# Modulator Extinction Ratio (optical pulses)

1. Mar 13, 2008

### n0_3sc

I am modulating a laser with 400ns pulses at 1MHz. Modulator is a Mach Zehnder.
Can anyone suggest a way of measuring either the peak power of my pulses or the extinction ratio of the modulator.

2. Mar 14, 2008

### Staff: Mentor

3. Mar 14, 2008

### n0_3sc

I havn't heard of a Laser Calorimeter but the pics make it look like a broadband thermal power meter.
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. Apr 2, 2008

### Uraptor

This measurement is fairly straight forward if you have an optical power meter and a high speed amplified optical detector that you can connect to an oscilloscope. The optical detector needs to be DC coupled.

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. Apr 2, 2008

### n0_3sc

Hey Uraptor,
The method I was suggested was too measure the difference in voltage from 'high' to 'low' on the oscilliscope $$V_1$$, and since I know my high speed optical detector's Responsivity $$R$$ and the relation:
$$I_{out} = RP_{in}$$
where $$I_{out}$$ is the current generated by the photodetector.
Then $$I_{out} = \frac{V_1}{Z}$$
where Z is the 50ohm impedance matched connection.
Substituting should give me the peak power $$P_{in}$$, right?

I am however curious about your method. You quote having to use a "high speed amplified optical detector", but wouldn't the amplification give a larger than expected peak power? I also find it hard to believe that the responsivity plays no role in your calculation.
Also, could you please explain how you established that formula for the peak power?

6. Apr 5, 2008

### Uraptor

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.

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7. Apr 6, 2008

### n0_3sc

Thanks Uraptor. That pdf was very helpful.
When my experiments are all up and running again, I will try both techniques for peak power measurement and list my result.

8. Apr 6, 2008

### n0_3sc

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

9. Apr 9, 2008

### n0_3sc

So i've been trying to measure P0 (The DC power level).
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. Apr 9, 2008

### n0_3sc

This is the voltage level on my high speed oscilloscope.

11. Apr 10, 2008

### Uraptor

Remember

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

12. Apr 10, 2008

### Uraptor

Can you turn down the modulation level to a low extinction ratio just to verify that your set up is working?

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

13. Apr 10, 2008

### n0_3sc

The extinction ratio of the setup I had was supposedly >>15...
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.