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borgqueenx
10-07-2011, 06:34
At the laser aparture the total output of my 650nm laser is 150mW according to my laserbee. i made a round in the lasershow editor so small it fitted the thermopile head of the laserbee LPM.
just cleaned the galvo's with first contact.
Can it be the dichro's destroy this big power?

If i put the LPM at the direct laser aparture of the 650nm module i get 810mW...

EDIT: just also tested before the galvo's. 150mW omg...
there is some leakage on the wooden case though...but not enough to make it burn a bit or hurt my finger.

Ideas?

-bart-
10-07-2011, 06:41
Does the beam fit your galvos exactly ?
Or does it spill over when scanned?

810 mw can't be right, be advised that other heat sources (such as your fingers) can influence your thermopile !

Jem
10-07-2011, 06:52
I guess first of all we need to know what the laser power is as specified as by the manufacturer. For instance, should the red laser have a specified power of 500mW, 800mW or perhaps a 1000mW?

Also, you can't accurately measure a modulated beam. Any pattern you project will be modulated and have blanking. To measure output accurately you will need something like Stanwax's 'ILDAGem', this will allow you to supply your lasers with an unmodulated 5 volt signal. Other than that in LD2000 (if you have it) you can go into the abstract generator and project a white circle, reduce this down to a single spot and aim it at your thermopile. This is the second best option.

Once you know you are dealing with an unmodulated beam we can work from there to diagnose your problem :)

Cheers

Jem

borgqueenx
10-07-2011, 06:55
guys i have a working LPM and never touched the spot where the beam hits on with my hands. it can have some dust on it though. But again i LPM'd the red laser before it hits the galvo's...150 mW. and when i tested the laser before building it in, it had 810mW.

According to lasever it should be 1000mW but for today im happy with 810mW....but ofcource i would rather have 810mW at beam aparture then at the laserhead.

@bart it spill just a very tiny bit. i do not believe this is the cause...

gashead
10-07-2011, 07:00
I'd measure this without any modulation to the head.. ie: making a small spot in your software could be what is reducing the power.

Ideally you need to remove the modulation connection from the head and then provide it +5v (full on) or find a way to make the head come up without any modulation input.

Just an idea. :)

N.

(What Jem said.) (learn to read other posts Nige)

badger1666
10-07-2011, 07:04
shove +5v on the red modulation input, it the only way to get a proper measurment of the beam on your power meter
as jem says is no good trying to get a reading off a modulated output..


oops as nige says :o while i was writing this he had replied

borgqueenx
10-07-2011, 07:08
okay il try that later but then come on guys....If i modulate it it loses SOOOOOO much power according to my LPM? that cant be right?
Or is modulating losing like 7/8 of the mW's normal when measuring?

borgqueenx
10-07-2011, 07:09
okay il try that later but then come on guys....If i modulate it it loses SOOOOOO much power according to my LPM? that cant be right?
Or is modulating losing like 7/8 of the mW's normal when measuring?

EDIT: will changing the modulation to TTL make the measurement any better? then it can only be 100% turned on or off?
ofcource measuring before it hits the galvo's?

gottaluvlasers
10-07-2011, 08:31
Ask Lasever what the input impedance of their drivers are. (or check it yourself). It should be an absolute bare minimum of ~100K ohms. Anything less than that and the input will never drive the modulation input to reach 5V. This is/was a problem with many chinese Lasers. Some impedances were so low they would only allow about 2V-3.5V max.

If it is not, it is an easy fix for you. Pretty much an op-amp and a supply and about 10 minutes of making a circuit.

But also, do as instructed above. Let us know what the UNMODULATED output power is.

-Marc

Solonar
10-07-2011, 09:18
When the beam is modulated, the LPM will only show the average of the on/off cycles. You are still peaking at high values, but the thermopile wont register the peaks as it takes time to equilibrate.

Its like measuring the DC component of your wall socket. The sine wave will average out to 0V.