I want to make a circuit with 20 LEDs here are the specs of the LEDs that I will be using:
Forward voltage:2.5V
Forward current max.:25mA
Reverse voltage max.:5V

I want to use two AA batteries to power the LEDs I have entered all the details into LED Center calculator. Here is what they said I would need to make the circuit:

Solution 0: 1 x 10 array uses 10 LEDs exactly

+3V

R = 22 ohms

R = 22 ohms

R = 22 ohms

R = 22 ohms

R = 22 ohms

R = 22 ohms

R = 22 ohms

R = 22 ohms

R = 22 ohms

R = 22 ohms
The wizard says: In solution 0:

• each 22 ohm resistor dissipates 13.75 mW
• the wizard thinks 1/4W resistors are fine for your application
• together, all resistors dissipate 137.5 mW
• together, the diodes dissipate 625 mW
• total power dissipated by the array is 762.5 mW
• the array draws current of 250 mA from the source.

Will I have to use that many resistors? will the two AA batteries be enough?

Many Thanks.

 

It is advised to use resistors with leds.
But 2 AA batteries can be used with certain leds without resistors.
Better to use resistors

 

An important but omitted parameter is the forward voltage. The calculator seems to "think" or was told to expect a forward voltage of about 2.5 volts, and allocates 2.5 of the 3 volts from the 2 x AA cells to the LED and another .5 volts for the resistor.

If you want to use half the resistors, you would need to increase the voltage, for example, 4 x AA = 6 volts so you could make five parallel strings of two LEDs (total of 5 volts) and one resistor (1 volt), thus cutting the number of resistors in half. You can still use 1/4watt resistors.

If you increase the voltage further, you can get down to one resistor when all of the LEDs are in series.

An alternative is to use a voltage converter that would boost the voltage from your battery pack to a higher voltage, but then it starts getting complicated.