Battery Can't Handle 3 LEDs in Series

I don't have a lot of room for wires. Is a pair of 3 and a pair of 2 better than 5 pairs?

The battery wasn't fresh.

How long will the 9v last? 6 AA might be a possibility.

sjaguar13 said:

I don't have a lot of room for wires. Is a pair of 3 and a pair of 2 better than 5 pairs?

The battery wasn't fresh.

How long will the 9v last? 6 AA might be a possibility.

Click to expand...

A pair of 3 and a pair of 2 is better than 5 pairs since there are only 4 strings of LEDs versus 5 strings. So, if you have 10 mA through each string, it will bw 40 mA versus 50 mA. Also, there will be 4 resistors rathen than 5 resistors.

It depends upon the capacity of the 9 V battery. If it has a capacity of 100 mAH, it will take about 100/40 = 2.5 Hours to discharge at 40 mA.

I copied this from another post in the Electronics Chat forum.

"Led resistor calculator" here is one of the many links
returned by the search engine "Google"

**broken link removed**

something like this...

The 555 circuit below is a flashing bicycle light powered with four C,D or AA cells (6 volts). Two sets of 20 LEDs will alternately flash at approximately 4.7 cycles per second using RC values shown (4.7K for R1, 150K for R2 and a 1uF capacitor). Time intervals for the two lamps are about 107 milliseconds (T1, upper LEDs) and 104 milliseconds (T2 lower LEDs). Two transistors are used to provide additional current beyond the 200 mA limit of the 555 timer. A single LED is placed in series with the base of the PNP transistor so that the lower 20 LEDs turn off when the 555 output goes high during the T1 time interval. The high output level of the 555 timer is 1.7 volts less than the supply voltage. Adding the LED increases the forward voltage required for the PNP transistor to about 2.7 volts so that the 1.7 volt difference from supply to the output is insufficient to turn on the transistor. Each LED is supplied with about 20 mA of current for a total of 220 mA. The circuit should work with additional LEDs up to about 40 for each group, or 81 total. The circuit will also work with fewer LEDs so it could be assembled and tested with just 5 LEDs (two groups of two plus one) before adding the others.

Original scheme edited by Bill Bowden, **broken link removed**

The calculator thing doesn't work for me because I have 10 red LEDs. If they are 2 volts each, I need 20 volts, but am only using a 9 volt battery.

The 40 alternate LEDs look cool. I was actually wondering how to make them alternate for another project I was thinking about doing, but that's a little complicated for me. All 40 are powered with only 6 volts?

sjaguar13 said:

All 40 are powered with only 6 volts?

Click to expand...

Yes, they are arranged in pairs, this is fine off 6 volts. Your 9V supply should be able to feed them in three's in series with a suitable resistor.

If 20 in pairs can be lit with 6v, why don't I just lit 5 pairs with my 9v? Wouldn't that be easier on the battery?

sjaguar13 said:

If 20 in pairs can be lit with 6v, why don't I just lit 5 pairs with my 9v? Wouldn't that be easier on the battery?

Click to expand...

No, it's heavier on the battery, you want them in as long a string as possible to avoid excessive waste (with suitable series resistors) - but 5 pairs (10 LED's) would take exactly the same power as 5 triplets (15 LED's) - 50% more light output for exactly the same power consumption.

With the pairs the extra power is wasted in the series resistors, the triplets waste less power as less voltage is dropped across the resistor, and it's a smaller value (and in the case of 10 LED's you would have less resistors).

Obviously for 10 LED's you have the problem that it's not divisible by three, do you need to have ten - or will 9 or 12 be OK.

There's 10 holes where to LEDs stick out. There will be an empty hole with 9 and no where for the extra 2 if I go with 12. So I guess a pair of 3 and a pair of 2 is the best way to go.

If 40 LEDs could be powered with the 6v, could I use 4 AA batteries and have them stay on longer?

sjaguar13 said:

There's 10 holes where to LEDs stick out. There will be an empty hole with 9 and no where for the extra 2 if I go with 12. So I guess a pair of 3 and a pair of 2 is the best way to go.

If 40 LEDs could be powered with the 6v, could I use 4 AA batteries and have them stay on longer?

Click to expand...

What battery are you planning using now?.

Right now, it's all set up for a 9v.

I was getting ready to re-wire the LEDs, but ran into another problem. I have the 555 timer and everything soldered to a resistor, and a wire from that running into the box. If I have the wires for the parallel from that, would I still need a resistor for each series, or since it already went through a resistor, is that good enough. Would I need to desolder the resistor, solder the wire where the resistor was, and then solder the resistor in front of just one set or LEDs in series? The resistor was originally put where it is because I had all 10 LEDs in one series.

You will need a resistor in each parallel leg because diodes in parallel will not share current properly.

j.

So I should take the resistor off the and move it to the right spot? Or could I just keep the resistor where it is, and add 4 other resistors?