- #1

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Can anyone suggest a way of measuring either the peak power of my pulses or the extinction ratio of the modulator.

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- Thread starter n0_3sc
- Start date

- #1

- 243

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Can anyone suggest a way of measuring either the peak power of my pulses or the extinction ratio of the modulator.

- #2

berkeman

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http://images.google.com/images?sou...___US232&q=laser+calorimeter&um=1&sa=N&tab=wi

How much laser power are we talking about? What other detectors do you have available?

- #3

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I have a thermal power meter but that off course measures the average power. We are talking about roughly 150microwatts average laser power. They are pulsed with 400ns pulses at a 5Mhz rep rate.

At my disposal the equipment I have access too are all sorts of high-speed photodiodes/oscilliscopes/spectral analyzers etc.

- #4

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First measure the average optical power (Pavg) in uW with the optical power meter.

Set up the high speed amplified optical detector and oscilloscope. You'll probably need to set the oscilloscope input impedance to 50ohms.

With no optical input to the detector measure the DC level, call this Vzero.

Then measure the AC waveform with the high speed amplified optical detector and the oscilloscope.

The extinction ratio is; e = (Vhigh - Vzero)/(Vlow - Vzero).

To determine the peak power you first need to measure the duty cycle (DC) of the ACwaveform. Then Ppeak = Pavg/(DC + (1-DC)/e)

- #5

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The method I was suggested was too measure the difference in voltage from 'high' to 'low' on the oscilliscope [tex]V_1[/tex], and since I know my high speed optical detector's Responsivity [tex]R[/tex] and the relation:

[tex] I_{out} = RP_{in}[/tex]

where [tex]I_{out}[/tex] is the current generated by the photodetector.

Then [tex] I_{out} = \frac{V_1}{Z}[/tex]

where Z is the 50ohm impedance matched connection.

Substituting should give me the peak power [tex]P_{in}[/tex], right?

I am however curious about your method. You quote having to use a "high speed

Also, could you please explain how you established that formula for the peak power?

- #6

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Your signal is fairly small that is the reason I suggested using an amplified optical detector.

Recall that an optical power meter is used to determine the average power of the signal.

Extinction ratio = V1/V0 = I1*Z/I0*Z = I1/I0 = P1*R/P0*R = P1/P0

The detector responsivity cancels out.

Now the peak power can be determined from the average power if you know extinction ratio and duty cycle -see attached file.

Assuming the detector responsivity you have been given is accurate your method will also work. As a sanity check it's probably a good idea to make the measurement both ways. Once you're convinced that both methods work just use the easiest one.

Recall that an optical power meter is used to determine the average power of the signal.

Extinction ratio = V1/V0 = I1*Z/I0*Z = I1/I0 = P1*R/P0*R = P1/P0

The detector responsivity cancels out.

Now the peak power can be determined from the average power if you know extinction ratio and duty cycle -see attached file.

Assuming the detector responsivity you have been given is accurate your method will also work. As a sanity check it's probably a good idea to make the measurement both ways. Once you're convinced that both methods work just use the easiest one.

- #7

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When my experiments are all up and running again, I will try both techniques for peak power measurement and list my result.

- #8

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The extinction ratio is; e = (Vhigh - Vzero)/(Vlow - Vzero).

Your pdf quotes [tex]\epsilon = \frac{P1}{P0}[/tex], now the extinction ratio in the quote is a voltage ratio, so do I need to square 'e' to get the same [tex]\epsilon[/tex] as in your pdf?

- #9

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One problem: The DC power/voltage level stays the same with no input signal and an input signal. I do expect the dc level to be very very low since I'm modulating the same pulse twice squaring the extinction ratio. But I know my dc level isn't 0 because that would make the extinction ratio = infinity! Impossible!

What else can I do?

- #10

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This is the voltage level on my high speed oscilloscope.One problem: The DC power/voltage level stays the same with no input signal and an input signal.

- #11

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Remember

Extinction ratio = V1/V0 = I1*Z/I0*Z = I1/I0 = P1*R/P0*R = P1/P0

Extinction ratio = V1/V0 = I1*Z/I0*Z = I1/I0 = P1*R/P0*R = P1/P0

- #12

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Once the extinction ratio is greater than 15 it becomes quite difficult to measure accurately.

- #13

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That setup is gone because the modulator failed (possibly too high powers).

I am using a simple setup now whose extinction ratio isn't that high. I compared your technique to measuring the peak power and the technique I was told - there is a factor of 2 difference. ie. Your technique yields an answer two times greater. I don't agree with my technique 100% since the Responsivity is wavelength dependent and I just approximated it.

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