photodiode in parallel with resistor and capacitor

kum111

hello ,

I am trying to detect a laser pulse with a photodiode , having a power of 100w max of the laser beam with pulse wave of 1microsecond . when i see the output in ocilloscope the wave do not have 1 microsencond this is without any resistor or capacitor. so my question is that what happens if there is a resistor in parallel to photodiode or a resistor and capacitor in parallel to photodiode . any information about the topic will be useful . thank you all

carlgrace

The resistor is probably going to load the photodiode. You should try a transimpedance amp.

kum111

but the wave which i measure at the load resistor do not have the same time as the laser pulse .. how could it be adjusted . laser pulse has 1 μs ... but here i just 1ms pulse per cm in oscilloscope with voltage of .5v . i have read about the transimpedance amp which is used for current to voltage . thanks for reply

carlgrace

the good thing about the transimpdance amp is that it puts a virtual ground at your photodiode, so all the photodiode current goes to the output. You probably have some time constant related to your resistor and capacitor that is slowing your system. Remember that your scope probe has a capacitance unless you tune it out.

berkeman

To measure a fast pulse like that, you need to use a fast photodiode+amp, preferably an integrated unit. Something like these:

http://www.osioptoelectronics.com/st...-overview.aspx

You can find more examples with a Google search on fast photodiode amp.

Since you aren't coupling all 100W of laser power into your photodiode (at least not more than once!), how are you splitting off some of the beam and coupling it to your photodiode? Modules are available with different coupling means (like fiberoptic, windowed, etc.).

kum111

hello carlgrace ,

Then how do i control this time constant , i will use the trans amp later if needed . first i have to find whats the effect of this only resistor in parallel , or resistor and capacitor in parallel to my photodiode . The scope probe capacitance changes with the x10 , x1 is it, am a computers student , i haven't worked much with this electronics equipment . thanks for reply

NascentOxygen

Why don't you try it for yourself? Place various resistances in parallel with your oscilloscope input and see whether the pulse is cleaned up. Start with 22MΩ, and work your way down. Tell us how you go. (I don't know how well this will work. Experiment.)

kum111

hello berkeman,

am using a si pin photodiode will has a pretty fast response , there is a glass (window may be am not sure )which is fixed in front of the laser to make it fall on diode .

kum111

i have done this and got the Gaussian shaped pulse to 50μs , which in laser is 1μs with not much of a variation on voltage with a 300Ω. i think there is sum effect on the charge time and discharge time of the diode , but am not sure what am getting here .. thank you

kum111

hello all ,

The datasheet of the photodiode says that a rise time and fall time of 100ns with Vr=10v , Rl=1kΩ , 820 nm . it i need fast rise time will this be only achieved in reverse bias ?

Integral

I have made power and pulse width measurements on a a 30W Coherent AviaX laser. We directed the beam onto a power puck and placed our PIN diode in the light reflected from the power puck to get Pulse width. If you indeed have a 100W laser then you may need to do something similar.

kum111

Here the laser has a 100w max power for the laser beam which is 70mm long and 2 mm wide . similar to you i have to measure the power and pulse of the laser to calibrate . so do you have anymore information will be helpful , thanks for reply

kum111

got it with 1k resistor and 10v of reverse voltlage I got pulse 1μs which was needed . thanks guys

Redbelly98

As an additional check, you should probably try some lower resistors and see that the pulse width is still 1 μs. (If it becomes less, then you were not measuring the true width at 1 μs.)

kum111

any information on integration of output from phototdiode will be helpful

Redbelly98

It's not clear, to me at least, what you are asking. What do you mean by "integration"?