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Newbie question about LEDs in Series/Parallel

Hi all. I apologise in advance for the extremely basic nature of this question.

Simply put I want to set up a circuit where I press a switch and a bunch of LEDs light up. Riveting I know.

My question is how many LEDs can I realistically run off 4 AA batteries (6 volts). From my Googling it would seem to me that, assuming a 2v drop value for the LEDs, I could run 3 LEDs in series without any resistors.

Having come across this site http://ledcalc.com/ which is telling me I would need to wire two sets in parallel, one of 2 LEDs and one of 1 LED.

In short I am asking what would be the optimal way and number of wiring up these LEDs. The switch would be on for short bursts, not continuous usage.
 
Nobody makes an LED with a voltage that is 2.0V, they have a range of voltage from 1.8V to 2.2V for a red LED. Some are 1.8V, some are 2.2V and some are in between. You get whatever is random. If you put three 1.8V LEDs ((they use 5.4V) to a new 6V battery then they will instantly burn out because they are not light bulbs that are simple resistors, instead they are diodes that draw a HUGE current if the voltage is a little higher than their random voltage drop.

Since LEDs are not white-hot resistors, they do not glow orange when the voltage is low, they simply do not light up. So you must calculate how low the battery voltage can drop to. If you have three 2.0V LEDs connected directly to a 6.0V battery then they will light. When the battery is used for a few minutes then its voltage has dropped to 5.6V then the LEDs might not light. A 6V battery will drop to 4V or less during its life.

Example: Two 1.8V LEDs in series with a 100 ohm resistor powered from a 6.2V new battery produces 26mA, bright. If the LEDs are actually 2.2V and the battery has dropped to 4.8V then the current is only 4mA, fairly dim.
If you test all your LEDs and do not use the ones that are more than 2V and replace the batteries when the total is 4.8V then two LEDs in series with a 100 ohm resistor will be fine.
 
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