I'm building an Arduino Mega 2560-based Throttle Quadrant to interface with my Flight Simulator system running FSX. The unit will have 6 potentiometers, 6 toggle switches [on-on and (on) -off-(on)], and 5 annunciators.
Each annunciator will have two LED's for illumination, and they will be either both on or both off, and their state is to be determined by the output of a single digital pin.
Here is the LED I plan to use alongside an annunciator housing with two such inserted. I have previously used such LED's singly with Arduino digital pin 5V outputs, in series with a 220 ohm resistor. I have no other information about these LED's:
Now I'm unsure of the best way to wire this up, but I see 3 options:
(1) Connect two LED's in parallel to each digital pin output. In this case I might need to attach a 220 ohm resistor in series to each LED. I'm uncertain as to whether I might be drawing too much current from the pin in this case.
(2) Connect two LED's in series to each pin, maybe without a series resistor? Will this work?
(3) Or to be safe, duplicate each digital output, and operate only one LED per pin, with 220 ohm in series. The tradeoff is that I would need twice the number of output pins (10 instead of 5), with a corresponding increase in the wiring mess.
I invite your comments on these options I am considering.
What color are your LEDs? Different LEDs have Different voltage drops.
If your LEDs red or yellow you may connect them in series with a small resistor (100-200 Ohm). Never connect LED without resistor.
If your LEDS blue or white connect each LED with separate resistor to give it current about 5-10 mA (option 1).
If you need high brightness (15-20 mA) use only option 3.
Atmega 2560 has an absolute limit of 40mA on each pin, and 200mA in total for all pins.
Thanks for your reply, which answers my question completely. My LED's are yellow, so I will consider connecting them in series (option 2), with a resistor between 100 and 200 ohm also in series, in accordance with your suggestion.
Thanks again for your help!
Forward voltage yellow led: 2.1 - 2.2 V, or 2.0 according to Sparkfun.
In series they need 4.2V. Assuming that your Mega 2560 board is not powered with a USB cable and is 5.0V, then there is only 0.8V for the resistor.
10 mA : 80 Ω
15 mA : 53 Ω
20 mA : 40 Ω
Solution: measure the 5V pin and the forward voltage of your leds (for 10, 15 and 20mA). Then decide which current is okay and calculate the resistor.
Suppose your computer has 4.95V at the USB port and there is 0.25V voltage drop in a cheap USB cable, and another 0.1V voltage drop in the connector and Mega board and the forward voltage of the Yellow led is 2.2V.
The resistor for 20 mA will be: (4.95 - 0.25 - 0.1 - (2 * 2.2)) / 20 mA = 10 Ω.
Let's assume that you use indeed 10 Ω, and then apply 7.5V to the barrel jack and the onboard voltage regulator makes 5.05V.
Then the two leds get 65 mA !
This is what happens when the margin is only 0.8V for the resistor.
Thank you so much for your detailed explanation! I always learn something when I come with questions to this community.
Since I am not considering using the barrel jack, or indeed anything other than the USB connection to the PC (which has to be there anyway for the data transmission to and from the simulator), it would appear from your numbers that I would be better off using a series resistor at the lower end of the range 40 - 80 ohm to get even moderate brightness. I can also see that some experimentation will be necessary to adjust matters.
A little practice. The circuit of two yellow LEDs and a 100 Ohm resistor in series showed the following results. Each LED has a voltage drop of exactly 2 V. When the supply voltage changes in the range from 4.75 V to 5.25 V, the current through the LEDs changes from 6.5 mA to 10 mA. As for me, the brightness is quite sufficient for the annunciator.
At 15mA one can read some text in a book at dusk.
20 ma is the absolute maximum rating for that LED, it will not last long time at this current.