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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.

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    \$\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 '18 at 16:05
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schematic

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.

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  • \$\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 '18 at 16:44

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