I need to run an LCD with the brightest possible led light without damaging arduino's digital pin.
The image is a part of the LCD's data sheet. It's somewhat unclear to me.
Does this mean that, if 5 volts are applied to the 15 & 16 pins with no external resistors, the led would draw 75mA?
And if that's true, is the following calculation correct?
Led Resistance = V/I = 5/0.075 = 66 Ohms
The total needed resistance to keep current below 40mA
Rt = V/I = 5/0.040 = 125 ohms
So we need to add external resistor,
Re = 125 - 66 = 59 ohms (minimum)
The default says J1 open, RE = 0R. RL will be (5 - 3.1) / 0.075 = 25R
No, you can not connect a 75mA load to an Arduino GPIO pin. But a backlight is normally always on. Hence you connect to 5V pin on Arduino.
If you want to connect to a GPIO pin, you have to limit the current to 20mA. e.g. RE=70R, RL=25R, J1 open..
Actually the 75mA LED will probably have a Vf = 2.9V @ 20mA.
If the backlight is designed for 75mA it will not be bright enough with 20mA.
Do not worry about exact maths for 5V. The series resistors are not very critical.
With a 3.3V supply and a 3.1V LED, you must use the correct calculation.
David.
You should follow the advice that David has provided.
Here is some additional background information:
(1) Georg Ohm discovered that for devices that came to be called 'resistors' there was a linear relationship between the current flowing through the device and the voltage appearing across the device. In other words if the voltage across the device went up (or down) the current also went up (or down) but their relationship was always the same. Similarly if external elements caused the current flowing through the device to go up (or down) the voltage across the device followed, at the same ratio.
(2) You cannot calculate the 'resistance' of an LED since it is a non-linear device. The voltage across an LED will vary as the current through it varies, but the ratio is not always the same and therefore it does not have the characteristics of a 'resistor' and Ohm's law does not apply.
(3) The Forward voltage specification (Vf) of an LED is not a 'rating', it is a characteristic. It is not a value that you 'apply' to get the specified current, it is the value you should ultimately measure if you choose the other circuit components to give the specified current.
(4) It's kind of a chicken vs egg dilemma since you need to know the Vf in order to calculate the component values needed to get the desired If but you don't know the actual Vf until after you have achieved the desired If.
(5) The bottom line is that you should use the nominal (typical) values given in the data sheet to calculate the required series resistance as David has done in his first sentence. When you connect everything up the actual circuit current and diode voltage will probably differ somewhat from your original assumptions but the LED will light up and the devices will not go up in smoke. If you should happen to be lucky and get exactly the desired 75 mA you should not expect to see exactly the specified 3.1 volts but you should expect to see a value between 2.9 and 3.3 volts.
Don