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For testing various laser diodes, I need a constant current source, adjustable up to 150 ... 200 mA or so, down to... well, maybe 1 or 2 mA? (low enough to measure where is the lasing threshold where the diode stops lasing and becomes an ordinary LED). The voltage on these diodes when lasing is pretty low, so I could power the whole thing from a 9 V battery, I guess.

I could do the classic simple design with a LM317 and a resistor, but there's something I don't like about it: the resistor (control pot) that controls the current is in the main current path. I don't want that. I want the control pot to be in a section of the circuit where the current is very low (way, way under 1 mA), so the power dissipated on the pot is minimal, so I could use a higher-resistance pot, and the pot stays healthy a long time.

The reason why I need to keep the pot healthy: with laser diodes it is very important to not have any kind of voltage/current spikes on the junction, because it's so incredibly easy to kill them (*) so a few small 100 nF capacitors should be placed at strategic places in the circuit, such as on the control pot and on the output (I could do those tweaks myself, I just need the basic idea for the circuit, I can only envision complex designs - I need something simple and small).

Any suggestion for a very simple circuit that doesn't suffer from the problem described?

(*) - It's so easy to fry them that most people never disconnect the diode from the driver (the constant current source). They only disconnect the driver from the battery. Otherwise the risk of bricking the laser is too high. I fried that way a very nice 100 mW blue laser, a few months ago.

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This is about as simple as it gets - and it meets your spec.
Replace zener with variable reference voltage. Transistor shown as bipolar can be FET.

enter image description here

That circuit comes from guess where :-)


Another.Diagram a bit small. Original was missing. Does high side monitoring which can be nice.

enter image description here

Our friends at Silicon Chip say at

  • Basically, this circuit is a conventional switchmode regulator adapted for constant current output and is specially designed for stepper motor drivers - although it could be used for other applications as well. The circuit works as follows: IC1 (LM2575T) and its associated components (D1, L1, C1, etc) operate as a switchmode power supply. Normally, for constant voltage operation, the output is connected - either directly or via a resistive divider - back to the feedback input (pin 4) of IC1.
  • In this circuit, however, Q1 senses the current flowing through R1 and produces a corresponding voltage across R3. This voltage is then fed to pin 4 of IC1. As a result, the the circuit regulates the current into a load rather than the voltage across the load. Only one adjustment is needed: you have to adjust VR1 for optimum stepper motor performance over the desired speed range. The simplest way to do this is to measure the motor current at its rated voltage at zero stepping speed and then adjust VR1 for this current. The prototype worked well with a stepper motor rated at 80O per winding and a 12V nominal input voltage. Some components might have to be modified for motors having different characteristics.

enter image description here

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  • \$\begingroup\$ Shouldn't this have some kind of smoothing between the switching and the diode? \$\endgroup\$ – endolith Sep 30 '11 at 2:02
  • \$\begingroup\$ @endolith - Probably. Note that he noted "so a few small 100 nF capacitors should be placed at strategic places in the circuit, such as on the control pot and on the output (I could do those tweaks myself, I just need the basic idea for the circuit, I can only envision complex designs - I need something simple and small)." \$\endgroup\$ – Russell McMahon Sep 30 '11 at 2:11
  • \$\begingroup\$ The second diagram feeds the reference current through the load - not a good idea if you want accurate currents down to a mA. \$\endgroup\$ – Wouter van Ooijen Sep 30 '11 at 8:13
  • \$\begingroup\$ @Wouter van Oojen - Potentially true. An idea starter - i thought he'd like to see a high sense circuit - not so common but can be useful. An eg TLV431 will need under 100 uA so say about 1% error from this cause at 10ma). Causes Say 10% drop in current at 1 mA. \$\endgroup\$ – Russell McMahon Sep 30 '11 at 8:58

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