guys, if i have 5 (5 mm white) led and 5 (3 mm white led) , and i got 10 (220ohm resistor) ,with power supply 5v,20mA and i need to connect them together , which circuit should i use, series or parallel?
connect all of the leds to gnd then use a 220 ohm resistor on the positive on each led then connect the 220 ohm resistors to your 5 supply. So basically connect them in parallel.
It won't work very well, the Vf is to much to put them in series and they do not like being in parallel. Use a second resistor and go from there.
One diode with one resistor in series is the only (good) way it will work with 5 volts and white led’s. Two diodes in series will not illuminate whatsoever due to the voltage required (~3.8 v each).
220 ohms on a white led with 5v should be 6 milliamps. Brightness should be okay with decent diodes. Bargain bin stuff may vary in brightness a bit.
In general, you cannot parallel led’s because brightness depends upon current. Placing them in parallel will result in visible differences in brightness because you cannot control the current in each diode when in parallel. One will always consume more current than others resulting in a different brightness. This changes with temperature as well.
Play with an online led calculator, for example: https://ledcalculator.net/
What happened to "try it and see what happens"?
Sounds like a school experiment/problem and @howardyamhowwei has been be given a set of parts, so each LED with its own resistor and across the power supply sounds like the solution.
Are you certain the power supply is 5 volts with a maximum of 20mA? If so, you will likely not get the LEDs to light properly, but it could be done by connecting all the LEDs in parallel, with the resistors in a series-parallel arrangement. With a voltage drop of approximately 3.8V, that would need a resistance of 60 ohms to limit the current to 20mA, the closest I can get by series-paralleling ten 220 ohm resistors is 88 ohms, allowing for a bit less than 14mA, and as was mentioned the LEDs will not evenly distribute that current.
That would give a total of almost 55mA, far in excess of the 20mA limit of the power source.
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