parallel in series

[Big Sammy]

Hey all...I need to wire 10 red led's (1.9V@20ma) from a 9 v battery. I was going to make them sequentially flash, but the flahsing circuit wouldn't cooperate no matter how many times I re-wired it.....I just want them to go on, and am struggling as to how to calculate the resistance needed, and the best way to do it.....2 parallel sets of 5 led's wired in series? If that is the best way to wire it, am I correct in figuring I need 52 ohm resistors on each led? (I realize I will have to get 56 ohm). Thanks for any info. on this one....

 

[crutschow]

5 x 1.9V = 9.5V so I don't see how you expect to light 5 LEDs in series with a 9V battery.

You can two sets of four in series for 7.6V with a (9-7.6) / 20mA = 70 ohms in series with each set, and one set of two in series for 3.8V with (9-3.9) / 20mA = 255 ohms in series.

Since the current draw would be the same (60mA total) you could just use three strings of 4 in series with a 70 ohm resistor in each string for a total of 12 LEDs.

 

[davenn]

Hey all...I need to wire 10 red led's (1.9V@20ma) from a 9 v battery. I was going to make them sequentially flash, but the flahsing circuit wouldn't cooperate no matter how many times I re-wired it.....I just want them to go on, and am struggling as to how to calculate the resistance needed, and the best way to do it.....2 parallel sets of 5 led's wired in series? If that is the best way to wire it, am I correct in figuring I need 52 ohm resistors on each led? (I realize I will have to get 56 ohm). Thanks for any info. on this one....

And apart from Carl's comments about lack of voltage .....
lets see your circuit you your were trying to use to do the sequential flashing

Dave

 

[Big Sammy]

I was thinking of putting 5 led's in parallel (9-1.9/.1)....(this would be the 20ma*5)....two of these parellel circuits wired together in series....but I'm not sure if I would have to do the following: [9-1.9-1.9(voltage drop for both parallel circuits wired in series)/.1]?

 

[Diver300]

You shouldn't put the LEDs in parallel.

If you put two in series, that's 3.8 V. That leaves 9 - 3.8 V = 5.2 V. For 20 mA you want 5.2 / 0.02 = 260 Ω (You can use 270 Ω as that is a standard value)

Then just make that same circuit with 2 LEDs and a 270 Ω resistor, four more times.

That way, if there is any variation in between the LED voltages, it will not affect the current in any LED significantly.

If you put 5 LEDs in parallel, and one of them happens to have a lower voltage, it will take more current than the others.

LEDs are not voltage devices they are current devices. You should never place 3.3 volts across a "3.3 volt" LED. It is a 350mA LED that just happens to need approximately 3.3 volts to work.

 

[audioguru]

There is no point discussing resistor values unless the output resistance of the flasher circuit is shown.