user82262: I have this circuit


And I want it to power all the led rgb (100 in total), I want to make a independent portable circuit, so I don't have to use the usb power. The thing is when I use this circuit it only power to 44 led.
WHile when I use the usb 5V it can full lights all led .


This is my code
#include <Adafruit_NeoPixel.h>


#define NUM_LEDS 100
#define LED_PIN 2
Adafruit_NeoPixel strip = Adafruit_NeoPixel(NUM_LEDS, LED_PIN, NEO_RGB+NEO_KHZ800);
void setup()
{
   

  strip.begin();
  strip.show(); 

}

void loop()
{
  

  for (int i = 0; i < NUM_LEDS; i = i+1)
  {
    strip.setPixelColor(i, strip.Color(255, 240, 255));

    strip.show();
    delay(1000);
  }
  

  
}


What may I be missing?

user313: The current available from a computer's USB port is 500 mA; some ports can supply more if the device negotiates for it.
You show AA cells in your diagram. Even assuming they're Alkaline cells, batteries differ in their capacity vs. discharge rates, and presumably in their internal resistances. Most any power-supply's voltage will sag as you draw more current from it. The likely answer is that as you add LED's the battery voltage falls below either what is needed to light the LEDs; or 2) the dropout voltage of Arduino's voltage regulator, which shuts down the Arduino (assuming you're connected to 'V_in', not '5v', which neither diagram shows).  Less-than-fresh batteries will exhibit this voltage drop sooner and to a greater degree.
This study of AA battery capacity vs. current draw might help understand the apparent loss of capacity at greater currents. But even without that phenomenon, the battery voltage will fall as (I * R_battery); capacity loss just makes it worse.

82262: <p>I have this circuit
<a href="https://i.stack.imgur.com/KdLGO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KdLGO.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/Ubypz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ubypz.png" alt="enter image description here" /></a></p>
<p>And I want it to power all the led rgb (100 in total), I want to make a independent portable circuit, so I don't have to use the usb power. The thing is when I use this circuit it only power to 44 led.</p>
<p>WHile when I use the usb 5V it can full lights all led .</p>
<p><a href="https://i.stack.imgur.com/WB6fI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WB6fI.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/GGbPV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GGbPV.png" alt="enter image description here" /></a></p>
<p>This is my code</p>
<pre class="lang-cpp prettyprint-override"><code>#include &lt;Adafruit_NeoPixel.h&gt;


#define NUM_LEDS 100
#define LED_PIN 2
Adafruit_NeoPixel strip = Adafruit_NeoPixel(NUM_LEDS, LED_PIN, NEO_RGB+NEO_KHZ800);
void setup()
{
   

  strip.begin();
  strip.show(); 

}

void loop()
{
  

  for (int i = 0; i &lt; NUM_LEDS; i = i+1)
  {
    strip.setPixelColor(i, strip.Color(255, 240, 255));

    strip.show();
    delay(1000);
  }
  

  
}

</code></pre>
<p>What may I be missing?</p>
 313: <p>The current available from a computer's USB port is 500 mA; some ports can supply more if the device negotiates for it.</p>
<p>You show AA cells in your diagram. Even assuming they're Alkaline cells, batteries differ in their capacity vs. discharge rates, and presumably in their internal resistances. Most any power-supply's voltage will sag as you draw more current from it. The likely answer is that as you add LED's the battery voltage falls below either what is needed to light the LEDs; or 2) the dropout voltage of Arduino's voltage regulator, which shuts down the Arduino (assuming you're connected to 'V_in', not '5v', which neither diagram shows).  Less-than-fresh batteries will exhibit this voltage drop sooner and to a greater degree.</p>
<p>This <a href="https://www.powerstream.com/AA-tests.htm" rel="nofollow noreferrer">study of AA battery capacity vs. current draw</a> might help understand the apparent loss of capacity at greater currents. But even without that phenomenon, the battery voltage will fall as (I * R_battery); capacity loss just makes it worse.</p>
