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Diode laser FM/AM modulation

Started by George Herold February 9, 2017
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 (<~1MHz.) of modulation changing the current changes 
the temperature, which changes the frequency.  At higher frequencies 
the thermal effects decrease.. (you get some average temperature) But the
change in carrier concentration in the diode still causes a modulation 
of the index of refraction* and hence frequency modulation.  This 
works for frequencies up beyond GHz.  (There's some weird light-field
electron relaxation oscillation in the ~10 GHz range, that I think might 
set the high frequency limit... but I'm way out of my depth.)  

We sell a Fabry-Perot cavity, by which users can see the laser sidebands.  
You can see a picture here.  (Figure 12 on page 12.)  Figure 12B 
shows the modulation index set at the first Bessel function zero... 
(Central carrier suppression... or whatever the right term is.)       

https://www.dropbox.com/s/6t994etv2nj3385/FP%20manual%20Ch2.pdf?dl=0

(Sorry for the long intro.)

So I've always just called this FM, and pretty much ignored the 
amplitude modulation aspects.  Now of course I have a question from 
a user, that is testing my understanding.  He wants to talk about 
this in terms of both AM and FM.  How do I compare the relative strengths 
of AM and FM modulation?  

Thanks, 
George H. 


*Hmm, I think that is mostly right, though I really should go read about it 
again from some expert.
On Thursday, February 9, 2017 at 9:27:03 AM UTC-5, George Herold wrote:
> 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 (<~1MHz.) of modulation changing the current changes > the temperature, which changes the frequency. At higher frequencies > the thermal effects decrease.. (you get some average temperature) But the > change in carrier concentration in the diode still causes a modulation > of the index of refraction* and hence frequency modulation. This > works for frequencies up beyond GHz. (There's some weird light-field > electron relaxation oscillation in the ~10 GHz range, that I think might > set the high frequency limit... but I'm way out of my depth.) > > We sell a Fabry-Perot cavity, by which users can see the laser sidebands. > You can see a picture here. (Figure 12 on page 12.) Figure 12B > shows the modulation index set at the first Bessel function zero... > (Central carrier suppression... or whatever the right term is.) > > https://www.dropbox.com/s/6t994etv2nj3385/FP%20manual%20Ch2.pdf?dl=0 > > (Sorry for the long intro.) > > So I've always just called this FM, and pretty much ignored the > amplitude modulation aspects. Now of course I have a question from > a user, that is testing my understanding. He wants to talk about > this in terms of both AM and FM. How do I compare the relative strengths > of AM and FM modulation? > > Thanks, > George H. > > > *Hmm, I think that is mostly right, though I really should go read about it > again from some expert.
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.&lsquo;&rdquo; 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.
On Thu, 09 Feb 2017 07:22:29 -0800, George Herold wrote:

> On Thursday, February 9, 2017 at 9:27:03 AM UTC-5, George Herold wrote: >> 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 (<~1MHz.) of modulation changing the current changes >> the temperature, which changes the frequency. At higher frequencies >> the thermal effects decrease.. (you get some average temperature) But >> the change in carrier concentration in the diode still causes a >> modulation of the index of refraction* and hence frequency modulation. >> This works for frequencies up beyond GHz. (There's some weird >> light-field electron relaxation oscillation in the ~10 GHz range, that >> I think might set the high frequency limit... but I'm way out of my >> depth.) >> >> We sell a Fabry-Perot cavity, by which users can see the laser >> sidebands. >> You can see a picture here. (Figure 12 on page 12.) Figure 12B shows >> the modulation index set at the first Bessel function zero... >> (Central carrier suppression... or whatever the right term is.) >> >> https://www.dropbox.com/s/6t994etv2nj3385/FP%20manual%20Ch2.pdf?dl=0 >> >> (Sorry for the long intro.) >> >> So I've always just called this FM, and pretty much ignored the >> amplitude modulation aspects. Now of course I have a question from a >> user, that is testing my understanding. He wants to talk about this in >> terms of both AM and FM. How do I compare the relative strengths of AM >> and FM modulation? >> >> Thanks, >> George H. >> >> >> *Hmm, I think that is mostly right, though I really should go read >> about it again from some expert. > > 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.&lsquo;&rdquo; 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.
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
>Well, there you go. &nbsp;Modulation index vs. modulation index. &nbsp;Two >dimensionless numbers, so no funny units in the answer.
Not so much. The modulation index is the peak phase deviation in radians, and so depends on both the frequency deviation and the modulation frequency. I think that the question "which is bigger" is pretty well meaningless except in the context of a specific measurement. Cheers Phil Hobbs (builder of tunable diode laser gizmos since 1990ish)
On Thursday, February 9, 2017 at 11:56:00 AM UTC-5, Tim Wescott wrote:
> On Thu, 09 Feb 2017 07:22:29 -0800, George Herold wrote: > > > On Thursday, February 9, 2017 at 9:27:03 AM UTC-5, George Herold wrote: > >> 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 (<~1MHz.) of modulation changing the current changes > >> the temperature, which changes the frequency. At higher frequencies > >> the thermal effects decrease.. (you get some average temperature) But > >> the change in carrier concentration in the diode still causes a > >> modulation of the index of refraction* and hence frequency modulation. > >> This works for frequencies up beyond GHz. (There's some weird > >> light-field electron relaxation oscillation in the ~10 GHz range, that > >> I think might set the high frequency limit... but I'm way out of my > >> depth.) > >> > >> We sell a Fabry-Perot cavity, by which users can see the laser > >> sidebands. > >> You can see a picture here. (Figure 12 on page 12.) Figure 12B shows > >> the modulation index set at the first Bessel function zero... > >> (Central carrier suppression... or whatever the right term is.) > >> > >> https://www.dropbox.com/s/6t994etv2nj3385/FP%20manual%20Ch2.pdf?dl=0 > >> > >> (Sorry for the long intro.) > >> > >> So I've always just called this FM, and pretty much ignored the > >> amplitude modulation aspects. Now of course I have a question from a > >> user, that is testing my understanding. He wants to talk about this in > >> terms of both AM and FM. How do I compare the relative strengths of AM > >> and FM modulation? > >> > >> Thanks, > >> George H. > >> > >> > >> *Hmm, I think that is mostly right, though I really should go read > >> about it again from some expert. > > > > 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.&lsquo;&rdquo; 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. > > Well, there you go. Modulation index vs. modulation index. Two > dimensionless numbers, so no funny units in the answer.
Grin... 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.
> > 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.
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 somewhere.) (Or measure it myself) I could then go measure the AM MI for the laser and get some number. George H.
> > "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
On Thu, 09 Feb 2017 09:23:32 -0800, pcdhobbs wrote:

>>Well, there you go. &nbsp;Modulation index vs. modulation index. &nbsp;Two >>dimensionless numbers, so no funny units in the answer. > > Not so much. The modulation index is the peak phase deviation in > radians, and so depends on both the frequency deviation and the > modulation frequency. > > I think that the question "which is bigger" is pretty well meaningless > except in the context of a specific measurement. > > Cheers > > Phil Hobbs (builder of tunable diode laser gizmos since 1990ish)
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!
On 02/09/2017 01:06 PM, Tim Wescott wrote:
> On Thu, 09 Feb 2017 09:23:32 -0800, pcdhobbs wrote: > >>> Well, there you go. Modulation index vs. modulation index. Two >>> dimensionless numbers, so no funny units in the answer. >> >> Not so much. The modulation index is the peak phase deviation in >> radians, and so depends on both the frequency deviation and the >> modulation frequency. >> >> I think that the question "which is bigger" is pretty well meaningless >> except in the context of a specific measurement. >> >> Cheers >> >> Phil Hobbs (builder of tunable diode laser gizmos since 1990ish) > > 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. >
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. <http://electrooptical.net/www/canceller/iodine.pdf> 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
On Thursday, February 9, 2017 at 2:10:35 PM UTC-5, Phil Hobbs wrote:
> On 02/09/2017 01:06 PM, Tim Wescott wrote: > > On Thu, 09 Feb 2017 09:23:32 -0800, pcdhobbs wrote: > > > >>> Well, there you go. Modulation index vs. modulation index. Two > >>> dimensionless numbers, so no funny units in the answer. > >> > >> Not so much. The modulation index is the peak phase deviation in > >> radians, and so depends on both the frequency deviation and the > >> modulation frequency. > >> > >> I think that the question "which is bigger" is pretty well meaningless > >> except in the context of a specific measurement. > >> > >> Cheers > >> > >> Phil Hobbs (builder of tunable diode laser gizmos since 1990ish) > > > > 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. > > > > Modulation index is not a term of art in AM, even with lasers. You talk > about "modulation depth".
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.
> > 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. > <http://electrooptical.net/www/canceller/iodine.pdf> > > 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
On 9.2.17 19:39, George Herold wrote:
> > 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 somewhere.) > (Or measure it myself) > > I could then go measure the AM MI for the laser and get some number. > > 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
On Friday, February 10, 2017 at 8:05:03 AM UTC-5, Tauno Voipio wrote:
> On 9.2.17 19:39, George Herold wrote: > > > > 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 somewhere.) > > (Or measure it myself) > > > > I could then go measure the AM MI for the laser and get some number. > > > > 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.