I need to drive a 650nm cheap 5mW laser diode.
It seems to have a current limiting resistor in series, but the laser still got damaged. I powered it directly from an 18650 at 4V and it was working fine. I left it on for an hour or so, but it was in a 3D-printed shell that I had made to position it. After this, the output of the diode became faint, even at the same voltage. I guess the plastic shell I made contained the heat the diode produced and let it runaway, I have no idea if such a small diode can even do that. I have connected a bare diode without the plastic shell to another 18650 and let it run for nearly 12 hours now, the voltage across the battery has dropped about 0.2 volts, but the diode's output doesn't look like it has changed. I will try to do the same with the shell and report back. I found a constant current circuit as in the image below, and I have a few questions:

1. Is this how the circuit works? The Zener diode provides a near-constant voltage, and as the resistance is fixed, the current flowing through the base of the transistor must be fixed; thus, the current flowing to the laser diode must also be fixed. Correct me if I'm wrong.
2. If it works as I think it does, then I can just change the resistances to fix the constant current. I also want this to run on 18650s, so the supply voltage would be around 4V, and I got 3.3V zener diodes. Will it work that way?

The reason to power the laser at all is to make a laser obstacle course with like 25 lasers. It only needs to work for a day, but I don't want my diodes to fry. If this circuit seems necessary then I will have to cook up 25 of these, which I think I can do. But I also want to know if this is overkill or useless.

I know this is one of the longer reads, Thanks much in advance to anyone who responds.

 

Switch the positions of the zener and the 1K resistor.

 

Switch the positions of the zener and the 1K resistor.

No don't do that: that would make the situation worse, and the current would vary by MORE than the variation in supply voltage. The circuit is already correct, but the explanation is not quite right,

The zener fixes the voltage at the base. The voltage at the emitter is 0.6V higher, so the voltage across the 220Ω resistor must always be Vzener - Vbe, about 4.3V. The current in the collector is the emitter current minus the base current, the base current is the emitter current divided by Hfe, which you can ignore if Hfe>50 as it will make less than 2% difference to the output. therefore, the collector current = 4.3V/220Ω
It will remain in regulation provided that Vzener+Vload < V supply

 

Here's two circuits to show how a zener can be used with either a NPN or PNP transistor configured as an "emitter follower" (common collector). The NPN version is more intuitive but they operate on the same principle just with inverted polarity.

In both cases, Vout = Vzener - ~0.7V.

It would be a good exercise to calculate the voltage at each node in both circuits. I highly recommend LTspice to help with this. I included the file, all you need to do is install the free software and click run.

 

Here's two circuits to show how a zener can be used with either a NPN or PNP transistor configured as an "emitter follower" (common collector). The NPN version is more intuitive but they operate on the same principle just with inverted polarity.

In both cases, Vout = Vzener - ~0.7V.

It would be a good exercise to calculate the voltage at each node in both circuits. I highly recommend LTspice to help with this. I included the file, all you need to do is install the free software and click run.

View attachment 339356

Neither of those is a constant current source. They are both voltage regulators. The TS needs a constant current circuit.

 

No don't do that: that would make the situation worse, and the current would vary by MORE than the variation in supply voltage. The circuit is already correct, but the explanation is not quite right,

The zener fixes the voltage at the base. The voltage at the emitter is 0.6V higher, so the voltage across the 220Ω resistor must always be Vzener - Vbe, about 4.3V. The current in the collector is the emitter current minus the base current, the base current is the emitter current divided by Hfe, which you can ignore if Hfe>50 as it will make less than 2% difference to the output. therefore, the collector current = 4.3V/220Ω
It will remain in regulation provided that Vzener+Vload < V supply

So i guess ill need higher supply voltage. Thanks a bunch

 

So i guess ill need higher supply voltage. Thanks a bunch

Or a lower voltage zener.

 

Here's two circuits to show how a zener can be used with either a NPN or PNP transistor configured as an "emitter follower" (common collector). The NPN version is more intuitive but they operate on the same principle just with inverted polarity.

In both cases, Vout = Vzener - ~0.7V.

It would be a good exercise to calculate the voltage at each node in both circuits. I highly recommend LTspice to help with this. I included the file, all you need to do is install the free software and click run.

View attachment 339356

I've heard about LTSpice, this gave me a strong reason to use it. Thanks

 

But still one thing remains. Is this much better than leaving it at just the built in series resistor? Does this circuit make a significant change in this scenario. I'll try to measure it too once I get an accurate enough ammeter.

EDIT: In case I choose to use the circuit, it is the 1K resistor that I will have to tune to get the right current right?

 

Neither of those is a constant current source. They are both voltage regulators. The TS needs a constant current circuit.

Oops. Here are constant current circuits:



Here we have Iout = Vzener - ~0.7V / R2

The difference between these and my previous circuits is that the collector (load) current is constant because all the transistor current must pass through R2. This feedback loop biases the base so the output current is fixed regardless of the load. A subtle but important detail.

 

EDIT: In case I choose to use the circuit, it is the 1K resistor that I will have to tune to get the right current right?

No, the 220Ω resistor determines the laser diode current.

 

Below is the LTspice sim of a two-transistor, low-voltage-drop, (near) constant-current circuit:

 

You can also operate the laser diode at less than the maximum rating!! A lower voltage and thus a bit less current will reduce the heating a fair amount. Simple, easy, and fairly reliable. But you may need to do a bit of math, or some experimenting. Laser diodes are not linear devices.

 

This is a neat circuit from Bob Pease.
The temperature coefficients of Vbe and the Schottky diode cancel out giving a current source that is stable with temperature.
The output current is about 0.4V/R1