I've got a small red laser pointer from a chinese ultrasonic distance meter, which has a small regulator circuit (6mm x 7mm) with SMD components.

I guess that is a constant current to drive the laser, but I was unable to understand the circuit.

laser circuit

I've used a microscope and a continuity tester to check all connections several times, as I was thinking I got some wire incorrectly attached. But I was not able to find any error.

The transistors are labeled L6 in a SOT-23 package, and I assume they are KST1623-L6 or 2SC1623-L6, both similar NPN transistors. I was not able to measure the capacitor without desoldering it, and I'm not confident in being able to take it apart and put it back being so small (~ .5mm x 1.4mm).

I powered the circuit using my Arduino regulated output at 3.3V during about 10 seconds, and the laser lights up, without any overheating observed.

I don't understand how Q2 works, as the base is attached to GND with 39 Ohms resistor, so as far as I know, ideally, it should be in cutoff mode. I'm curious how this circuit can work and why they decided to use it.

  • 1
    \$\begingroup\$ Can you post a photo of the PCB? It almost looks like a constant current "sink" driver but I would expect a resistor in the emitter of Q1. Having R2 connected to GND as shown doesn't make any sense. \$\endgroup\$
    – Transistor
    Aug 25, 2018 at 16:05

1 Answer 1



simulate this circuit – Schematic created using CircuitLab

Figure 1. A possible correct version of the circuit.

Have another look and see if it's actually the circuit of Figure 1.

How it works:

  • R1 provides bias to Q1 which starts to turn on. D1 starts to light up.
  • When the D1 current through R2 reaches about \$ I = \frac {V}{R} = \frac {0.7}{39} = 18 \ \text {mA} \$ Q2's base will be forward biased, it will turn on and steal the bias away from Q1. (The turn-on point is 0.7 V.)
  • The circuit will stabilise at 18 mA.

This seems reasonable and will result in consistent operation across a range of battery voltage.

  • \$\begingroup\$ Yes! I find out that my cheap multimeter was measuring the 39 Ohms resistor as connected between the Q1 emitter and GND. A track under Q2 was the real connection to the base of Q2. Thanks a lot, now it makes sense. \$\endgroup\$
    – ram
    Aug 25, 2018 at 16:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.