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After my previous question I noticed that the laser diode I have is anode-grounded, meaning that the case (which is the diode's metal case and often connected to circuit ground to reduce noise, I guess) is tied to the anode of the laser diode.

Anode-grounded drivers work from a negative supply as shown below:

enter image description here

In my case I don't have a negative supply. Can I just use a positive supply if I make sure the the diode's metal case is electrically isolated from ground?

How important is to connect the diode's metal case to the ground for its protection? Should I buy a negative -5 V voltage regulator?

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  • \$\begingroup\$ Please see my previous question and particularly last comments of an answer: electronics.stackexchange.com/questions/651603/… I noticed the diode is anode grounded after the answer. \$\endgroup\$
    – cm64
    Jan 27 at 0:36
  • \$\begingroup\$ The anode can connect to a positive supply if you want but, make sure the body of the laser (that may also be connected to the anode and hence the positive supply) is not accessible to some user who might want to ground it via some optical connection. \$\endgroup\$
    – Andy aka
    Jan 27 at 10:54
  • \$\begingroup\$ @Andyaka Million thanks for the comment. The diode is grounded anode type. But one more thing wanna have your comment. If someone insists to ground the body(metal case), then can I use the circuit this way with a -5V linear regulator? : i.stack.imgur.com/kmTnj.png \$\endgroup\$
    – cm64
    Jan 27 at 12:17
  • \$\begingroup\$ Yes, that circuit looks OK but you should simulate it. Maybe the base resistor could be 100 ohms and the op-amp looks suitable @cm64 \$\endgroup\$
    – Andy aka
    Jan 27 at 12:46

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How important is to connect the diode's metal case to the ground for its protection? Should I buy a negative -5 V voltage regulator?

Employing a negative voltage regulator is the safest route when the laser diode anode is galvanically connected to its metal body. If you fed the anode from a positive bias generator and the metal body got grounded by a user making an optical connection to the laser, it would short out the laser (not harm it of course) and short out the bias. However, if this isn't seen as a big deal then use a positive bias to the anode.

If this might be a problem then use a negative supply and bias circuit as per what you posted in a comment: -

enter image description here

You should reference the place that you found the above circuit to abide by SE rules BTW. Anyway, the circuit looks good to me except I might employ a higher value base resistor. You might have to mess around with the value of C1 and R2 in case there is instability. The op-amp you picked is quite low bandwidth (200 kHz) so the integration circuit might need C1 increasing maybe 10 to 100 times. The op-amp is also only capable of a 20 mA drive current at the output so it might just run out of steam as you push the laser current towards 500 mA.

Or, you might decide to use a low-gate-threshold-voltage MOSFET instead of the BJT.

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  • \$\begingroup\$ I didnt find it I draw myself. \$\endgroup\$
    – cm64
    Jan 27 at 13:31
  • \$\begingroup\$ OK, no problemo @cm64 \$\endgroup\$
    – Andy aka
    Jan 27 at 13:31
  • \$\begingroup\$ Btw some add R C in parallel with laser diodes: encrypted-tbn0.gstatic.com/… Its a mystery to me why. Do you think for transient protection? \$\endgroup\$
    – cm64
    Jan 27 at 13:32
  • \$\begingroup\$ @cm64 a laser is a very non-linear device and, with a closed-loop feedback, the R and C might be there to stop oscillations. I'd probably put 100 nF across the laser if I were you. I mean it's not used for modulated data transmission so it won't spoil anything. Regarding the resistor, I'd be tempted to place a 10k ohm across it like the circuit you linked. And, if those components were not present AND, the circuit oscillated, then you might get unholy transient currents so, I guess you could call it a form of transient protection. \$\endgroup\$
    – Andy aka
    Jan 27 at 13:35
  • \$\begingroup\$ @cm64 are we done here now? \$\endgroup\$
    – Andy aka
    Feb 2 at 17:55

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