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I am doing a project where I have three Laser LEDs. They need to send three different signal lengths. I am doing this with three NE555D Timer, one for each signal.
The problem is: I am not coming to the needed current (25mA) for the LED.
Should I do it with a Transistor? How should I change the circuit?

Thank you :) enter image description here

At Trig I am using a button. At Output (Out) I am not getting the 25mA.

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  • \$\begingroup\$ We don't know what your circuit looks like so we don't know what changes it would need to drive your LEDs that are unknown too. NE555 chips can generally go to 200mA so the 25mA current cannot be the issue. \$\endgroup\$ Commented Dec 16, 2025 at 10:33
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    \$\begingroup\$ How long are the pulses, at what duty cycle / frequency, and how are you measuring the current? \$\endgroup\$ Commented Dec 16, 2025 at 10:34
  • \$\begingroup\$ What is the reason for R1? All current to operate the circuit and light the LED will have to pass through R1 - with a 9 volt supply, R1 will allow a maximum of 0.13 mA. \$\endgroup\$ Commented Dec 16, 2025 at 17:08

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The circuit presented has multiple issues.

First of all, it cannot work if NE555 supply does not come from battery, but via R1.

The transistor is also useless as it is just used as a diode which provides a voltage drop before the load.

The diode should not have that many volts over it, so with 9V supply the 3 ohm resistor can already drive more than 1A current which exceeds NE555 output and the diode current.

Depending on what the supply or battery is, it might be incapable of producing the current you want.

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  • \$\begingroup\$ I have a 9V battery. \$\endgroup\$ Commented Dec 16, 2025 at 15:29
  • \$\begingroup\$ I would take a look at some published manufacturers application circuiits. If you asked an AI to help you , it let you down badly. Or you transcribed this from a bad photocopy. The original bipolar NE555 can deliver 200mA from its output pin. The modern CMOS verision somewhat less. But only if its Vcc power pin is connected to the power supply rather than somewhere inside the triming circuit. \$\endgroup\$ Commented Jan 15 at 16:03

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