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schematic

simulate this circuit – Schematic created using CircuitLab

My circuit is like this, 40 lines of R - LED - LED (total 80 LEDs). When the count of LEDs is less, LED's brightness is stable. As the count of LEDs increases brightness becomes unstable (But brightness of all LEDs is same at any point). So I measured voltage around resistor when the count of LEDs is many, and it was also unstable. Why this occur?

Edit: Power supply is Apple iPad(second model) Charger. LED's color is red. 'unstable' here means 'brightness goes up and down'. There is no shutdown. And unstability is not severe. I can observe it only when in darkness(all other lights off) and the front of LED.

Edit(2): I tried other power supplies(phone charger(5V 1.2A), 3x AA alkarine battery) but result is same

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    \$\begingroup\$ What kind of power supply do you use for your 5V? \$\endgroup\$ – Edesign Jul 19 '16 at 9:33
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    \$\begingroup\$ What do you really mean by "unstable"? Does it mean that it is lower than expected and the LEDs are dimmer or does it mean that it varies randomly, making the LEDs flicker? And I second @Edesign: please, provide the specs of the power supply you are using. \$\endgroup\$ – Lorenzo Donati Jul 19 '16 at 9:45
  • \$\begingroup\$ Power supply is Apple charger for iPad(second model). And 'unstable' means 'somewhat flicker'. \$\endgroup\$ – firia2000 Jul 19 '16 at 10:23
  • \$\begingroup\$ Please edit your question to include this information. Additionally, what color LEDs are you using? \$\endgroup\$ – user2943160 Jul 19 '16 at 10:54
  • \$\begingroup\$ LED's color is all red. \$\endgroup\$ – firia2000 Jul 19 '16 at 10:57
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Simple: Because your power supply can't reliably deliver the current for the high number of LEDs.

The current through a single LTL-307EE is, according to the datasheet of that LED, 20 mA. Hence, 40 branches will draw a current of 0.8 A.

Note that the resistor you're using isn't really the optimal choice for maximum brightness, here – at 20 mA, it should drop 1V, but instead, it drops 20 mA * 100 Ohm = 2 V, so you end up with significantly less current – it's hard to say, because the I/V curve of a diode is nonlinear, but you're probably getting half as much brightness as you could. Try using 50 Ohm! (Hint: two parallel 100 Ohm resistors form one 50 Ohm resistor)

When you draw more current than what your supply can deliver, power supplies must drop their voltage – that's a physical necessity.

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  • \$\begingroup\$ I did not intend max birghtness because it was too bright. I intended 10mA so selected 100 Ohm resistor. My power supply is Apple charger for iPad(second model) and it is 5.1V when I measure power supply with no load, 4.7V with 80 LEDs. \$\endgroup\$ – firia2000 Jul 19 '16 at 10:27
  • \$\begingroup\$ you're probably doing less than 10mA, because a diode is not a resistor. \$\endgroup\$ – Marcus Müller Jul 19 '16 at 10:28
  • \$\begingroup\$ I'd assume that a USB charger would deliver <0.8 A reliably – but, this is apple hardware, so I can't say whether they're in a low-power mode until the device has authenticated or so. The flickering points to something getting overloaded and frequently shut down. Simply try a different power supply. It shouldn't be hard to find a cheap 5V USB charger. \$\endgroup\$ – Marcus Müller Jul 19 '16 at 10:30
  • \$\begingroup\$ OK, I'll try another power supply. But regardless of that, switching regulator (DC to DC converter) can help? \$\endgroup\$ – firia2000 Jul 19 '16 at 10:35
  • \$\begingroup\$ I don't understand your question? Yes, any 5V power supply you'll find cheaply nowadays is a switch mode power supply. No, adding a DC/DC converter to your iPad charger won't help, probably, because it's more complex and costly to do so than to get a better power supply, and all it could ideally do is save 20–30% of the energy that your circuit uses, if it had a really high efficiency, by getting rid of the resistors – but really, that's not a desirable solution here. \$\endgroup\$ – Marcus Müller Jul 19 '16 at 10:40

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