? :roll:

- - - Updated - - -

Bottom of the line, choosing the limiting resistor must take into account the necessary illumination effect and keeping a safe margin to the theoretically limit of 30mA maximum admitted.
In practice we may choose to do small experiments by trying various resistors values (47 ohms up to 220 ohms) and check the optical results.
 

C'mon, have you ever found any Li-Ion battery charged for more than 4.5V? :roll:
No, but he says he has. Maybe it's one of them special explodey ones. Either that or his multimeter's bust. Something's not right. Maybe it's not Li-Ion, but some other type?
 

Thanks for all of the answers. I guess im going to have to try to check another battery to see if my dmm isnt acting funny. but when the controller was hooked to the charger the battery stayed around 5.2-5.3v, could be defective.

this is the battery
**broken link removed**
 

Well it appears that my dmm had a low battery so it was throwing off the readings. the voltage of the battery in the controller was 4.1v
 

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