# Trigger PIC-Laser

I'm having trouble to figure out one way to trigger one laser, the signal must come for one dsPIC that have 3.3V output. In my circuit I have power supply's of 3.3V,6V and 12V. The problem is that the laser works with 6V and consumes 100mA. So i don't know what can i do to preserve the 6V,because if i use transistor I lost like 0.7V.

Some ideas how can i do this circuit?

• Can we have a part number for the laser please? – Oli Glaser Feb 5 '13 at 16:14
• the laser part number is DLSC-660S-50-SD – Daniel Seerig Feb 5 '13 at 16:19
• I'm not sure where you are getting a loss of 0.7V from using a transistor. A BJT can usually saturate with a collector-emitter voltage of about 0.2V, and you can get a MOSFET with an even lower voltage drop. – Phil Frost Feb 5 '13 at 16:36

## 1 Answer

Okay, according to the datasheet and instruction manual, there is a power supply lead and a modulation input lead. This means you don't need to switch the power directly.

• Note on below info - see the comments on supply voltage and make sure your device can operate with both supply and modulation inputs up to 6V. If you are not sure read the datasheets carefully, if there is no solid info in those e-mail the manufacturer. If you get no reply it may be safer to run from a 5V rail.

The voltage range is 5-6V, so connect the power leads to your 6V supply, and connect the modulation input to the collector of transistor switched from your dsPIC something like the circuit below:

Your laser is the S variety, so it's off when the input is high:

• Synchro option (S): Laser is ON if modulation signal is at 0.0V, Laser is OFF if modulation signal is at 5.0V; Maximum frequency is at least 10KHz (except for the Green Laser, generally less than 100 Hz)

Since the circuit shown inverts, this means that when your dsPIC input is high the laser will be on. The transistor can be any general purpose NPN, and the resistor values are not fixed either.

Also, a note on the "0.7V loss" from the transistor. I think you are confusing the base emitter drop (which is around 0.7V typically) with the typical collector-emitter saturation voltage (which is <200mV typically) so even if you were switching the power directly it wouldn't be an issue as long as you provide enough base current to switch the transistor on fully. As Phil says in the comments, a good MOSFET can be even lower as you can easily achieve a typical on-resistance of e.g. < 100mΩ (which at 100mA would be 10mV drop) Here is an example part.

• Note the text you quote explicitly says the digital logic level for "off" is 5 V, not 6 V. Applying 6 V like your circuit does could be violating the ratings of the modulation circuit inside the laser module (although the linked datasheet doesn't seem to give any information about this). – The Photon Feb 5 '13 at 16:56
• @ThePhoton - Yes I noticed this - the datasheet(s) seem to vary between 6V and 5V for the optimum voltage. It said "depending on part" somewhere, but didn't say which part. I think all parts have a range of 5-6V for supply if I read correctly, but it gives no details on the modulation input (although I'd assume this would be e.g. a max of VDD + 0.3V) The OP needs to thoroughly read the (not very helpful at a glance) datasheets and if necessary contact the manufacturer. I'll make a note in the answer. – Oli Glaser Feb 5 '13 at 17:00
• oh thank you,yes I was looking and really confuse me about the 0.7V. But that is what I need. Thank you guys – Daniel Seerig Feb 5 '13 at 17:03