So we modulate the light from a diode laser by wiggling the diode current. This has two effects, the most obvious is that it changes the amplitude of the light. (AM) Less obvious (though I think more dominant in changing the frequency.) is that changing the laser current has two other effects on the frequency. At low frequencies (
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OK this copied from C. Weiman Rev. Sci. Instrum. 62 (l), January 1991 "Using diode laser in atomic physics."
One of the important advantages of diode lasers over other optical sources is that their amplitude and frequency can be modulated very easily and rapidly by changing the injection current. Unfortunately, when the injection current is modulated, one obtains both AM and FM, and these are not independent. The simplest useful picture of the modulation response of diode lasers is that the AM and FM are both present but with different sensitivities.?? Also modulation can be complicated by the fact that the relative phase between the AM and the FM changes as a function of modulation frequency. To a good approximation, the AM and the FM are linear with the injection current but the FM modulation index can be more than ten times the AM index. This means that the amplitude change can be ignored in many atomic physics applications. Thus the laser can be scanned over spectroscopic features and/or jumped back and forth to specific frequencies just by applying the appropriate modulation to the injection current. Applications of various modulation capabilities are discussed in Sec. V B.
end quote.
Well, there you go. Modulation index vs. modulation index. Two dimensionless numbers, so no funny units in the answer.
This thread reminds me of an article in an amateur radio magazine from the 1980's, about converting a microwave oven to an FM transmitter at
2.4GHz, using much the same technique -- modulate the plate voltage of a microwave oven magnetron, and you modulate the frequency more than the amplitude. So (since it's _amateur_ radio), you just call it FM with incidental AM, and go out in the field with a microwave oven that has a horn antenna sticking out where the door used to be."Of course it's safe, dear, I saw it in a magazine!"
-- Tim Wescott Control systems, embedded software and circuit design I'm looking for work! See my website if you're interested http://www.wescottdesign.com
Not so much. The modulation index is the peak phase deviation in radians, a nd so depends on both the frequency deviation and the modulation frequency.
I think that the question "which is bigger" is pretty well meaningless exce pt in the context of a specific measurement.
Cheers
Phil Hobbs (builder of tunable diode laser gizmos since 1990ish)
n M d n h
icated by the
xGrin... there was a reference to another article, I'm sure someone has studied it in detail. I sent the customer a copy of Weiman's article he can chase down the references if he's interested.
Hmm can I ask a silly radio question. If I've got (say) 50% AM, modulation index = 0.5, how big are the side bands compared to the central carrier? (I guess I should be able to go look that up somew here.) (Or measure it myself)
I could then go measure the AM MI for the laser and get some number.
George H.
Hmm. On the other hand, for modulations under half a radian or so, the total power in the modulated signal will be roughly equal to the total power in an AM signal with the same modulation index.
Because of the whole cos(theta + phi) ~ cos(theta) + sin(phi) thing.
-- Tim Wescott Wescott Design Services http://www.wescottdesign.com I'm looking for work -- see my website!
Modulation index is not a term of art in AM, even with lasers. You talk about "modulation depth".
The modulation sensitivity of a diode laser is so large compared with anything in RF that you're rarely in that situation unless your modulation frequency is very high. Typically the laser power will go up and down by roughly 10%-30% between mode jumps, which limits the attainable continuous tuning range. Depending on the modulation frequency, the modulation index can vary over many orders of magnitude for the same frequency deviation (and hence the same modulation depth). You can compare sideband amplitudes as a function of f_mod, but that's it. Of course there will be some frequency where the amplitudes are the same, but only one. That's mostly what I mean when I say that there's no point comparing the two except in the context of some specific measurement.
One of my early noise canceller papers is on fixing up a tunable diode laser measurement of the iodine spectrum to get rid of the AM problem.
Another common application is the so-called "modulation-generated carrier" approach, where you interrogate an interferometer with a modulated diode. The trick is to choose the modulation index to be near
2.6, where the first and second sidebands are of equal amplitude and together account for 85% of the total power. From the Bessel expansion, they're 90 degrees out of phase, so that's a cute method of getting an I/Q measurement for nearly free.Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
OK throwing out some numbers.
100% AM will give side bands that are 1/2 the amplitude. (6dB down) (I measured with my 'scope FFT. I guess that has to come out the the math too.)I think a reasonable current modulation number is +/-2 mA. (200 mVp-p into 50 ohms.. is sorta a typical RF amplitude.) And the laser is typically run ~20 mA above threshold so ~10% AM depth, which means a side band of ~1/20 And it's lost in the much stronger FM....
Thanks guys, (diode lasers are magical.)
George H.
There's no such thing as modulation index in AM. MI belongs to angle modulation (PM, FM).
An AM signal modulated with a single sine wave is
AM(t) = A(1 + M cos(omega_m t)) cos(omega_c t),
where
A = carrier amplitude M = modulation depth (0 .. 1) omega_m = modulation signal angle frequency omega_c = carrier signal angle frequency t = time.
The expansion of the expression and cranking out the sideband signals from the cos(omega_m t) cos(omega_c t) is left as homework.
-- -TV
Right, Thanks Tauno. (I scribbled down the math...(I missed the 1+ term.. so thanks for that.)
I should let you all know that when I post questions like this, I'll often send a link to whomever asked me the question... saves me having to repeat what others say here.
A quote from my last customer.
"cheers George - impressive backup you have there! "
I just wanted to second that!
George h.
If you leave the 1+ away, you'll get what is called DSB-SC or, with square wave modulation, BPSK.
-- -TV
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