I am new to electronics and I wanted to make a LED dimmer. I have connected 2 LED and a 100 ohm resistor in series. I have connected 24 of those [2 LED and resistor] in parallel to the output pin of the 555 timer. The total LED configuration only draws about 100mA so I thought of driving those LED directly using 555 timer. Here is the circuit diagram.

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I plan to drive the LED configuration using 4 1.5V AA batteries in series making total of 6V. As I did that, the LED configuration drew only about 0.25 mA of current, at the voltage of 4.33V. The LED configuration was very dim due to this reason. (please ignore the cheap equipment and the mess)Output PIN Voltage of 555 timer

The above picture shows the Output pin voltage of 555 timer. It is 4.33 V and the current is 0.25 mA shown in below picture.

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I tried to power the same LED configuration by using a 12V 18A power supply. The voltage on the 555 timer output pin was around 3.8V and the current draw by the total configuration was around 73mA. The LED glowed very bright.

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Above picture shows the Voltage of output pin of 555 timer and the below picture shows the current draw by the total LED configuration when powered by 12V 18A Power supply.

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I don't understand why there is different current draw by the same LED configuration even though the voltage of the 12V power supply is lowered by 555 timer. What would be a good way to dim my LED configuration using 555 timer and 4 1.5V AA batteries? Is it necessary to use MOSFET to dim this configuration? please enlighten me! Thank you for your time!

  • \$\begingroup\$ I don't understand why there is different current draw <-- different current draw to what? \$\endgroup\$
    – Andy aka
    Mar 1 at 9:41
  • \$\begingroup\$ by the LED configuration \$\endgroup\$
    – izack
    Mar 1 at 9:44
  • \$\begingroup\$ Welcome! Please simulate your circuit. \$\endgroup\$
    – winny
    Mar 1 at 10:13
  • \$\begingroup\$ Some versions of the 555 might not have a very strong output pin. 100mA is quite a lot Iif the signal is only designed to activate a logic gate. I found this Texas Instruments datasheet - their NE555/NA555/SE555/SA555 chips can power 100mA on the output pin, but maybe yours isn't the exact same chip. \$\endgroup\$
    – user253751
    Mar 1 at 10:32
  • \$\begingroup\$ @winny how do i do that? \$\endgroup\$
    – izack
    Mar 1 at 10:37

2 Answers 2


The fact that you only get 4.33V on the OUT pin tells us your 555 chip isn't able to output so much current and keep the voltage up. You might need to use a transistor (BJT or MOSFET) to drive more current.

There are many manufacturers and variants of 555 chips, so I can't just Google the datasheet and find out the current limit for your chip. (I did Google it, and I found a Texas Instruments 555 chip which says it can output 100mA, but evidently this isn't the one you have because your one can't.)


You never say what LEDs you have and what are their specifications.

But it is clear that they are blue LEDs and your circuit then behaves as expected.

Blue LEDs are rate at about 3V or even higher.

The two LEDs in series then require 6V or more supply voltage.

So a 6V battery does not have enough voltage to turn on two blue LEDs in series very brightly.

The difference is that if you use a 12V supply, the output of NE555 has either 12V or 0V, as it outputs pulses. It can't have any other value even if your meter averages it to some value like 7V. The 12V is enough voltage to drive the LEDs.


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